AP Statistics Quiz: Difference Of Two Means Setup

20 questions · exam conditions

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Difference Of Two Means SetupQuestion 1 of 20

A researcher is comparing mean reaction time (in milliseconds) between people who drank caffeinated coffee and people who drank decaf. A random sample of 36 caffeinated participants had a mean reaction time of 240 ms, and a random sample of 34 decaf participants had a mean reaction time of 255 ms. The research claim is that caffeine reduces mean reaction time. Which hypotheses are appropriate?

H_0: \mu_{\text{caf}}-\mu_{\text{decaf}}=0

;

H_a: \mu_{\text{caf}}-\mu_{\text{decaf}}<0

H_0: \mu_{\text{caf}}-\mu_{\text{decaf}}=0

;

H_a: \mu_{\text{caf}}-\mu_{\text{decaf}}>0

H_0: \mu_{\text{decaf}}-\mu_{\text{caf}}=0

;

H_a: \mu_{\text{decaf}}-\mu_{\text{caf}}<0

H_0: \bar{x}_{\text{caf}}-\bar{x}_{\text{decaf}}=0

;

H_a: \bar{x}_{\text{caf}}-\bar{x}_{\text{decaf}}<0

H_0: \mu_{\text{caf}}=240

;

H_a: \mu_{\text{caf}}<240

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AP Statistics Quiz

AP Statistics Quiz: Difference Of Two Means Setup

Practice Difference Of Two Means Setup in AP Statistics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Difference Of Two Means Setup, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Statistics.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$H_0: \mu_{\text{caf}}-\mu_{\text{decaf}}=0$ ; $H_a: \mu_{\text{caf}}-\mu_{\text{decaf}}<0$ (correct answer)
$H_0: \mu_{\text{caf}}-\mu_{\text{decaf}}=0$ ; $H_a: \mu_{\text{caf}}-\mu_{\text{decaf}}>0$
$H_0: \mu_{\text{decaf}}-\mu_{\text{caf}}=0$ ; $H_a: \mu_{\text{decaf}}-\mu_{\text{caf}}<0$
$H_0: \bar{x}_{\text{caf}}-\bar{x}_{\text{decaf}}=0$ ; $H_a: \bar{x}_{\text{caf}}-\bar{x}_{\text{decaf}}<0$
$H_0: \mu_{\text{caf}}=240$ ; $H_a: \mu_{\text{caf}}<240$

Explanation: This AP Statistics question tests hypothesis formulation for two-sample mean differences, claiming caffeine reduces reaction time, so $\mu_{\text{caf}} < \mu_{\text{decaf}}$ or $H_a: \mu_{\text{caf}} - \mu_{\text{decaf}} < 0$ . Choice A matches this with population parameters and correct order. Distractors include choice D, using sample means, which doesn't test populations. Choice E tests one group against its sample mean, missing the comparison. Mini-lesson: define $\mu_{\text{caf}}$ and $\mu_{\text{decaf}}$ , null is $\mu_{\text{caf}} - \mu_{\text{decaf}} = 0$ , alternative $< 0$ since caffeine is claimed to lower (faster) time compared to decaf.

Question 2

A nutritionist compares the mean LDL cholesterol level (mg/dL) for adults following a Mediterranean diet versus adults following their usual diet. A sample of $n=32$ Mediterranean-diet adults had a mean LDL of $\bar{x}=112.3$ , and a sample of $n=30$ usual-diet adults had a mean LDL of $\bar{x}=119.7$ . The nutritionist claims the Mediterranean diet reduces the population mean LDL level. Which hypotheses are appropriate?

$H_0: \mu_M-\mu_U=0$ ; $H_a: \mu_M-\mu_U\ne 0$
$H_0: \mu_U-\mu_M=0$ ; $H_a: \mