AP Statistics

Lesson 1.1 · Unit 1 · CED topics 1.1–1.2

Individuals, variables, and types of data

Every statistics problem starts with a spreadsheet, even if you never see it. Rows are the things being measured; columns are what's recorded about each one. Before you pick a graph or a summary number, you have to know which kind of column you're looking at: a mean of bus routes is nonsense.

Exam tip: when asked to "identify the variable," name it with its units and its type: "the quantitative variable is hours of sleep per night." Half-answers lose the point.

Lesson 1.2 · Unit 1 · CED topics 1.3–1.4

Representing and comparing categorical data

A categorical variable has no shape, center, or spread. All you can do is count how many individuals fall into each category, and then convert those counts into proportions so that groups of different sizes can be compared fairly.

Exam tip: with groups of different sizes, graders expect percentages, not counts, and comparative words; "higher than," "the same as," "roughly twice."

Lesson 1.3 · Unit 1 · CED topics 1.5–1.6

Graphs for quantitative data and describing shape

A quantitative variable does have a shape, and you can only see it in a picture. Three graphs do the job: a dotplot (one dot per value), a stemplot (digits sorted into rows), and a histogram (values grouped into equal-width bins).

Lesson 1.4 · Unit 1 · CED topic 1.7

Measures of center and spread

Once you can see a distribution, you want two numbers for it: where it sits (center) and how wide it is (spread). Each comes in a resistant version that ignores extreme values and a non-resistant version that uses every data point.

Transformations. Adding a constant shifts center and position by that constant and leaves spread unchanged (everyone sleeps 30 minutes more: mean 7.5, SD still 1.22). Multiplying by \(k\) multiplies center and spread by \(k\): in minutes, the mean is 420 and the SD is \(60 \times 1.2247 \approx 73.5\).

Lesson 1.5 · Unit 1 · CED topics 1.8–1.9

Boxplots and comparing distributions

A boxplot compresses a distribution into five numbers, which makes it the best tool for putting two groups side by side. The exam's favorite Unit 1 question is exactly that: "Compare the distributions." Your answer needs shape, outliers, center, and spread, each stated with comparative language and in context.

Exam tip: every sentence must name both groups and use a comparison word ("higher than," "similar to," "more variable than"). Listing each group's statistics without comparing them earns partial credit at best. Use median and IQR when either group is skewed or has outliers.

Lesson 1.6 · Unit 1 · CED topic 1.10

The normal distribution: z-scores, percentiles, and the empirical rule

Many measurements (heights, test scores, measurement errors) pile up in a symmetric, bell-shaped pattern. When a distribution is approximately normal, the mean and standard deviation tell you everything, and one calculation converts any value into a percentile.

Assessing normality. Before using the normal model, check that a histogram is roughly symmetric and bell-shaped with no outliers and that about 68% of the data fall within one SD of the mean. Skewed data (income, commute times) don't belong in normalcdf.

Unit 1 practice · 10 problems

Unit 1 practice: Exploring One-Variable Data

Ten problems covering the whole unit, in roughly exam order. Work each one on paper before revealing the answer: the reveal shows the key steps, not just the number. A calculator is fine for problems 4, 8, and 9; show the z-score and a sketch anyway.

1. A fitness app records, for each user: user ID number, age (years), average daily steps, hours of sleep per night, favorite activity (run, bike, swim, or walk), and number of workouts last week. Identify the individuals and classify each variable as categorical, quantitative discrete, or quantitative continuous.

   Show answer

   The individuals are the app's users. User ID is categorical (an identifier: averaging IDs means nothing). Favorite activity is categorical. Age and hours of sleep are quantitative and continuous (any value in a range is possible). Average daily steps and number of workouts last week are quantitative and discrete: they are counts. The test: if the average of the values would mean something, the variable is quantitative.

2. Sixty freshmen and ninety seniors were asked which type of club they belong to.

   Group	Sports	Arts	STEM	None	Total
   Freshmen	27	15	12	6	60
   Seniors	30	18	27	15	90

   A student says, "More seniors than freshmen are in sports clubs, so sports are more popular with seniors." Compute the relative frequencies for each group and write two sentences comparing the distributions.

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   Freshmen: sports \(27/60 = 0.45\), arts \(0.25\), STEM \(0.20\), none \(0.10\). Seniors: sports \(30/90 = 0.333\), arts \(0.20\), STEM \(0.30\), none \(0.167\). The student is wrong: the groups differ in size, so compare proportions, not counts. Sports is the most common club type in both grades, but a higher proportion of freshmen (45%) than seniors (33%) belong to one. STEM clubs are more popular among seniors (30% vs. 20%), and seniors are more likely to belong to no club at all (17% vs. 10%).

3. A histogram of the sale prices (thousands of dollars) of 40 homes in a town has bars of height 14, 12, 7, 4, 2, and 1 over the intervals 100–200, 200–300, 300–400, 400–500, 500–600, and 600–700. Describe the shape of the distribution, state which interval contains the median, and explain whether the mean is greater than, less than, or about equal to the median.

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   The distribution is unimodal and skewed right: the peak is in the 100–200 interval and the frequencies trail off in a long tail toward the expensive homes, with no gaps. The first bar holds 14 homes and the first two hold 26, so the 20th and 21st values, the median for \(n = 40\), fall in the 200–300 interval. The mean is greater than the median: the long right tail pulls the mean toward the high values (using bar midpoints, the mean is roughly \(278\) thousand dollars), while the median is resistant.

4. The numbers of emails five employees received yesterday were 8, 11, 12, 15, and 19. By hand, find the mean, median, range, IQR, and sample standard deviation, and interpret the standard deviation in context.

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   Mean \(\bar x = 65/5 = 13\) emails; median = 12 (the third value). Range \(= 19 - 8 = 11\). Lower half 8, 11 gives \(Q_1 = 9.5\); upper half 15, 19 gives \(Q_3 = 17\); IQR \(= 7.5\) emails. Deviations from 13 are \(-5, -2, -1, 2, 6\); squares 25, 4, 1, 4, 36 sum to 70, so \[s_x = \sqrt{\frac{70}{4}} = \sqrt{17.5} \approx 4.18 \text{ emails}.\] An employee's email count typically differs from the mean of 13 by about 4.18 emails.

5. Wait times (minutes) for 11 patients at a clinic, in order: 14, 18, 20, 21, 22, 23, 25, 26, 28, 31, 45. Find the five-number summary, use the 1.5×IQR rule to check for outliers, and describe the boxplot you would draw.

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   Median = 23 (the 6th value). Lower half 14, 18, 20, 21, 22 gives \(Q_1 = 20\); upper half 25, 26, 28, 31, 45 gives \(Q_3 = 28\). Five-number summary: 14, 20, 23, 28, 45. IQR \(= 8\); fences \(20 - 1.5(8) = 8\) and \(28 + 1.5(8) = 40\). Since \(45 \gt 40\), the 45-minute wait is an outlier; 14 is not. The boxplot has a box from 20 to 28 with a line at 23, a lower whisker to 14, an upper whisker to 31 (the largest non-outlier), and a separate point at 45.

6. Daily high temperatures in a city over a month have mean 20 °C, median 19 °C, standard deviation 4 °C, and IQR 6 °C. (a) Find the mean, median, standard deviation, and IQR in degrees Fahrenheit, using \(F = 1.8C + 32\). (b) A thermometer was found to read 2 °C too low; every reading is corrected by adding 2 °C. What are the corrected mean and standard deviation in °C?

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   (a) Multiplying by 1.8 scales both center and spread; adding 32 shifts center only. Mean \(= 1.8(20) + 32 = 68\) °F; median \(= 1.8(19) + 32 = 66.2\) °F; SD \(= 1.8(4) = 7.2\) °F; IQR \(= 1.8(6) = 10.8\) °F. (b) Adding a constant shifts every value and the center by 2 but leaves spread unchanged: mean 22 °C, SD still 4 °C.

7. Final exam scores for two sections of the same course are summarized by five-number summaries. Section A (in person): 52, 64, 72, 80, 96. Section B (online): 44, 58, 66, 84, 100. Neither section has outliers. Compare the two distributions.

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   Shape: Section A is roughly symmetric (the median sits in the middle of the box and the whiskers are similar in length), while Section B is skewed right: the upper half of its box (66 to 84) and upper whisker are longer than the lower ones. Outliers: neither section has any. Center: the median score in Section A (72) is higher than in Section B (66). Spread: Section B's scores are more variable than Section A's: an IQR of 26 points versus 16, and a range of 56 versus 44. Every sentence names both groups and uses a comparison word; that is what the rubric rewards.

8. Composite ACT scores are approximately \(N(21, 5.4)\) and SAT total scores are approximately \(N(1050, 200)\). Priya scored 30 on the ACT; Marcus scored 1350 on the SAT. Who did relatively better? Use z-scores and percentiles to justify.

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   Priya: \(z = \dfrac{30 - 21}{5.4} \approx 1.67\); Marcus: \(z = \dfrac{1350 - 1050}{200} = 1.50\). Priya's score is 1.67 standard deviations above the mean versus 1.50 for Marcus, so Priya did relatively better. Her score is at about the 95th percentile (area to the left of 1.67 is 0.952); his is at about the 93rd (area 0.933).

9. Finishing times in a large 5K race are approximately normal with mean 28 minutes and standard deviation 5 minutes. (a) What proportion of runners finish in under 20 minutes? (b) What proportion finish between 25 and 35 minutes? (c) How fast must a runner be to finish in the fastest 5%?

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   (a) \(z = \dfrac{20 - 28}{5} = -1.6\); area to the left is \(0.0548\), so about 5.5% finish under 20 minutes (normalcdf(−1E99, 20, 28, 5)). (b) \(z = -0.6\) and \(z = 1.4\); area between is \(0.9192 - 0.2743 = 0.6449\), about 64%. (c) The fastest 5% have area 0.05 to the left: \(z = -1.645\), so \(x = 28 - 1.645(5) \approx 19.8\) minutes (invNorm(0.05, 28, 5) = 19.78). Runners must finish in under about 19.8 minutes.

10. IQ scores are approximately \(N(100, 15)\). Using the 68–95–99.7 rule, approximately what percent of people have IQ scores between 85 and 130?

    47.5% 68% 81.5% 95%

    47.5% is only the piece from 100 to 130 (half of the 95% within two standard deviations). It leaves out the 34% between 85 and the mean, so it counts just one side of the interval.

    68% is the percent within one standard deviation of the mean, 85 to 115, but this interval reaches 130, two standard deviations above the mean. Because the upper half is wider than the lower half, the answer must fall between 68% and 95%.

    From 85 to 100 is one standard deviation below the mean, holding half of 68%, or 34%; from 100 to 130 is two standard deviations above the mean, holding half of 95%, or 47.5%. Together, \(34\% + 47.5\% = 81.5\%\) (the exact normal area is 0.8186).

    95% is the percent within two standard deviations on both sides, 70 to 130. The interval here starts at 85, not 70, so the 13.5% between 70 and 85 has to be removed.

Lesson 2.1 · Unit 2 · CED topics 2.1–2.3

Two categorical variables: two-way tables and conditional distributions

Unit 1 handled one variable at a time. Now the question is whether knowing one thing about an individual tells you anything about another. For two categorical variables the tool is a two-way table, and the trick is to read it by rows or columns, not by raw cell counts.

Exam tip: to argue for an association, quote two conditional proportions and say they differ. To argue against it, show they are about equal. Never argue from raw counts when the groups have different sizes.

Lesson 2.2 · Unit 2 · CED topics 2.4–2.5

Scatterplots and correlation

For two quantitative variables, the picture is a scatterplot: the explanatory variable on the horizontal axis, the response on the vertical, one point per individual. You describe it in four words (form, direction, strength, unusual features), and then, if the form is linear, one number summarizes the strength.

Lesson 2.3 · Unit 2 · CED topics 2.6–2.7

The least-squares regression line

When a scatterplot is linear, you want the line that fits it best so you can predict \(y\) from \(x\). "Best" has a precise meaning: the line that makes the sum of the squared vertical distances from the points as small as possible.

Lesson 2.4 · Unit 2 · CED topic 2.8

Residuals, residual plots, r², and s

A regression line is a model, and every model needs a report card. Residuals tell you how far each point misses; a residual plot tells you whether a line was the right shape at all; and two numbers, \(r^2\) and \(s\), grade the fit overall.

Lesson 2.5 · Unit 2 · CED topic 2.9

Influential points and transforming to achieve linearity

Two things break a regression: a single point that drags the line around, and a relationship that was never straight to begin with. You need to recognize both from a plot.

Unit 2 practice · 10 problems

Unit 2 practice: Exploring Two-Variable Data

Ten problems covering the whole unit, in roughly exam order. Work each one on paper before revealing the answer: the reveal shows the key steps, not just the number. Use LinReg(a+bx) for problem 5; everything else is by hand.

1. A counselor asked 100 juniors and 120 seniors about their plans after graduation.

   Grade	Four-year college	Work or trade school	Gap year	Total
   Juniors	55	25	20	100
   Seniors	84	24	12	120
   Total	139	49	32	220

   (a) What proportion of all students surveyed plan to attend a four-year college? (b) Give the conditional distribution of plans for each grade. (c) Is there an association between grade and plans? Justify.

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   (a) Marginal: \(139/220 \approx 0.632\). (b) Juniors: college \(55/100 = 0.55\), work \(0.25\), gap year \(0.20\). Seniors: college \(84/120 = 0.70\), work \(0.20\), gap year \(0.10\). (c) Yes. The conditional distributions differ: 70% of seniors plan on a four-year college versus 55% of juniors, and juniors are twice as likely to plan a gap year (20% vs. 10%). Knowing a student's grade changes the likely plan, so the variables are associated in this sample. A segmented bar chart would show the two bars split differently.

2. A scatterplot of asking price (dollars) against age (years) for 30 used cars shows points running from upper left to lower right in a fairly narrow, straight band, with one 3-year-old car priced far above the band; \(r = -0.87\). (a) Describe the association. (b) If price were recorded in thousands of dollars instead, what would \(r\) be? (c) A classmate says \(r = -0.87\) proves the relationship is strong and linear. What's wrong with that?

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   (a) There is a strong, negative, linear association between age and asking price: older cars tend to have lower prices. One unusual point, a 3-year-old car priced well above the trend, is an outlier in the \(y\) direction. (b) Still \(-0.87\); correlation has no units and is unchanged by a change of units in either variable. (c) \(r\) measures the strength of a linear association only: it cannot tell you the form. A strongly curved pattern can also produce \(|r|\) near 0.87, so you must look at the scatterplot (or a residual plot) to judge linearity.

3. For 40 test plots, the amount of fertilizer applied (kg per hectare) has \(\bar x = 40\) and \(s_x = 12\); crop yield (tonnes per hectare) has \(\bar y = 6.5\) and \(s_y = 0.9\); the correlation is \(r = 0.85\). Find the least-squares regression line, interpret its slope in context, and predict the yield for a plot that received 55 kg/ha.

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   \(b = r\dfrac{s_y}{s_x} = 0.85\cdot\dfrac{0.9}{12} = 0.06375\) and \(a = \bar y - b\bar x = 6.5 - 0.06375(40) = 3.95\), so \(\widehat{\text{yield}} = 3.95 + 0.0638(\text{fertilizer})\). Slope: for each additional kilogram of fertilizer per hectare, the predicted yield increases by about 0.064 tonnes per hectare. At 55 kg/ha: \(\hat y = 3.95 + 0.06375(55) \approx 7.46\) tonnes per hectare. (55 is within about 1.25 SDs of \(\bar x\), so this is not an extrapolation.)

4. The regression line for predicting a music student's audition score from weekly practice hours is \(\widehat{\text{score}} = 12.5 + 3.2(\text{hours})\). One student practiced 6 hours per week and scored 30. Find and interpret the residual.

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   Predicted score \(= 12.5 + 3.2(6) = 31.7\). Residual \(= \text{actual} - \text{predicted} = 30 - 31.7 = -1.7\). The student scored 1.7 points lower than the line predicted for someone practicing 6 hours per week; the model overpredicted this student's score. A negative residual means the point lies below the line.

5. The ages (years) and resale values (hundreds of dollars) of six laptops of the same model:

   Age (x)	2	4	5	7	8	10
   Value (y)	31	27	26	20	19	14

   Use your calculator to find the LSRL, \(r\), and \(r^2\). Then predict the value of a 6-year-old laptop and find the residual for the 5-year-old laptop.

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   LinReg(a+bx) with age in L1 and value in L2: \(\widehat{\text{value}} = 35.69 - 2.143(\text{age})\), \(r = -0.995\), \(r^2 = 0.990\). Each additional year of age is associated with a predicted drop of about $214 in resale value. At age 6: \(\hat y = 35.69 - 2.143(6) \approx 22.83\), or about $2,280. For the 5-year-old laptop: predicted \(35.69 - 2.143(5) = 24.98\), so the residual is \(26 - 24.98 = +1.02\) hundred dollars: it sold for about $100 more than the line predicted.

6. Software output for predicting a home's monthly heating cost (dollars) from the month's average outside temperature (°F), based on 18 months:

   Predictor	Coef	SE Coef	T	P
   Constant	210.40	9.80	21.47	0.000
   Temp	−3.850	0.420	−9.17	0.000
   S = 14.2    R-Sq = 84.6%

   Write the equation, find \(r\), and interpret the slope, \(r^2\), and \(s\) in context. Predict the heating cost for a month averaging 30 °F.

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   \(\widehat{\text{cost}} = 210.40 - 3.85(\text{temp})\). Since the slope is negative, \(r = -\sqrt{0.846} \approx -0.92\). Slope: for each 1 °F increase in average outside temperature, the predicted monthly heating cost decreases by about $3.85. \(r^2\): about 84.6% of the variation in monthly heating cost is explained by the linear relationship with outside temperature. \(s\): actual heating costs are typically about $14.20 away from the cost predicted by the line. At 30 °F: \(210.40 - 3.85(30) = 94.90\), so about $95.

7. Two residual plots are described. Plot I: residuals are negative for the smallest and largest \(x\)-values and positive in the middle, forming an arch. Plot II: residuals scatter randomly above and below zero with no visible pattern. For each, state whether a linear model is appropriate and explain what the plot tells you.

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   Plot I: a linear model is not appropriate. The arch means the line systematically overpredicts at the ends and underpredicts in the middle: the true relationship is curved, and a transformation or nonlinear model should be considered. Plot II: a linear model is appropriate. Random scatter about zero means the line has captured the pattern and what remains is unexplained noise. A residual plot magnifies departures from linearity that are hard to see on the scatterplot itself.

8. A scatterplot of weekly earnings against hours worked for 20 part-time employees shows a positive linear pattern for those working 10 to 30 hours. One employee worked 60 hours but reported unusually low earnings, landing well below the trend. Is this point an outlier, high-leverage, influential, or some combination? Describe how removing it would change the slope, \(r\), and \(s\).

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   The point is high-leverage, its \(x\)-value (60 hours) is far from \(\bar x\), and because it also lies off the trend, it is influential. Removing it would make the slope steeper (the low point at far right drags the right end of the line down), move \(r\) closer to 1 (the remaining points hug the line more tightly), and decrease \(s\) (the residuals shrink). Note that the point's own residual may not look huge, precisely because it pulls the line toward itself; influence is judged by what changes when the point is removed.

9. A town's population (thousands) was recorded each year since 2000. A linear model of population on years-since-2000 has a U-shaped residual plot. Fitting \(\log(\text{population})\) instead gives \(\widehat{\log(\text{pop})} = 1.70 + 0.045(\text{years})\) with \(r^2 = 0.99\) and a patternless residual plot. Predict the population in 2015, explain why the log model was chosen, and interpret the slope 0.045.

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   For 2015, years \(= 15\): \(\log\hat y = 1.70 + 0.045(15) = 2.375\), so \(\hat y = 10^{2.375} \approx 237\) thousand people. The log model was chosen because the linear model's residual plot was curved (a systematic misfit) while the transformed data are linear with random residuals and a very high \(r^2\). A straight line in \(\log y\) versus \(x\) means exponential growth: each year multiplies the predicted population by \(10^{0.045} \approx 1.109\), about 10.9% growth per year.

10. Across a large number of fires, the correlation between the number of firefighters sent and the dollar amount of damage is \(r = 0.92\). Which statement is correct?

    Sending more firefighters causes more damage. About 92% of the variation in damage is explained by the number of firefighters. A lurking variable such as the size of the fire likely drives both variables; correlation does not imply causation. Measuring damage in thousands of dollars instead of dollars would reduce \(r\).

    This confuses correlation with causation. A strong association in observational data never establishes that one variable causes the other, and here the causal story is backwards anyway: bigger, more damaging fires attract more firefighters.

    The fraction of variation explained is \(r^2\), not \(r\): \(0.92^2 = 0.846\), so about 85% of the variation in damage is explained, not 92%. Reading \(r\) itself as a percent of variation is one of the most common Unit 2 errors.

    Bigger fires call for more firefighters and cause more damage, so fire size is a lurking variable that explains the association without either measured variable causing the other.

    Correlation has no units and is unchanged by any linear change of units: dividing every damage figure by 1,000 rescales the variable but leaves \(r\) exactly 0.92. Only the slope of a regression line would change.

Lesson 3.1 · Unit 3 · CED topics 3.1–3.2

Populations, samples, and types of studies

Units 1 and 2 described data you already had. Unit 3 asks where the data came from, no amount of clever analysis rescues a badly collected sample.

Why not always take a census? Cost, time, and sometimes impossibility (testing every light bulb to failure leaves none to sell). A well-chosen sample of a few thousand estimates a national parameter very accurately.

Lesson 3.2 · Unit 3 · CED topic 3.3

Random sampling methods

Letting people choose whether to be in a sample, or letting the researcher pick convenient ones, produces samples that systematically differ from the population. The cure is chance: when a random mechanism picks the sample, nothing can tilt the selection.

Lesson 3.3 · Unit 3 · CED topics 3.4–3.5

Bias, confounding, and why observational studies can't prove cause

Even a random sample goes wrong if some people can't be reached, refuse, or don't answer truthfully. And even a perfect survey can't answer "does X cause Y?" if the people who did X differ from the rest in other ways.

Lesson 3.4 · Unit 3 · CED topics 3.5–3.6

Designing experiments

An experiment is the only design that can support a cause-and-effect claim, and it earns that power from one move: letting chance decide who gets which treatment. Everything else is about making the comparison sharper.

Exam tip: you stratify when sampling; you block when assigning treatments. Blocks are formed on a variable known to affect the response, never randomly.

Lesson 3.5 · Unit 3 · CED topic 3.7

Scope of inference: what each design lets you conclude

Every study allows some conclusions and forbids others. The rules come down to two questions: were the individuals chosen at random, and were the treatments assigned at random? Each "yes" unlocks one kind of claim.

Exam tip: medical experiments use volunteers, so "cause and effect, but not generalizable beyond similar volunteers" is the most common correct answer. A large sample doesn't justify generalizing: size isn't randomness.

Unit 3 practice · 10 problems

Unit 3 practice: Collecting Data

Ten problems covering the whole unit, in roughly exam order. Most answers here are sentences, not numbers: write them out in full before revealing the answer, and check that every design you describe includes the random step. No calculator needed.

1. A city's transportation department wants to know what proportion of the 84,000 vehicles registered in the city would fail an emissions test. Inspectors randomly select 500 registered vehicles and test them; 35 fail. Identify the population, the sample, the parameter, and the statistic.

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   Population: all 84,000 vehicles registered in the city. Sample: the 500 vehicles that were tested. Parameter: \(p\), the true proportion of all registered vehicles in the city that would fail the emissions test (unknown). Statistic: \(\hat p = 35/500 = 0.07\), the proportion of the sampled vehicles that failed. Always write the parameter in words as well as in symbols.

2. Which of the following is an experiment?

   Researchers compare the lung function of 300 smokers and 300 nonsmokers. A company emails a satisfaction survey to 1,000 randomly chosen customers. Researchers randomly assign 200 volunteers to take either vitamin D or a placebo daily and record how many colds each gets over the winter. A county attempts to count every resident.

   This is an observational study: the people chose whether to smoke, so the researchers did not impose the explanatory variable. Comparing two existing groups looks like treatment versus control, but smoking is confounded with other lifestyle differences.

   A survey is a sampling study: it measures the opinions of a random sample but imposes no treatment on anyone. Random selection of subjects is not the same thing as random assignment of treatments.

   In an experiment the researchers impose the treatment: they decided who took vitamin D and who took the placebo, using random assignment. The one question to ask is whether the researchers assigned the conditions: only here is the answer yes.

   This is a census, an attempt to measure every individual in the population. It collects data but imposes no treatment and compares no groups, so it is neither an experiment nor even a sample.

3. Name the sampling method in each case. (a) A principal numbers all 1,200 students and uses randIntNoRep(1, 1200, 100) to choose 100. (b) She selects an SRS of 25 students from each of the four grades. (c) She randomly selects 5 of the 40 homerooms and surveys every student in them. (d) She posts a survey link on the school website for anyone to answer. (e) She stands at the front door and surveys every 12th student who enters, starting with a randomly chosen student among the first 12.

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   (a) Simple random sample: every group of 100 students is equally likely. (b) Stratified random sample: the grades are strata, and an SRS is taken from every stratum. (c) Cluster sample: homerooms are clusters, some are chosen at random, and everyone in the chosen clusters is included. (d) Voluntary response sample, not random; people with strong opinions are overrepresented. (e) Systematic random sample: a random start, then every 12th individual.

4. A manager wants to choose 4 of her 30 employees for a focus group. Describe how to select an SRS using this line from a table of random digits, and give the sample it produces: 41072 50733 18960 27411 96085

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   Label the employees 01 through 30. Read two-digit groups from left to right, skipping any number above 30 (and 00) and any repeat, until four distinct labels are chosen. The pairs are 41 (skip), 07 ✓, 25 ✓, 07 (repeat, skip), 33 (skip), 18 ✓, 96 (skip), 02 ✓. The sample is employees 02, 07, 18, and 25. On the exam, all four pieces are graded: the labels, how you read the digits, the skip rule, and the stopping rule.

5. A university with 6,000 undergraduates and 2,000 graduate students wants to survey 200 students about extending library hours. (a) Explain why a stratified random sample by student level is likely to give a more precise estimate than an SRS of 200, and describe how to select it. (b) Describe a cluster sample the university could use instead, and give one advantage and one disadvantage.

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   (a) Undergraduates and graduate students probably use the library differently (graduate students more often at night), so their opinions are likely to differ. Stratifying guarantees both groups are represented in the right proportion, an SRS of \(200 \times 6000/8000 = 150\) undergraduates and a separate SRS of 50 graduate students, and removes the between-group variation from the estimate, reducing its variability. An SRS might, by chance, badly over- or underrepresent graduate students. (b) Randomly select, say, 10 of the university's course sections and survey every student in those sections. Advantage: cheaper and faster; you visit 10 classrooms instead of tracking down 200 individuals. Disadvantage: students in the same section tend to be similar, so the estimate is less precise than an SRS of the same size.

6. (a) A TV station invites viewers to text "YES" or "NO" on a proposed stadium tax; of 3,000 texts, 78% say NO. Name the type of bias and its likely direction. (b) A polling firm calls landline numbers between 10 a.m. and 2 p.m. on weekdays to estimate the proportion of adults who work full time. Name two sources of bias and the direction of each.

   Show answer

   (a) Voluntary response bias. People who feel strongly, especially those angry about a new tax, are far more likely to bother texting, so 78% probably overestimates the proportion of all viewers (let alone all residents) who oppose the tax. (b) Undercoverage: adults without landlines (younger and often working adults) have no chance of being selected. Nonresponse: full-time workers are unlikely to be home to answer between 10 and 2, so they are selected but not reached. Both push the estimate the same way: underestimating the proportion who work full time. Neither is fixed by calling more numbers.

7. An observational study of 2,000 elementary students finds that those who eat breakfast every day score higher on standardized tests than those who don't. A headline reads, "Breakfast boosts test scores." Name a specific confounding variable, explain how it is confounded with eating breakfast, and explain why the headline is not justified.

   Show answer

   Household income is a plausible confounder. Higher-income families are more likely to provide breakfast every day (associated with the explanatory variable) and more likely to provide books, tutoring, and quiet study space that raise test scores (affects the response). Because the breakfast eaters and non-eaters differ in income, the effect of breakfast cannot be separated from the effect of income. The students chose (or their families chose) whether they ate breakfast, nothing was randomly assigned, so this observational study can show an association but cannot establish that breakfast causes higher scores. Full credit requires naming the variable and linking it to both the explanatory variable and the response.

8. Sixty volunteers agree to take part in a study of whether a 10-minute guided meditation before an exam reduces test anxiety, measured by a standard 0–40 anxiety questionnaire completed immediately after the exam. Describe a completely randomized design, including how you would carry out the random assignment and what comparison you would make.

   Show answer

   Number the volunteers 01 to 60. Use randIntNoRep(1, 60, 30): the 30 volunteers whose numbers appear do the 10-minute guided meditation immediately before the exam; the other 30 sit quietly for 10 minutes instead (the control, so both groups get the same 10-minute pause and only the meditation differs). All 60 take the same exam under identical conditions and then complete the anxiety questionnaire. Compare the mean anxiety scores of the two groups; if the meditation group's mean is lower by more than random assignment alone would plausibly produce, conclude that the meditation reduced anxiety. Random assignment balances hidden variables (baseline anxiety, preparation) across the groups; replication (30 per group) keeps chance variation from hiding a real effect.

9. Suppose 24 of the 60 volunteers in the previous problem have a diagnosed anxiety disorder and 36 do not. Explain why a randomized block design would be better than the completely randomized design, describe how to carry it out, and state how blocking differs from stratifying.

   Show answer

   A diagnosed anxiety disorder strongly affects the response (anxiety score) regardless of treatment. If the completely randomized design happened to put more of those 24 volunteers in one group, the treatment effect would be blurred. Block on diagnosis: within the 24 diagnosed volunteers, randomly assign 12 to meditation and 12 to control (label 01–24, randIntNoRep(1, 24, 12)); within the 36 without a diagnosis, randomly assign 18 and 18 the same way. Compare meditation with control within each block, then combine. Blocking removes the diagnosis-related variation from the comparison, making a real meditation effect easier to detect. Stratifying is the analogous idea when sampling from a population; blocking is done when assigning treatments in an experiment. Blocks are always formed on a variable known to affect the response, never at random.

10. (a) A random sample of 300 adults in one state finds that those who drink two or more cups of coffee daily report lower rates of depression. (b) In a separate study, 80 volunteers are randomly assigned to drink two cups of regular or decaffeinated coffee daily for eight weeks; the regular-coffee group reports lower depression scores. For each study, state whether the results can be generalized to a larger population and whether a cause-and-effect conclusion is justified. Justify each answer.

    Show answer

    (a) Random selection from the state's adults permits generalizing the association between heavy coffee drinking and lower depression rates to all adults in the state. But coffee consumption was not randomly assigned, people chose it, so a confounding variable (sleep habits, social activity, income) could explain the link, and we cannot conclude that coffee reduces depression. (b) Random assignment permits a cause-and-effect conclusion: for people like these volunteers, drinking regular coffee reduced depression scores relative to decaf. But the volunteers were not randomly selected from any population, so the result generalizes only to people similar to them. Two random steps, two separate conclusions.

Lesson 4.1 · Unit 4 · CED topics 4.1–4.2

Randomness, simulation, and the law of large numbers

A random process is unpredictable in the short run but has a stable pattern in the long run, and that long-run pattern is what probability measures. When the math is hard, you can find the pattern by imitating the process many times: a simulation.

Lesson 4.2 · Unit 4 · CED topics 4.3–4.4

Probability rules: sample spaces, complements, and the addition rule

Once you can list what might happen, probability becomes bookkeeping. A few rules connect the probabilities of related events, and a Venn diagram keeps the accounts straight.

Exam tip: the phrases "at least one" and "or" are signals: complement rule for the first, addition rule for the second. Define your events in symbols before computing, and show the rule you're applying.

Lesson 4.3 · Unit 4 · CED topics 4.5–4.6

Conditional probability, tree diagrams, and independence

Knowing that one event happened often changes the odds of another. Conditional probability makes that precise, gives you a rule for "and," and provides the test for whether two events are truly independent.

Lesson 4.4 · Unit 4 · CED topics 4.7–4.8

Discrete random variables: expected value and standard deviation

A random variable attaches a number to each outcome of a chance process: the number of toppings on the next pizza ordered, your winnings on a raffle ticket. Its probability distribution is a table, and from the table you compute a mean and a standard deviation just as you did for data, with probabilities playing the role of relative frequencies.

Lesson 4.5 · Unit 4 · CED topic 4.9

Transforming and combining random variables

Real quantities are built from other random quantities: a bill is a fixed fee plus a per-item charge, a delivery time is prep time plus driving time. You can find the mean and standard deviation of the combination without listing every outcome, as long as you remember the one rule everyone gets wrong.

Exam tip: write the variance step explicitly; "\(\sigma_T^2 = 4 + 16 = 20\), so \(\sigma_T = 4.47\)". Skipping to the square root hides whether you added variances or SDs.

Lesson 4.6 · Unit 4 · CED topics 4.10–4.11

The binomial distribution

Count the successes in a fixed number of identical yes/no trials (free throws made out of 10, defective parts in a batch of 20, correct guesses on 12 questions), and you have a binomial random variable. Its distribution has a formula, a mean, and a standard deviation you can write down without a table.

Common error: "at least 6" is \(X \ge 6\), so subtract binomcdf up to 5, not 6. Draw a quick number line if the boundary is confusing.

Lesson 4.7 · Unit 4 · CED topic 4.12

The geometric distribution

The binomial counts successes in a fixed number of trials. Flip the question: how many trials until the first success?, and the number of trials is no longer fixed; it's the random variable. That's the geometric setting, and it has the simplest formulas in the unit.

Unit 4 practice · 10 problems

Unit 4 practice: Probability, Random Variables, and Probability Distributions

Ten problems covering the whole unit, in roughly exam order. Work each one on paper before revealing the answer. Use binompdf, binomcdf, geometpdf, and geometcdf for problems 8 and 9, but write the formula with numbers substituted as well.

1. Thirty percent of customers at a coffee shop pay with the shop's loyalty app. Describe how to use a random number generator to simulate one trial that estimates the probability that at least 4 of the next 10 customers pay with the app, and explain how you would use many trials to estimate the probability.

   Show answer

   Describe: use randInt(1, 10, 10) to generate ten integers, one per customer. Let 1, 2, or 3 represent a customer who pays with the app (probability 0.3) and 4 through 10 a customer who doesn't. Count how many of the ten integers are 1–3 and record whether the count is at least 4. Run: repeat for many trials, say 200. Count: the proportion of trials in which at least 4 of the 10 used the app is the estimate. (This assumes customers' payment choices are independent.)

2. An American roulette wheel has 38 equally likely slots, 18 of them red. After the ball lands on black seven spins in a row, a gambler bets heavily on red, saying red is "due." Explain what is wrong with the gambler's reasoning and what the law of large numbers actually says.

   Show answer

   Each spin is independent, so the probability of red on the next spin is still \(18/38 \approx 0.474\): the wheel has no memory and does not compensate for past results. The law of large numbers says that over a very large number of spins the proportion of reds will get close to 0.474; it makes no promise about the next few spins and does not require a short-run streak to be balanced out. Believing that a streak makes the opposite outcome more likely is the gambler's fallacy.

3. At a high school, 55% of students have a part-time job, 35% play a school sport, and 20% do both. A student is chosen at random. (a) Find the probability that the student has a job or plays a sport. (b) Find the probability the student does neither. (c) Are "has a job" and "plays a sport" mutually exclusive? Explain.

   Show answer

   Let \(J\) = has a job, \(S\) = plays a sport. (a) General addition rule: \(P(J \cup S) = 0.55 + 0.35 - 0.20 = 0.70\). (b) Complement: \(P(\text{neither}) = 1 - 0.70 = 0.30\). (c) No: mutually exclusive events cannot both occur, but \(P(J \cap S) = 0.20 \ne 0\). In a Venn diagram: job only 0.35, both 0.20, sport only 0.15, neither 0.30, summing to 1.

4. A random sample of 150 students was asked whether they had used a phone during class that day.

   Grade	Used phone	Did not	Total
   Freshmen	36	24	60
   Seniors	54	36	90
   Total	90	60	150

   One student is selected at random. Find \(P(\text{used phone})\), \(P(\text{used phone} \mid \text{freshman})\), and \(P(\text{freshman} \mid \text{used phone})\). Are the events "used phone" and "freshman" independent? Justify.

   Show answer

   \(P(\text{used phone}) = 90/150 = 0.60\). \(P(\text{used phone} \mid \text{freshman}) = 36/60 = 0.60\) (restrict to the 60 freshmen). \(P(\text{freshman} \mid \text{used phone}) = 36/90 = 0.40\) (restrict to the 90 phone users). Since \(P(\text{used phone} \mid \text{freshman}) = 0.60 = P(\text{used phone})\), knowing a student is a freshman does not change the probability of phone use: the events are independent. (Check: \(P(\text{used phone} \mid \text{senior}) = 54/90 = 0.60\) as well.)

5. A factory makes 60% of its bolts on Machine A, which produces 2% defective bolts, and 40% on Machine B, which produces 5% defective. A bolt is chosen at random from the day's output. (a) Find the probability it is defective. (b) Given that it is defective, find the probability it came from Machine A. Show a tree diagram or the rules you used.

   Show answer

   Tree: first branches \(P(A) = 0.6\), \(P(B) = 0.4\); second branches are defective or not given the machine. Multiply along branches: \(P(A \cap D) = 0.6(0.02) = 0.012\) and \(P(B \cap D) = 0.4(0.05) = 0.020\). (a) \(P(D) = 0.012 + 0.020 = 0.032\). (b) Conditional probability: \[P(A \mid D) = \frac{P(A \cap D)}{P(D)} = \frac{0.012}{0.032} = 0.375.\] Although Machine A makes most of the bolts, only 37.5% of the defective ones come from it, because its defect rate is so much lower.

6. Let \(X\) = the number of cars owned by a randomly selected household in a town.

   x	0	1	2	3
   P(X = x)	0.10	0.35	0.40	0.15

   Find and interpret \(\mu_X\). Find \(\sigma_X\). Find \(P(X \ge 2)\).

   Show answer

   \(\mu_X = 0(0.10) + 1(0.35) + 2(0.40) + 3(0.15) = 1.6\) cars. Over many randomly selected households, the average number of cars would be about 1.6. Variance: \[\sigma_X^2 = (0 - 1.6)^2(0.10) + (1 - 1.6)^2(0.35) + (2 - 1.6)^2(0.40) + (3 - 1.6)^2(0.15) = 0.256 + 0.126 + 0.064 + 0.294 = 0.74,\] so \(\sigma_X = \sqrt{0.74} \approx 0.86\) cars: a household's car count typically differs from 1.6 by about 0.86. \(P(X \ge 2) = 0.40 + 0.15 = 0.55\).

7. A bakery's daily number of online orders \(X\) has mean 40 and standard deviation 5; its daily number of phone orders \(Y\) has mean 25 and standard deviation 3. The two are independent. (a) Find the mean and standard deviation of the total number of orders \(T = X + Y\). (b) Find the mean and standard deviation of \(X - Y\). (c) A delivery service charges the bakery $3 per online order plus a flat $2 per day, so the daily cost is \(C = 3X + 2\). Find the mean and standard deviation of \(C\).

   Show answer

   (a) \(\mu_T = 40 + 25 = 65\) orders; \(\sigma_T^2 = 5^2 + 3^2 = 34\), so \(\sigma_T = \sqrt{34} \approx 5.83\) orders (variances add because \(X\) and \(Y\) are independent). (b) \(\mu_{X-Y} = 40 - 25 = 15\); the variance of a difference is still the sum \(25 + 9 = 34\), so \(\sigma_{X-Y} \approx 5.83\) as well. (c) \(\mu_C = 3(40) + 2 = \$122\); \(\sigma_C = 3(5) = \$15\): the flat $2 shifts the mean but does not change the spread. Never add standard deviations.

8. Fifteen percent of flights at an airport are delayed, independently of one another. Let \(X\) = the number of delayed flights among 12 randomly selected flights. (a) Explain why \(X\) is binomial. (b) Find \(P(X = 2)\). (c) Find \(P(X \ge 3)\). (d) Find and interpret the mean and standard deviation of \(X\).

   Show answer

   (a) Binary (delayed or not), Independent (given), fixed Number of trials \(n = 12\), Same probability \(p = 0.15\) each flight: \(X \sim B(12, 0.15)\). (b) \(P(X = 2) = \binom{12}{2}(0.15)^2(0.85)^{10} = 66(0.0225)(0.1969) \approx 0.292\) (binompdf(12, 0.15, 2)). (c) \(P(X \ge 3) = 1 - P(X \le 2) = 1 - 0.736 = 0.264\) (1 − binomcdf(12, 0.15, 2)). (d) \(\mu_X = 12(0.15) = 1.8\) flights and \(\sigma_X = \sqrt{12(0.15)(0.85)} \approx 1.24\) flights: over many sets of 12 flights, about 1.8 would be delayed on average, typically within about 1.24 of that.

9. A basketball player makes 40% of her three-point attempts, independently. She shoots until she makes one. Let \(X\) = the number of the attempt on which she makes her first shot. (a) Explain why \(X\) is geometric rather than binomial. (b) Find \(P(X = 3)\). (c) Find \(P(X \le 3)\) and \(P(X \gt 4)\). (d) Find the expected number of attempts.

   Show answer

   (a) The trials are binary, independent, with the same \(p = 0.4\), but there is no fixed number of trials: the count of attempts until the first success is the random variable. (b) \(P(X = 3) = (0.6)^2(0.4) = 0.144\): miss, miss, make. (c) \(P(X \le 3) = 1 - (0.6)^3 = 1 - 0.216 = 0.784\) (geometcdf(0.4, 3)); \(P(X \gt 4) = (0.6)^4 = 0.1296\), the probability the first four attempts all miss. (d) \(\mu_X = 1/p = 1/0.4 = 2.5\) attempts.

10. Events \(A\) and \(B\) have \(P(A) = 0.4\), \(P(B) = 0.5\), and \(P(A \cap B) = 0.2\). Which statement is true?

    \(A\) and \(B\) are mutually exclusive. \(A\) and \(B\) are independent. \(P(A \cup B) = 0.9\). \(P(A \mid B) = 0.5\).

    Mutually exclusive means the events cannot happen together, which requires \(P(A \cap B) = 0\). Here \(P(A \cap B) = 0.2\), so the events overlap; mutually exclusive and independent are different ideas, and these events are the second, not the first.

    \(P(A \mid B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{0.2}{0.5} = 0.4 = P(A)\), so knowing \(B\) occurred does not change the chance of \(A\); equivalently \(P(A)P(B) = (0.4)(0.5) = 0.2 = P(A \cap B)\).

    0.9 is \(P(A) + P(B)\) with the overlap forgotten, which counts the intersection twice. The general addition rule subtracts it: \(P(A \cup B) = 0.4 + 0.5 - 0.2 = 0.7\).

    0.5 is \(P(B)\), not \(P(A \mid B)\). The conditional probability is \(P(A \cap B)/P(B) = 0.2/0.5 = 0.4\); mixing up the conditioning event with the event of interest is the usual slip.

Lesson 5.1 · Unit 5 · CED topics 5.1–5.2

Normal calculations revisited and combining normal random variables

Unit 4 gave you the rules for the mean and variance of a sum or difference of random variables. One more fact makes those rules powerful: if the variables are normal and independent, their sum or difference is normal too. That means you can answer probability questions about totals and differences with the same z-score machinery you used in Unit 1.

Lesson 5.2 · Unit 5 · CED topics 5.3–5.4

Sampling distributions, bias, and variability

Take a random sample, compute a statistic, and you get one number. Take a different random sample and you get a different number. Inference is possible only because that sample-to-sample variation has a predictable pattern, and that pattern is the sampling distribution.

Exam tip: when asked to "describe the sampling distribution," give shape, center, and spread: all three, in context.

Lesson 5.3 · Unit 5 · CED topic 5.5

Sampling distribution of a sample proportion

A sample proportion \(\hat{p}\) is a count of successes divided by \(n\), and that count is binomial. Divide the binomial mean and standard deviation by \(n\) and you have the center and spread of \(\hat{p}\). When the counts are large enough, the shape is approximately normal, and probability questions about \(\hat{p}\) become z-score problems.

Exam tip: graders look for all three conditions-and-parameters in words. Write "approximately normal because 70 and 130 are both at least 10," not just "normal."

Lesson 5.4 · Unit 5 · CED topic 5.6

Sampling distribution of a difference of two proportions

Most interesting questions compare two groups: do seniors at School A work more than seniors at School B? The statistic is \(\hat{p}_1 - \hat{p}_2\), and its sampling distribution follows directly from the rules for differences of independent random variables.

Common error: subtracting the standard deviations, or forgetting to add the variances for a difference. The variance of a difference is a sum.

Lesson 5.5 · Unit 5 · CED topic 5.7

Sampling distribution of a sample mean and the Central Limit Theorem

Averages are less variable than individual observations, and, this is the remarkable part, averages of enough observations are approximately normal no matter what shape the population has. That second fact is the Central Limit Theorem, and it's the reason so much of inference works.

Lesson 5.6 · Unit 5 · CED topic 5.8

Sampling distribution of a difference of two means

The last piece of the unit combines two ideas you already have: the sampling distribution of \(\bar{x}\), and the rule that variances of independent variables add. Put them together and you can describe \(\bar{x}_1 - \bar{x}_2\), the statistic behind every two-sample comparison of means in Unit 7.

Exam tip: the formula sheet gives \(\sigma_{\bar x_1 - \bar x_2}\), but you must still state the conditions that justify using a normal model. "Both sample sizes are at least 30, so the CLT applies" is the sentence graders want.

Unit 5 practice · 10 problems

Unit 5 practice: Sampling Distributions

Ten problems covering the whole unit, in roughly exam order. For every probability, write the distribution of the statistic (shape, center, spread) and the z-score before you reach for normalcdf: that's where the points are.

1. A machine cuts rods of type A with lengths \(N(50.0, 0.4)\) mm and rods of type B with lengths \(N(30.0, 0.3)\) mm, independently. One rod of each type is joined end to end. What is the probability the assembled length exceeds 80.8 mm?

   Show answer

   Let \(L = A + B\). \(\mu_L = 50.0 + 30.0 = 80.0\) mm and \(\sigma_L = \sqrt{0.4^2 + 0.3^2} = \sqrt{0.25} = 0.5\) mm; since \(A\) and \(B\) are independent normals, \(L \sim N(80.0, 0.5)\). \[z = \frac{80.8 - 80.0}{0.5} = 1.6, \qquad P(L \gt 80.8) = P(Z \gt 1.6) \approx 0.0548.\] normalcdf(80.8, 1E99, 80, 0.5) = 0.0548.

2. Heights of adult men are approximately \(N(70, 3)\) inches and heights of adult women approximately \(N(64.5, 2.5)\) inches. A man and a woman are selected independently at random. What is the probability the woman is taller than the man?

   Show answer

   Let \(D = M - W\). \(\mu_D = 70 - 64.5 = 5.5\) in; \(\sigma_D = \sqrt{3^2 + 2.5^2} = \sqrt{15.25} \approx 3.905\) in (variances add for a difference). \(D\) is normal because \(M\) and \(W\) are independent normals. "Woman taller" means \(D \lt 0\): \[z = \frac{0 - 5.5}{3.905} \approx -1.41, \qquad P(D \lt 0) \approx 0.0795.\] About 8% of randomly paired couples would have the woman taller.

3. A population consists of three values: 3, 6, and 9, so \(\mu = 6\) and the population maximum is 9. List all samples of size 2 drawn without replacement, compute the sample mean and sample maximum for each, and determine whether each statistic is an unbiased estimator of the corresponding parameter.

   Show answer

   Samples: {3, 6}, {3, 9}, {6, 9}. Sample means: 4.5, 6, 7.5; their mean is \((4.5 + 6 + 7.5)/3 = 6 = \mu\), so \(\bar x\) is an unbiased estimator of \(\mu\). Sample maximums: 6, 9, 9; their mean is \(24/3 = 8 \ne 9\), so the sample maximum is a biased estimator that systematically underestimates the population maximum (it can never exceed 9 and usually falls short).

4. Four statistics are proposed for estimating a parameter. Their sampling distributions: I is centered at the parameter with a wide spread; II is centered at the parameter with a narrow spread; III is centered well above the parameter with a narrow spread; IV is centered well above the parameter with a wide spread. Which statistic is unbiased but has high variability?

   I II III IV

   “Unbiased” means the sampling distribution is centered at the parameter (true for I and II), and “high variability” means a wide spread (true for I and IV); only I has both. A larger sample would shrink its spread without moving its center.

   II is unbiased, but it has low variability: it is the ideal estimator, precise and centered on the target. The question asks for the one that is centered correctly but widely scattered.

   III has low variability but is biased: its center sits well above the parameter, so it is consistently wrong in the same direction. Precision is not accuracy, and a larger sample would not fix the bias.

   IV does have the high variability the question describes, but it is also biased, since it is centered above the parameter. It combines the worst of both properties rather than being the unbiased-but-variable case.

5. About 12% of adults are left-handed. A researcher takes an SRS of 300 adults from a large city. (a) Describe the sampling distribution of \(\hat p\), the sample proportion who are left-handed. (b) Find \(P(\hat p \ge 0.15)\). (c) Explain why part (b) could not be done the same way with an SRS of 50 adults.

   Show answer

   (a) Shape: approximately normal, because \(np = 300(0.12) = 36 \ge 10\) and \(n(1-p) = 264 \ge 10\). Center: \(\mu_{\hat p} = 0.12\). Spread: 300 is under 10% of the city's adults, so \(\sigma_{\hat p} = \sqrt{0.12(0.88)/300} \approx 0.0188\). (b) \(z = \dfrac{0.15 - 0.12}{0.0188} \approx 1.60\), so \(P(\hat p \ge 0.15) \approx 0.0549\) (normalcdf(0.15, 1E99, 0.12, √(0.12·0.88/300))). (c) With \(n = 50\), \(np = 6 \lt 10\); the Large Counts condition fails, the sampling distribution is skewed right, and a normal approximation is not justified: you would use the binomial distribution directly.

6. Among the customers of Store 1, 45% are members of its rewards program; at Store 2, 40% are. Independent SRSs of 200 customers from Store 1 and 250 from Store 2 are taken. What is the probability that the sample from Store 2 shows a higher proportion of members than the sample from Store 1?

   Show answer

   Let the statistic be \(\hat p_1 - \hat p_2\). Shape: the counts \(200(0.45) = 90\), \(110\), \(250(0.40) = 100\), and \(150\) are all at least 10, so approximately normal. Center: \(0.45 - 0.40 = 0.05\). Spread (both samples under 10% of their stores' customers): \[\sigma_{\hat p_1 - \hat p_2} = \sqrt{\frac{0.45(0.55)}{200} + \frac{0.40(0.60)}{250}} \approx 0.0469.\] "Store 2 higher" means \(\hat p_1 - \hat p_2 \lt 0\): \(z = \dfrac{0 - 0.05}{0.0469} \approx -1.07\), so the probability is about \(0.143\). Even with a real 5-point gap, about 1 sample pair in 7 would point the wrong way.

7. A filling machine puts \(N(20.0, 0.3)\) ounces of juice in each bottle. (a) Find the probability that one randomly selected bottle contains less than 19.8 oz. (b) Find the probability that the mean contents of an SRS of 9 bottles is less than 19.8 oz. (c) Explain why the normal calculation in (b) is valid even though \(n = 9 \lt 30\).

   Show answer

   (a) \(z = \dfrac{19.8 - 20.0}{0.3} \approx -0.67\), so \(P \approx 0.2525\). (b) \(\bar x\) has mean 20.0 and \(\sigma_{\bar x} = 0.3/\sqrt{9} = 0.1\) oz; \(z = \dfrac{19.8 - 20.0}{0.1} = -2\), so \(P(\bar x \lt 19.8) \approx 0.0228\). (c) When the population itself is normal, the sampling distribution of \(\bar x\) is exactly normal for any sample size; the \(n \ge 30\) rule (Central Limit Theorem) is only needed when the population shape is unknown or non-normal. One light bottle is common (25%); a light average of nine is rare (2%).

8. Tips left at a restaurant are strongly right-skewed with mean $5.20 and standard deviation $3.10. A server waits on 40 randomly selected tables in a week. (a) Find the probability that the server's mean tip exceeds $6.00, and justify the method. (b) Explain why you cannot use the same method to find the probability that a single tip exceeds $6.00.

   Show answer

   (a) \(n = 40 \ge 30\), so by the Central Limit Theorem \(\bar x\) is approximately normal despite the skewed population, with mean $5.20 and \(\sigma_{\bar x} = 3.10/\sqrt{40} \approx 0.490\). \[z = \frac{6.00 - 5.20}{0.490} \approx 1.63, \qquad P(\bar x \gt 6.00) \approx 0.0513.\] (b) A single tip comes from the population distribution, which is strongly skewed, not normal: the CLT says nothing about individual observations. Without knowing the population's shape, \(P(X \gt 6)\) cannot be computed.

9. Scores on a district math assessment have mean 68 and standard deviation 10 at School 1, and mean 65 with standard deviation 12 at School 2. Independent random samples of 50 students from School 1 and 40 from School 2 are taken. Find the probability that the School 1 sample mean exceeds the School 2 sample mean by more than 5 points.

   Show answer

   Shape: both \(n \ge 30\), so \(\bar x_1 - \bar x_2\) is approximately normal by the CLT. Center: \(68 - 65 = 3\) points. Spread: \[\sigma_{\bar x_1 - \bar x_2} = \sqrt{\frac{10^2}{50} + \frac{12^2}{40}} = \sqrt{2 + 3.6} \approx 2.366 \text{ points}.\] \(z = \dfrac{5 - 3}{2.366} \approx 0.85\), so \(P(\bar x_1 - \bar x_2 \gt 5) \approx 0.199\).

10. A researcher plans to estimate the mean commute time of a city's 400,000 workers with an SRS. (a) If the sample size is increased from 100 to 400, what happens to the mean and to the standard deviation of the sampling distribution of \(\bar x\)? (b) Would the answer change if the city had 4,000,000 workers instead? (c) Commute times are right-skewed; what can you say about the shape of the sampling distribution of \(\bar x\) for \(n = 400\)?

    Show answer

    (a) The mean stays at \(\mu\): \(\bar x\) is unbiased regardless of \(n\). The standard deviation \(\sigma/\sqrt{n}\) is cut in half, since \(\sqrt{400} = 2\sqrt{100}\); quadrupling the sample halves the spread. (b) No. As long as the sample is under 10% of the population, the population size has essentially no effect on \(\sigma_{\bar x}\): 400 out of 400,000 and 400 out of 4,000,000 give the same spread. (c) With \(n = 400 \ge 30\), the Central Limit Theorem says the sampling distribution of \(\bar x\) is approximately normal even though individual commute times are skewed.

Lesson 6.1 · Unit 6 · CED topics 6.1–6.2

Confidence intervals for a population proportion

A sample proportion is a single guess at \(p\), and Unit 5 told you how far off that guess typically is. A confidence interval turns that knowledge into a range of plausible values for \(p\), with a stated level of confidence. Every interval in this course has the same shape: estimate ± (critical value)(standard error).

Lesson 6.2 · Unit 6 · CED topic 6.3

Interpreting confidence intervals and choosing a sample size

Computing an interval is the easy part. The exam cares far more about whether you can say what it means, and what it doesn't. Two different sentences are involved: one interprets the interval, the other interprets the confidence level.

The margin of error covers only random sampling variability. It says nothing about undercoverage, nonresponse, or badly worded questions: a biased sample of 10,000 has a tiny margin of error and a wrong answer.

Lesson 6.3 · Unit 6 · CED topics 6.4–6.6

Significance tests for a population proportion

A confidence interval estimates a parameter. A significance test weighs the evidence against a specific claim about it. The logic: assume the claim is true, ask how surprising the sample would be, and if it's surprising enough, reject the claim.

Lesson 6.4 · Unit 6 · CED topic 6.7

Conclusions, errors, and power

A significance test is a decision made under uncertainty, so it can be wrong in two ways. Knowing which error matters more tells you how to choose \(\alpha\); knowing what "power" means tells you how to design a study that can actually detect what you're looking for.

Wording that loses points: never write "accept \(H_0\)": failing to reject means the data are consistent with \(H_0\), not that it's true. Conclusions are about the alternative: "there is (or is not) convincing evidence that …".

Lesson 6.5 · Unit 6 · CED topics 6.8–6.9

Confidence intervals for a difference of two proportions

Comparing two groups is the most common inference task on the exam. The parameter is now a difference, \(p_1 - p_2\), and the standard error is Lesson 5.4's formula with the sample proportions standing in for the unknown \(p_1\) and \(p_2\).

Lesson 6.6 · Unit 6 · CED topics 6.10–6.11

Significance tests for a difference of two proportions

The two-sample test asks whether an observed gap between two sample proportions is too large to be chance. One twist: the null hypothesis says the two population proportions are equal, so the best estimate of that common proportion pools both samples together.

Unit 6 practice · 10 problems

Unit 6 practice, Inference for Categorical Data: Proportions

Ten problems covering the whole unit, in roughly exam order. Problems 2, 5, and 8 are full four-step write-ups: State, Plan (with conditions), Do, Conclude in context. Use 1-PropZInt, 1-PropZTest, 2-PropZInt, and 2-PropZTest to check your arithmetic, but write the formulas with numbers substituted.

1. Find the critical value \(z^*\) for a 98% confidence interval, and explain why the input to invNorm is 0.99 rather than 0.98.

   Show answer

   A 98% interval leaves 2% in the two tails combined, 1% in each. The critical value is the z-score with area 0.99 to its left (0.98 in the middle plus 0.01 in the lower tail): \(z^* = \text{invNorm}(0.99, 0, 1) \approx 2.326\). Entering 0.98 would give the value with 2% in the upper tail alone, which corresponds to a 96% interval. For reference: 1.645 (90%), 1.960 (95%), 2.576 (99%).

2. In a random sample of 300 residents of a city of 60,000 adults, 174 said they favor a proposed park levy. Construct and interpret a 95% confidence interval for the proportion of all adult residents who favor the levy.

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   State: \(p\) = the true proportion of all adult residents of the city who favor the park levy. We want a 95% confidence interval for \(p\).

   Plan: one-sample z interval for \(p\). Random: random sample ✓. 10%: \(300 \le 6{,}000\) ✓. Large Counts: 174 successes and 126 failures are both at least 10 ✓.

   Do: \(\hat p = 174/300 = 0.58\). \[0.58 \pm 1.960\sqrt{\frac{0.58(0.42)}{300}} = 0.58 \pm 1.960(0.0285) = 0.58 \pm 0.056 \;\Rightarrow\; (0.524,\, 0.636).\] (1-PropZInt with x = 174, n = 300 gives (0.5241, 0.6359).)

   Conclude: We are 95% confident that the true proportion of all adult residents of the city who favor the park levy is between 0.524 and 0.636. Because the whole interval is above 0.50, the data give convincing evidence that a majority favor the levy.

3. Refer to the interval (0.524, 0.636) from the previous problem. Which statement is a correct interpretation of the 95% confidence level?

   There is a 95% probability that the true proportion is between 0.524 and 0.636. About 95% of the city's residents have between 52.4% and 63.6% support for the levy. If many random samples of 300 residents were taken and a 95% confidence interval computed from each, about 95% of those intervals would capture the true proportion who favor the levy. About 95% of all sample proportions would fall between 0.524 and 0.636.

   The population proportion \(p\) is a fixed number, so this particular interval either contains it or it doesn't: there is no 95% probability about it. The 95% belongs to the method over many samples, not to this one interval.

   A population proportion is a single number describing the whole city; individual residents don't each “have” a support percentage. This statement misreads a parameter as a property of individuals and makes no sense.

   The confidence level describes the long-run capture rate of the method: in repeated sampling, about 95% of the intervals produced this way would contain the true proportion, and about 5% would miss it.

   This describes the sampling distribution of \(\hat p\), which is centered at the true \(p\), not at this sample's 0.58. The confidence level is about intervals capturing \(p\), not about other sample proportions landing inside this particular interval.

4. A state agency wants to estimate the proportion of drivers who use a phone while driving, with a margin of error of at most 2 percentage points at 95% confidence. (a) What sample size is needed if no prior estimate is available? (b) A previous study found about 30% of drivers use a phone. What sample size does that estimate suggest?

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   (a) With \(\hat p = 0.5\) (the conservative choice, since \(\hat p(1-\hat p)\) is largest there): \(n \ge \left(\dfrac{1.960}{0.02}\right)^2(0.5)(0.5) = 2401\). (b) With \(\hat p = 0.3\): \(n \ge \left(\dfrac{1.960}{0.02}\right)^2(0.3)(0.7) = 2016.84\), so \(n = 2017\): always round up. The prior estimate saves nearly 400 drivers. Note the 2-point margin is expensive: halving it to 1 point would quadruple the sample.

5. A manufacturer claims that only 5% of its LED bulbs fail within the first year. A consumer group suspects the true failure rate is higher and tests an SRS of 400 bulbs from the manufacturer's large output; 30 fail within a year. Is there convincing evidence that the failure rate exceeds 5%? Use \(\alpha = 0.05\).

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   State: \(p\) = the true proportion of the manufacturer's LED bulbs that fail within the first year. \(H_0\!: p = 0.05\) (the claim is accurate); \(H_a\!: p \gt 0.05\) (the failure rate is higher than claimed). \(\alpha = 0.05\).

   Plan: one-sample z test for \(p\). Random: SRS ✓. 10%: 400 bulbs is far less than 10% of the manufacturer's output ✓. Large Counts (using \(p_0\)): \(400(0.05) = 20\) and \(400(0.95) = 380\), both at least 10 ✓.

   Do: \(\hat p = 30/400 = 0.075\). \[z = \frac{0.075 - 0.05}{\sqrt{\dfrac{0.05(0.95)}{400}}} = \frac{0.025}{0.0109} \approx 2.29, \qquad \text{p-value} = P(Z \ge 2.29) \approx 0.0109.\]

   Conclude: Because the p-value of 0.0109 is less than \(\alpha = 0.05\), we reject \(H_0\). There is convincing evidence that more than 5% of the manufacturer's LED bulbs fail within the first year.

6. For the bulb test in the previous problem: (a) Describe a Type I error and a Type II error in context, and give a consequence of each. (b) Name two changes that would increase the power of the test, and state the cost of each.

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   (a) Type I: the consumer group concludes the failure rate exceeds 5% when it really is 5%, the manufacturer is wrongly accused, perhaps facing a recall or bad publicity it doesn't deserve. Type II: the failure rate really is higher than 5%, but the sample doesn't provide convincing evidence, so the group fails to reject \(H_0\); consumers keep buying bulbs that fail more often than advertised. (b) Testing more bulbs (larger \(n\)) shrinks the standard deviation of \(\hat p\), making a real excess easier to detect; the cost is time and money. Using a larger \(\alpha\), say 0.10, makes rejection easier; the cost is a higher probability of a Type I error. (A larger true failure rate also raises power, but that isn't under the group's control.)

7. A community college compares pass rates for an online section and an in-person section of the same statistics course. Of 200 randomly sampled students who took the course online, 152 passed; of 240 randomly sampled in-person students, 168 passed. Construct a 95% confidence interval for the difference in pass rates (online minus in-person) and state whether it gives convincing evidence of a difference.

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   \(\hat p_1 = 152/200 = 0.76\), \(\hat p_2 = 168/240 = 0.70\). Conditions: independent random samples; each sample is under 10% of the students who take the course; successes and failures 152, 48, 168, 72 are all at least 10. \[(0.76 - 0.70) \pm 1.960\sqrt{\frac{0.76(0.24)}{200} + \frac{0.70(0.30)}{240}} = 0.06 \pm 1.960(0.0423) = 0.06 \pm 0.083 \;\Rightarrow\; (-0.023,\, 0.143).\] We are 95% confident that the true difference in pass rates (online minus in-person) is between −0.023 and 0.143. Because the interval contains 0, a difference of zero is plausible: there is not convincing evidence that the pass rates differ.

8. A nursery randomly assigned 300 tomato seedlings to receive a fungicide treatment (150 plants) or no treatment (150 plants). After six weeks, 120 treated seedlings and 102 untreated seedlings had survived. Is there convincing evidence at \(\alpha = 0.05\) that the fungicide increases the survival rate?

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   State: \(p_1\) = the true proportion of seedlings like these that would survive with the fungicide; \(p_2\) = the same without it. \(H_0\!: p_1 - p_2 = 0\) (the fungicide has no effect on survival); \(H_a\!: p_1 - p_2 \gt 0\) (the fungicide increases survival). \(\alpha = 0.05\).

   Plan: two-sample z test for \(p_1 - p_2\). Random: treatments were randomly assigned ✓ (so the 10% condition does not apply). Large Counts, using the pooled proportion \(\hat p_c = 222/300 = 0.74\): \(150(0.74) = 111\) and \(150(0.26) = 39\) for each group, all at least 10 ✓.

   Do: \(\hat p_1 = 120/150 = 0.80\), \(\hat p_2 = 102/150 = 0.68\). \[z = \frac{0.80 - 0.68}{\sqrt{0.74(0.26)\left(\dfrac{1}{150} + \dfrac{1}{150}\right)}} = \frac{0.12}{0.0506} \approx 2.37, \qquad \text{p-value} = P(Z \ge 2.37) \approx 0.0089.\]

   Conclude: Because the p-value of 0.0089 is less than \(\alpha = 0.05\), we reject \(H_0\). There is convincing evidence that the fungicide increases the proportion of seedlings like these that survive six weeks. Because treatments were randomly assigned, this is a cause-and-effect conclusion.

9. The LED bulb test in problem 5 produced a p-value of 0.0109. (a) Interpret this p-value in context. (b) Suppose instead the sample had produced a p-value of 0.18. Write the conclusion, and explain why "we accept \(H_0\); the failure rate is 5%" would be wrong.

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   (a) Assuming the true failure rate is 0.05, there is a 0.0109 probability of getting a sample proportion of 0.075 or greater in a random sample of 400 bulbs purely by chance. (b) Because 0.18 is greater than \(\alpha = 0.05\), we fail to reject \(H_0\): the data do not provide convincing evidence that more than 5% of the bulbs fail within a year. Failing to reject means the data are consistent with a 5% failure rate: it does not show that the rate is exactly 5%. Many other values (4%, 6%) would also be consistent with the data; a confidence interval would show which. "Accept \(H_0\)" claims more than the test can deliver and loses the conclusion point.

10. A pollster wants a narrower 95% confidence interval for a population proportion. Which change will make the interval narrower?

    Raising the confidence level to 99%. Reducing the sample size from 1,000 to 500. Increasing the sample size from 500 to 2,000. Using \(\hat p = 0.5\) in the standard error instead of the observed \(\hat p = 0.3\).

    Raising the confidence level raises the critical value from \(z^* = 1.960\) to \(2.576\), which widens the interval. More confidence always costs width; you cannot get both narrower and more confident without more data.

    Halving \(n\) multiplies the standard error by \(\sqrt{2}\), so the interval gets wider, not narrower. Sample size sits in the denominator of the margin of error, so less data means less precision.

    The margin of error is \(z^*\sqrt{\hat p(1-\hat p)/n}\), so quadrupling \(n\) divides it by \(\sqrt{4} = 2\) and the interval is half as wide.

    \(\hat p(1-\hat p)\) is largest at 0.5, so substituting 0.5 gives the widest possible interval, which is exactly why 0.5 is the conservative choice when planning a sample size, not a way to tighten an interval.

Lesson 7.1 · Unit 7 · CED topics 7.1–7.2

The t-distribution and confidence intervals for a mean

To build an interval for \(\mu\) you need \(\sigma_{\bar x} = \sigma/\sqrt{n}\), but in real life you never know \(\sigma\). You substitute the sample standard deviation \(s\), and that substitution adds uncertainty. The statistic \((\bar x - \mu)/(s/\sqrt{n})\) no longer follows the standard normal curve; it follows a t-distribution.

Lesson 7.2 · Unit 7 · CED topic 7.3

Interpreting intervals for a mean and choosing a sample size

The interpretation rules from Unit 6 carry over word for word: the parameter is just \(\mu\) instead of \(p\). What's new is thinking about what controls the width of a t interval, and how to plan a sample size before collecting data.

Lesson 7.3 · Unit 7 · CED topics 7.4–7.6

Significance tests for a population mean

Same logic as the test for a proportion: assume the claimed value of \(\mu\), measure how many standard errors away the sample mean landed, and convert that to a p-value. The only changes are the t statistic and its degrees of freedom.

Lesson 7.4 · Unit 7 · CED topic 7.7

Paired data: the matched-pairs t procedure

Before-and-after measurements on the same people are not two independent samples: each "after" belongs to a specific "before." The right move is to subtract within each pair and analyze the single list of differences with the one-sample t procedures you already know.

Lesson 7.5 · Unit 7 · CED topics 7.7–7.8

Confidence intervals for a difference of two means

When two groups are genuinely independent (different people, different batteries, no pairing) you estimate \(\mu_1 - \mu_2\) directly. The standard error comes from Lesson 5.6 with the sample standard deviations substituted, and there's a wrinkle about degrees of freedom.

Lesson 7.6 · Unit 7 · CED topics 7.9–7.10

Significance tests for a difference of two means

The two-sample t test is the workhorse of experimental comparisons: did the treatment group's mean differ from the control group's by more than chance would produce? Unlike the two-proportion test, there is no pooling: each group keeps its own standard deviation.

Unit 7 practice · 10 problems

Unit 7 practice, Inference for Quantitative Data: Means

Ten problems covering the whole unit, in roughly exam order. Problems 2, 5, 6, and 9 are full four-step write-ups. Use invT, TInterval, T-Test, 2-SampTInt, and 2-SampTTest (Pooled: No) to check the arithmetic; always report the degrees of freedom.

1. (a) Find the critical value \(t^*\) for a 95% confidence interval for a mean based on a sample of size 12. (b) Explain why a t critical value is used instead of \(z^* = 1.960\). (c) What happens to \(t^*\) as the sample size grows?

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   (a) df \(= 12 - 1 = 11\); \(t^* = \text{invT}(0.975, 11) \approx 2.201\). (b) The population standard deviation \(\sigma\) is unknown and is estimated by the sample standard deviation \(s\). That substitution adds variability to the standardized statistic \((\bar x - \mu)/(s/\sqrt{n})\), which therefore follows a t-distribution with heavier tails than the standard normal, so a larger critical value is needed to reach 95% confidence. (c) As \(n\) (and df) increases, the t-distribution approaches the standard normal and \(t^*\) decreases toward 1.960, at df = 99 it is already 1.984.

2. A consumer lab measured the caffeine content (mg) of 8 randomly selected 12-ounce cups of coffee from a national chain: 95, 102, 88, 110, 97, 105, 92, 99. Construct and interpret a 95% confidence interval for the mean caffeine content of the chain's 12-ounce coffees.

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   State: \(\mu\) = the true mean caffeine content (mg) of the chain's 12-ounce cups of coffee. 95% confidence interval.

   Plan: one-sample t interval for \(\mu\). Random: random sample of cups ✓. 10%: 8 cups is far under 10% of all cups the chain serves ✓. Normal/Large Sample: \(n = 8 \lt 30\), but a dotplot of the data (88, 92, 95, 97, 99, 102, 105, 110) is roughly symmetric with no outliers ✓.

   Do: \(\bar x = 98.5\) mg, \(s = 7.11\) mg, df = 7, \(t^* = \text{invT}(0.975, 7) = 2.365\). \[98.5 \pm 2.365\,\frac{7.11}{\sqrt{8}} = 98.5 \pm 2.365(2.514) = 98.5 \pm 5.95 \;\Rightarrow\; (92.55,\, 104.45).\]

   Conclude: We are 95% confident that the true mean caffeine content of the chain's 12-ounce coffees is between 92.6 mg and 104.4 mg.

3. The chain in the previous problem advertises 100 mg of caffeine per 12-ounce cup. (a) Does the interval (92.6, 104.4) give convincing evidence that the true mean differs from 100 mg? Explain. (b) A student writes, "95% of the chain's cups contain between 92.6 and 104.4 mg of caffeine." Explain why this is incorrect.

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   (a) No. 100 mg is inside the interval, so it is a plausible value for the true mean; at the 5% significance level the data are consistent with the advertised amount. (b) The interval estimates the mean caffeine content, not the contents of individual cups. Individual cups vary far more than the sample mean does, the sample itself ranged from 88 to 110 mg, so far fewer than 95% of cups would fall in such a narrow range. Confidence intervals are statements about parameters, never about individual observations or about \(\bar x\).

4. A hospital wants to estimate the mean time patients spend in its emergency-room waiting area to within ±1.5 minutes with 95% confidence. A pilot study suggests \(\sigma \approx 8\) minutes. How many patients should be sampled?

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   Since \(n\) (and therefore df) is unknown in advance, use \(z^*\): \[n \ge \left(\frac{z^*\sigma}{ME}\right)^2 = \left(\frac{1.960 \times 8}{1.5}\right)^2 = 109.3,\] so sample at least 110 patients. Round up: 109 patients would leave the margin slightly above 1.5 minutes.

5. A pharmacy claims its average prescription wait time is 10 minutes. A reporter times a random sample of 35 customers and finds \(\bar x = 11.2\) minutes with \(s = 3.6\) minutes. Is there convincing evidence at \(\alpha = 0.05\) that the true mean wait time exceeds 10 minutes?

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   State: \(\mu\) = the true mean prescription wait time at this pharmacy. \(H_0\!: \mu = 10\) (the claim is accurate); \(H_a\!: \mu \gt 10\) (waits are longer than claimed on average). \(\alpha = 0.05\).

   Plan: one-sample t test for \(\mu\). Random: random sample of customers ✓. 10%: 35 is far under 10% of the pharmacy's customers ✓. Normal/Large Sample: \(n = 35 \ge 30\), so the sampling distribution of \(\bar x\) is approximately normal by the CLT ✓.

   Do: \[t = \frac{11.2 - 10}{3.6/\sqrt{35}} = \frac{1.2}{0.6085} \approx 1.97, \qquad \text{df} = 34, \qquad \text{p-value} = P(t_{34} \ge 1.97) \approx 0.0284.\] (T-Test with Stats: μ₀ = 10, \(\bar x\) = 11.2, Sx = 3.6, n = 35, μ > μ₀.)

   Conclude: Because the p-value of 0.0284 is less than \(\alpha = 0.05\), we reject \(H_0\). There is convincing evidence that the true mean prescription wait time at this pharmacy is greater than 10 minutes.

6. Eight randomly selected volunteers completed a reaction-time test (milliseconds) after a normal night's sleep and again after 24 hours without sleep.

   Volunteer	1	2	3	4	5	6	7	8
   Rested	245	268	232	290	255	240	275	262
   Sleep-deprived	262	280	239	301	271	238	296	274

   Is there convincing evidence at \(\alpha = 0.05\) that sleep deprivation increases mean reaction time? Explain why a paired procedure is appropriate.

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   Each volunteer supplies both measurements, so the two lists are linked pair by pair; analyze the differences \(d = \) sleep-deprived − rested: 17, 12, 7, 11, 16, −2, 21, 12.

   State: \(\mu_d\) = the true mean increase in reaction time (sleep-deprived minus rested) for volunteers like these. \(H_0\!: \mu_d = 0\) (no change on average); \(H_a\!: \mu_d \gt 0\) (reaction time increases). \(\alpha = 0.05\).

   Plan: paired t test for \(\mu_d\). Random: volunteers were randomly selected ✓. 10%: 8 is far under 10% of the population of interest ✓. Normal/Large Sample: \(n = 8\), but a dotplot of the eight differences shows no outliers or strong skewness ✓.

   Do: \(\bar d = 11.75\) ms, \(s_d = 7.01\) ms. \[t = \frac{11.75 - 0}{7.01/\sqrt{8}} = \frac{11.75}{2.477} \approx 4.74, \qquad \text{df} = 7, \qquad \text{p-value} = P(t_7 \ge 4.74) \approx 0.0010.\]

   Conclude: Because the p-value of 0.0010 is less than \(\alpha = 0.05\), we reject \(H_0\). There is convincing evidence that 24 hours of sleep deprivation increases mean reaction time for volunteers like these.

7. For each study, state whether a paired t procedure or a two-sample t procedure is appropriate, and explain why. (a) Ten cars are each driven on a fixed course once with regular gasoline and once with premium, in random order, and fuel economy is recorded each time. (b) Random samples of 40 freshmen and 40 seniors report their hours of sleep last night. (c) Twenty pairs of identical twins are recruited; within each pair, one twin is randomly assigned to a memory-training program and the other is not, and both take a memory test afterward.

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   (a) Paired: each car is measured twice, so each premium result is naturally matched to a regular result from the same car; analyze the ten differences. Pairing removes car-to-car variation from the comparison. (b) Two-sample: the freshmen and seniors are separate, independent groups with no natural link between any freshman and any senior. (c) Paired (matched pairs): the twins within a pair are alike, so the difference within each pair isolates the training effect; analyze the twenty differences. The question to ask: is there a natural one-to-one link between the observations in the two groups?

8. Independent random samples of two brands of rechargeable batteries were tested. Brand 1: \(n_1 = 20\), \(\bar x_1 = 34.2\) hours, \(s_1 = 3.1\) hours. Brand 2: \(n_2 = 25\), \(\bar x_2 = 31.8\) hours, \(s_2 = 3.9\) hours. Neither sample shows outliers or strong skewness. Construct a 95% confidence interval for \(\mu_1 - \mu_2\) using the conservative degrees of freedom, interpret it, and state what it says about the brands.

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   Conditions: independent random samples; each sample is a tiny fraction of production; both \(n \lt 30\) but neither sample shows outliers or strong skew, so Normal/Large Sample is met. Conservative df = \(20 - 1 = 19\), \(t^* = \text{invT}(0.975, 19) = 2.093\). \[(34.2 - 31.8) \pm 2.093\sqrt{\frac{3.1^2}{20} + \frac{3.9^2}{25}} = 2.4 \pm 2.093(1.044) = 2.4 \pm 2.18 \;\Rightarrow\; (0.22,\, 4.58).\] We are 95% confident that the true mean lifetime of Brand 1 exceeds that of Brand 2 by between 0.22 and 4.58 hours. Because the entire interval is above 0, there is convincing evidence that Brand 1 batteries last longer on average. (2-SampTInt with df = 43.0 gives (0.30, 4.50): same conclusion.)

9. Fifty students in an introductory course were randomly assigned to learn a chapter with a new interactive study technique (24 students) or with the standard textbook approach (26 students), then took the same test. Interactive: \(\bar x_1 = 78.3\), \(s_1 = 8.1\). Standard: \(\bar x_2 = 72.9\), \(s_2 = 9.4\). Both groups' scores are roughly symmetric with no outliers. Is there convincing evidence at \(\alpha = 0.05\) that the mean test scores differ between the two techniques?

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   State: \(\mu_1 - \mu_2\), where \(\mu_1\) and \(\mu_2\) are the true mean test scores for students like these using the interactive and standard techniques. \(H_0\!: \mu_1 - \mu_2 = 0\) (the techniques produce the same mean score); \(H_a\!: \mu_1 - \mu_2 \ne 0\) (the mean scores differ). \(\alpha = 0.05\).

   Plan: two-sample t test for \(\mu_1 - \mu_2\). Random: students were randomly assigned to techniques ✓ (so no 10% condition). Normal/Large Sample: both \(n \lt 30\), but both groups are roughly symmetric with no outliers ✓.

   Do: \[t = \frac{78.3 - 72.9}{\sqrt{\dfrac{8.1^2}{24} + \dfrac{9.4^2}{26}}} = \frac{5.4}{2.476} \approx 2.18.\] Calculator df = 47.8: two-sided p-value \(\approx 0.034\). Conservative df = 23: p-value \(\approx 0.040\).

   Conclude: Because the p-value of 0.034 is less than \(\alpha = 0.05\), we reject \(H_0\). There is convincing evidence that the mean test scores differ between the two techniques for students like these: and since the interactive group scored higher, that the interactive technique raised scores. Random assignment permits this cause-and-effect conclusion.

10. A 2-SampTTest with \(H_a\!: \mu_1 \ne \mu_2\) (Pooled: No) reports t = 2.35, p = 0.0236, df = 41.3. At \(\alpha = 0.01\), which conclusion is correct?

    Reject \(H_0\); there is convincing evidence the means differ. Fail to reject \(H_0\); there is not convincing evidence at the 1% level that the means differ. Accept \(H_0\); the two means are equal. Reject \(H_0\) because t is greater than 2.

    Rejecting requires \(p \le \alpha\), and \(0.0236 \gt 0.01\). This conclusion would be correct at \(\alpha = 0.05\), so a student who picks it has probably compared p to the default 5% rather than the stated 1%.

    The p-value 0.0236 is greater than \(\alpha = 0.01\), so we fail to reject \(H_0\): the data do not provide convincing evidence at the 1% level that the means differ. (The non-integer df = 41.3 confirms the unpooled procedure was used.)

    A significance test can never prove \(H_0\) true; failing to reject means the evidence against it is insufficient, not that the means are equal. “Accept \(H_0\)” is language graders specifically penalize.

    The decision compares the p-value with \(\alpha\), not \(t\) with an arbitrary cutoff like 2. A \(t\) of 2.35 is fairly large, but at this df the two-sided 1% threshold is about 2.70, so it falls short.

Lesson 8.1 · Unit 8 · CED topics 8.1–8.3

Chi-square goodness-of-fit

Unit 6 handled a categorical variable with two outcomes. When there are three or more categories (flavors, colors, days of the week) a single proportion won't do. The chi-square goodness-of-fit test compares the whole set of observed counts with the counts a claimed distribution predicts.

Lesson 8.2 · Unit 8 · CED topics 8.4–8.5, 8.7

Chi-square test for homogeneity

Now compare several populations on one categorical variable: do students at three schools prefer the same lunch options? The data form a two-way table, and the expected counts come from the table itself rather than from a claimed distribution.

Lesson 8.3 · Unit 8 · CED topics 8.6–8.7

Chi-square test for independence

Same table, same arithmetic, different question. When a single random sample is classified by two categorical variables, you're not comparing populations: you're asking whether the two variables are associated in the one population you sampled.

Compare the two conclusion sentences. Homogeneity: "the distribution of lunch preference differs among the three schools." Independence: "there is an association between age group and news source." Using the wrong one signals that you misread the design: a common deduction.

Lesson 8.4 · Unit 8 · CED topic 8.7

Choosing the right chi-square test and following up

The three chi-square tests share one formula, so the exam tests whether you can tell them apart by design, and whether you can say something useful after rejecting \(H_0\). "The distributions differ" is a start; a grader wants to know how.

Exam tip: a chi-square test never proves causation and never tells you the strength of an association. For that, compare the conditional distributions (row percentages), as in Unit 2.

Unit 8 practice · 10 problems

Unit 8 practice, Inference for Categorical Data: Chi-Square

Ten problems covering the whole unit, in roughly exam order. Problems 2, 4, and 5 are full four-step write-ups. Use χ²GOF-Test and χ²-Test (with matrices) to check the arithmetic, but show the expected counts and at least the first terms of the sum.

1. A candy company claims its bags contain 24% blue, 20% orange, 16% green, 14% yellow, 13% red, and 13% brown candies. A random sample of 250 candies will be used to test the claim. (a) Find the expected count for each color and the degrees of freedom. (b) Is the Large Counts condition met? (c) A student objects that an expected count of 32.5 candies is impossible. Respond.

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   (a) Expected counts \(E_i = 250p_i\): blue 60, orange 50, green 40, yellow 35, red 32.5, brown 32.5 (they sum to 250). df \(= 6 - 1 = 5\). (b) Yes: every expected count is at least 5. (c) An expected count is a long-run average, not a prediction for one bag: if the claim is true, samples of 250 would average 32.5 red candies. Expected counts are never rounded to whole numbers.

2. A video game's publisher states that its loot boxes contain a common item 60% of the time, a rare item 30%, and an epic item 10%. Players who opened a random sample of 300 boxes recorded 165 common, 100 rare, and 35 epic items. Is there convincing evidence at \(\alpha = 0.05\) that the actual distribution differs from the publisher's claim?

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   State: \(H_0\): the distribution of loot-box items is 60% common, 30% rare, 10% epic. \(H_a\): at least one of these proportions is different. \(\alpha = 0.05\).

   Plan: chi-square goodness-of-fit test. Random: random sample of boxes ✓. 10%: 300 is under 10% of all boxes opened ✓. Large Counts: expected counts \(300(0.6) = 180\), \(300(0.3) = 90\), \(300(0.1) = 30\) are all at least 5 ✓.

   Do: \[\chi^2 = \frac{(165 - 180)^2}{180} + \frac{(100 - 90)^2}{90} + \frac{(35 - 30)^2}{30} = 1.25 + 1.11 + 0.83 = 3.19,\] df = 2, p-value \(= P(\chi^2_2 \ge 3.19) \approx 0.202\) (χ²cdf(3.19, 1E99, 2)).

   Conclude: Because the p-value of 0.202 is greater than \(\alpha = 0.05\), we fail to reject \(H_0\). There is not convincing evidence that the distribution of loot-box items differs from the publisher's claim.

3. In a two-way table with 3 rows and 4 columns and a grand total of 400, one cell lies in a row with total 130 and a column with total 90. (a) Find the expected count for that cell under the null hypothesis. (b) Find the degrees of freedom for the chi-square test. (c) Explain in one sentence where the expected-count formula comes from.

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   (a) \(E = \dfrac{(\text{row total})(\text{column total})}{\text{grand total}} = \dfrac{130 \times 90}{400} = 29.25\). (b) df \(= (3 - 1)(4 - 1) = 6\). (c) If the variables were unrelated, the column's share of the whole table, \(90/400\), would apply equally to every row, so that row's 130 individuals would be expected to include \(130 \times 90/400\) in this column.

4. Independent random samples of 120 urban and 80 rural residents of a state were asked their primary means of getting to work.

   Residence	Drive	Public transit	Bike or walk	Total
   Urban	60	40	20	120
   Rural	60	8	12	80
   Total	120	48	32	200

   Is there convincing evidence at \(\alpha = 0.05\) that the distribution of commuting method differs between urban and rural residents of the state?

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   State: \(H_0\): the distribution of primary commuting method is the same for urban and rural residents of the state. \(H_a\): the distributions differ. \(\alpha = 0.05\).

   Plan: chi-square test for homogeneity (two separate samples, one categorical variable). Random: two independent random samples ✓. 10%: each sample is under 10% of its population ✓. Large Counts: expected counts; urban 72, 28.8, 19.2; rural 48, 19.2, 12.8 (e.g., \(120 \times 120/200 = 72\)): are all at least 5 ✓.

   Do: \[\chi^2 = \frac{(60 - 72)^2}{72} + \frac{(40 - 28.8)^2}{28.8} + \frac{(20 - 19.2)^2}{19.2} + \frac{(60 - 48)^2}{48} + \frac{(8 - 19.2)^2}{19.2} + \frac{(12 - 12.8)^2}{12.8} \approx 15.97,\] df \(= (2 - 1)(3 - 1) = 2\), p-value \(\approx 0.0003\).

   Conclude: Because the p-value of 0.0003 is less than \(\alpha = 0.05\), we reject \(H_0\). There is convincing evidence that the distribution of primary commuting method differs between urban and rural residents of the state.

5. A random sample of 300 adults in a county was asked whether they exercise at least three times a week and whether they rate their sleep quality as good or poor.

   Exercise 3+ times/week	Good sleep	Poor sleep	Total
   Yes	90	50	140
   No	60	100	160
   Total	150	150	300

   Is there convincing evidence at \(\alpha = 0.01\) of an association between regular exercise and sleep quality among adults in this county?

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   State: \(H_0\): there is no association between regular exercise and sleep quality among adults in the county. \(H_a\): there is an association. \(\alpha = 0.01\).

   Plan: chi-square test for independence (one sample, two variables). Random: random sample ✓. 10%: 300 is under 10% of the county's adults ✓. Large Counts: expected counts \(140 \times 150/300 = 70\), 70, 80, 80 are all at least 5 ✓.

   Do: \[\chi^2 = \frac{(90 - 70)^2}{70} + \frac{(50 - 70)^2}{70} + \frac{(60 - 80)^2}{80} + \frac{(100 - 80)^2}{80} = 5.71 + 5.71 + 5.00 + 5.00 = 21.43,\] df \(= (2 - 1)(2 - 1) = 1\), p-value \(\approx 0.000004\) (report as \(\lt 0.0001\)).

   Conclude: Because the p-value is far less than \(\alpha = 0.01\), we reject \(H_0\). There is convincing evidence of an association between regular exercise and sleep quality among adults in this county: 64% of regular exercisers reported good sleep versus 38% of non-exercisers. (This is an observational study, so exercise cannot be said to cause better sleep.)

6. Name the appropriate chi-square test and its degrees of freedom for each study. (a) A random sample of 500 registered voters is classified by party (three categories) and opinion on a ballot measure (favor, oppose, undecided). (b) A blood bank compares the blood-type distribution (O, A, B, AB) of 200 randomly selected donors with the known national percentages. (c) Random samples of 100 residents from each of three cities are asked which of four coffee chains they prefer.

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   (a) Test for independence: one sample, two categorical variables measured on each voter; df \(= (3 - 1)(3 - 1) = 4\). (b) Goodness-of-fit: one sample, one variable, compared with a claimed distribution; df \(= 4 - 1 = 3\). (c) Test for homogeneity: three separate samples (row totals fixed by the researcher), one categorical variable; df \(= (3 - 1)(4 - 1) = 6\). Homogeneity and independence share the arithmetic; the design decides the name and the wording of the conclusion.

7. Refer to the urban/rural commuting test in problem 4 (\(\chi^2 = 15.97\)). Identify the two cells that contribute most to the chi-square statistic and describe, in context and with conditional percentages, what they reveal about how the distributions differ.

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   The components \((O - E)^2/E\) are: urban drive 2.00, urban transit 4.36, urban bike/walk 0.03, rural drive 3.00, rural transit 6.53, rural bike/walk 0.05. The largest is rural/public transit: only 8 rural residents used transit versus 19.2 expected. Next is urban/public transit: 40 observed versus 28.8 expected. In context, 33% of urban residents (40/120) commute by public transit compared with just 10% of rural residents (8/80), while rural residents are much more likely to drive (75% vs. 50%). The bike/walk proportions are similar (17% vs. 15%) and contribute almost nothing. A significant \(\chi^2\) says the distributions differ; the components say how.

8. A researcher plans a goodness-of-fit test with a random sample of 60 items across five categories whose claimed proportions are 0.50, 0.24, 0.12, 0.08, and 0.06. Check the Large Counts condition, and explain what the researcher could do if it fails.

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   Expected counts are \(60(0.50) = 30\), 14.4, 7.2, 4.8, and \(60(0.06) = 3.6\). The last two are below 5, so the Large Counts condition fails and the chi-square distribution would not approximate the sampling distribution of the statistic well. Options: collect a larger sample (with \(n = 100\), the smallest expected count would be 6), or combine the two smallest categories into one "other" category with expected count \(60(0.14) = 8.4\), which reduces df from 4 to 3. Never drop the data or adjust the observed counts.

9. A chi-square test is carried out on a two-way table with 3 rows and 4 columns. The degrees of freedom are

   12 11 6 7

   12 is the number of cells in the table (\(3 \times 4\)), not the degrees of freedom. Once the row and column totals are fixed, only \((r-1)(c-1)\) of the cells are free to vary.

   11 is cells minus one: the goodness-of-fit formula (categories − 1) misapplied to a two-way table. Goodness-of-fit uses \(k - 1\); tests of homogeneity and independence do not.

   For a two-way table, df \(= (r - 1)(c - 1) = (3 - 1)(4 - 1) = 6\): fix the marginal totals and only six cell counts can be chosen freely before the rest are determined.

   7 is \(3 + 4\), adding the dimensions instead of multiplying the reduced dimensions. Degrees of freedom for a two-way table always come from the product \((r-1)(c-1)\).

10. The loot-box test in problem 2 gave \(\chi^2 = 3.19\) and a p-value of 0.202. (a) Interpret the p-value in context. (b) Explain why the p-value for a chi-square test is always the area to the right of the statistic, even though the alternative hypothesis isn't "one-sided" in the usual sense.

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    (a) Assuming the publisher's claimed distribution (60/30/10) is correct, there is a 0.202 probability of getting a chi-square statistic of 3.19 or larger from a random sample of 300 boxes purely by chance. That's not surprising, so the data are consistent with the claim. (b) The chi-square statistic measures total discrepancy between observed and expected counts, and squaring makes every deviation, above or below expected, contribute positively. Any departure from \(H_0\) in any direction makes \(\chi^2\) larger, so "at least as extreme as observed" always means "at least this large," a right-tail area. A tiny \(\chi^2\) means the observed counts match expectations unusually well, not evidence against \(H_0\).

Lesson 9.1 · Unit 9 · CED topics 9.1–9.2

Regression output and the sampling distribution of the slope

The least-squares line you fit in Unit 2 was computed from a sample. A different random sample of cars would give a slightly different line, so the slope \(b\) is a statistic with a sampling distribution, and that makes it a candidate for confidence intervals and tests, just like \(\bar x\) and \(\hat p\).

Exam tip: ignore the Constant row's T and P unless a question asks about the intercept. Inference questions are almost always about the slope.

Lesson 9.2 · Unit 9 · CED topics 9.2–9.3

Confidence intervals for the slope

The interval for \(\beta\) has the familiar shape, estimate ± critical value × standard error, with the estimate and standard error read straight off the output. The skill the exam tests is interpreting it: an interval for a slope is a statement about a rate.

Because the interval lies entirely below 0, it also gives convincing evidence at the 5% level that \(\beta \ne 0\): the same conclusion the test in Lesson 9.3 reaches. A 99% interval (\(t^* = 3.012\)) would be (−8.99, −6.25); more confidence, wider range.

Lesson 9.3 · Unit 9 · CED topics 9.4–9.6

Significance tests for the slope

"Is there a linear relationship at all?" is a question about \(\beta\): if \(\beta = 0\), the mean response doesn't change with \(x\) and the line is useless for prediction. The test asks whether the sample slope is far enough from 0, in standard-error units, to rule out chance.

Unit 9 practice · 10 problems

Unit 9 practice, Inference for Quantitative Data: Slopes

Ten problems covering the whole unit, in roughly exam order. Problems 4 and 6 are full four-step write-ups. Most of the unit is reading computer output, so the calculator only helps with invT and tcdf here.

1. A dealer recorded the age (years) and asking price (thousands of dollars) of a random sample of 20 used cars of one model. Software output:

   Predictor	Coef	SE Coef	T	P
   Constant	24.83	1.12	22.17	0.000
   Age	−1.940	0.210	−9.24	0.000
   S = 2.15    R-Sq = 82.5%

   Write the equation of the least-squares line, interpret the slope in context, explain what \(SE_b = 0.210\) measures, and state the degrees of freedom for inference about the slope.

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   \(\widehat{\text{price}} = 24.83 - 1.94(\text{age})\). Slope: each additional year of age is associated with a predicted decrease of about $1,940 in asking price. \(SE_b = 0.210\) estimates the standard deviation of the sampling distribution of \(b\): in repeated random samples of 20 cars, the sample slope would typically differ from the true slope \(\beta\) by about 0.21 thousand dollars per year. Degrees of freedom \(= n - 2 = 18\). (Check: \(T = -1.94/0.210 \approx -9.24\), as printed.)

2. Using the output in problem 1, find the correlation \(r\) and interpret \(r^2\) and \(s\) in context.

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   \(r^2 = 0.825\), and the slope is negative, so \(r = -\sqrt{0.825} \approx -0.91\): a strong negative linear association between age and price. \(r^2\): about 82.5% of the variation in the asking prices of these cars is explained by the linear relationship with age. \(s = 2.15\): actual asking prices are typically about $2,150 away from the price predicted by the regression line.

3. Before doing inference for the slope, an analyst examines the used-car data. The scatterplot shows a roughly linear pattern. The residual plot shows no curve, but the vertical spread of the residuals clearly increases as age increases. A histogram of the residuals is roughly symmetric with no outliers. The cars were a random sample from a very large inventory. Check each condition for inference about the slope and identify any that are violated.

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   Linear: met: the scatterplot is linear and the residual plot shows no curved pattern. Independent: met: random sample of 20 from a very large inventory, so the 10% condition holds. Normal: met: the residual histogram is roughly symmetric with no outliers. Equal SD: violated: the fan shape (residual spread growing with age) means the scatter about the line is not constant across \(x\). Random: met. With Equal SD violated, the standard error \(SE_b\) and the t-based p-values and intervals are not fully trustworthy; a transformation of the response (such as \(\log(\text{price})\)) often fixes both the fan shape and any curvature.

4. Assume all conditions are met for the used-car data in problem 1. Construct and interpret a 95% confidence interval for the slope of the population regression line.

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   State: \(\beta\) = the true slope of the population regression line relating asking price (thousands of dollars) to age (years) for used cars of this model. 95% confidence.

   Plan: t interval for the slope. Conditions (Linear, Independent, Normal, Equal SD, Random) are given as met; \(n = 20\) is under 10% of the dealer's inventory.

   Do: df \(= 20 - 2 = 18\), \(t^* = \text{invT}(0.975, 18) = 2.101\). \[-1.94 \pm 2.101(0.210) = -1.94 \pm 0.441 \;\Rightarrow\; (-2.38,\, -1.50).\]

   Conclude: We are 95% confident that the true slope of the population regression line relating asking price to age is between −2.38 and −1.50 thousand dollars per year: that is, each additional year of age is associated with a decrease in predicted price of between $1,500 and $2,380.

5. (a) Does the interval (−2.38, −1.50) from problem 4 give convincing evidence, at the 5% significance level, of a linear relationship between age and price? Explain. (b) A student concludes, "Aging causes a car's price to fall by about $1,940 per year." Comment on the word causes.

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   (a) Yes. The interval does not contain 0, so \(\beta = 0\) is not a plausible value; a two-sided test of \(H_0\!: \beta = 0\) at \(\alpha = 0.05\) would reject. There is convincing evidence of a negative linear relationship between age and asking price. (b) The data are observational, the dealer didn't randomly assign ages to cars, so the regression establishes an association, not causation. Older cars also tend to have more mileage and wear, and those variables are confounded with age. Say "is associated with," not "causes." (It also matters that \(b = -1.94\) is a sample estimate; the population slope could plausibly be anywhere in the interval.)

6. For a random sample of 16 houses in a county, a regression of monthly electricity cost (dollars) on living area (square feet) produced slope \(b = 0.0412\) with \(SE_b = 0.0135\). The scatterplot is linear, the residual plot shows no pattern and constant spread, and a dotplot of residuals has no outliers. Is there convincing evidence at \(\alpha = 0.05\) of a linear relationship between living area and monthly electricity cost?

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   State: \(\beta\) = the true slope of the population regression line relating monthly electricity cost to living area for houses in this county. \(H_0\!: \beta = 0\) (no linear relationship); \(H_a\!: \beta \ne 0\) (there is a linear relationship). \(\alpha = 0.05\).

   Plan: t test for the slope. Linear: scatterplot linear, no curve in the residual plot ✓. Independent: 16 houses is under 10% of the county's houses ✓. Normal: residual dotplot shows no outliers or strong skew ✓. Equal SD: residual spread roughly constant ✓. Random: random sample ✓.

   Do: \[t = \frac{0.0412 - 0}{0.0135} \approx 3.05, \qquad \text{df} = 14, \qquad \text{p-value} = 2P(t_{14} \ge 3.05) \approx 0.0086.\]

   Conclude: Because the p-value of 0.0086 is less than \(\alpha = 0.05\), we reject \(H_0\). There is convincing evidence of a linear relationship between living area and monthly electricity cost for houses in this county; since \(b \gt 0\), larger houses tend to have higher costs.

7. Regression output for predicting a plant's height from weekly hours of sunlight shows a positive slope with P = 0.036 in the slope row. (a) What p-value should you report for \(H_a\!: \beta \gt 0\)? (b) For \(H_a\!: \beta \lt 0\)? (c) What would you report for \(H_a\!: \beta \ne 0\)? Explain.

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   Software prints the two-sided p-value. (a) The sample slope is positive, matching the direction of \(H_a\), so the one-sided p-value is half of 0.036: \(0.018\). (b) The sample slope points the wrong way for this alternative, so the p-value is \(1 - 0.018 = 0.982\), no evidence at all for a negative slope. (c) Report 0.036 as printed. Decide the alternative from the research question before looking at the output; you cannot switch to one-sided after seeing that it halves the p-value.

8. Part of a regression output for 22 randomly selected employees, predicting annual sales (thousands of dollars) from years of experience, has been smudged: the Experience row reads Coef 3.42, SE Coef ____, T 2.85. Recover \(SE_b\), state the degrees of freedom, and write the interpretation of the slope.

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   Since \(T = b/SE_b\), \(SE_b = b/T = 3.42/2.85 = 1.20\). df \(= 22 - 2 = 20\). Slope: each additional year of experience is associated with a predicted increase of about $3,420 in annual sales. (The same trick recovers \(SE_b\) from a TI-84 LinRegTTest, which prints \(b\) and \(t\) but not \(SE_b\).)

9. A test of \(H_0\!: \beta = 0\) versus \(H_a\!: \beta \ne 0\) for a regression based on 400 randomly selected adults gives \(t = 4.1\) and a p-value less than 0.0001, with \(r^2 = 0.04\). Which statement is correct?

   The explanatory variable causes changes in the response. There is a strong linear relationship between the variables. There is convincing evidence of a linear relationship between the variables in the population, though it explains little of the variation in the response. About 96% of the variation in the response is explained by the line.

   The adults were randomly selected, not randomly assigned to values of the explanatory variable, so this is observational data and no causal claim is justified, however small the p-value.

   This confuses statistical significance with strength. With \(n = 400\), even a weak relationship produces a tiny p-value; \(r^2 = 0.04\) (so \(|r| = 0.2\)) says the linear relationship is weak.

   A tiny p-value says the slope is distinguishable from 0 in the population, the relationship is real, but \(r^2 = 0.04\) means the line explains only 4% of the variation in the response, so it is weak. Significance and strength are separate questions.

   This reverses \(r^2\): the line explains 4% of the variation, and 96% is the unexplained fraction, \(1 - r^2\), that remains in the residuals.

10. For the used-car data in problem 1, the standard deviation of the residuals is \(s = 2.15\) and the standard deviation of the ages is \(s_x = 2.35\) years, with \(n = 20\). (a) Use the formula \(SE_b = \dfrac{s}{s_x\sqrt{n - 1}}\) to verify the printed value of \(SE_b\). (b) Explain how each of \(s\), \(s_x\), and \(n\) affects the precision of the slope estimate.

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    (a) \(SE_b = \dfrac{2.15}{2.35\sqrt{19}} = \dfrac{2.15}{10.24} \approx 0.210\), matching the output. (b) Less scatter about the line (smaller \(s\)) makes the slope more precise. A wider range of \(x\)-values (larger \(s_x\)) also makes it more precise: points spread far apart in \(x\) pin down the tilt of the line better than points bunched together. And a larger \(n\) shrinks \(SE_b\) in proportion to \(1/\sqrt{n - 1}\). To estimate a slope well, sample cars across a wide range of ages, not just 3- and 4-year-olds.

Unit recap

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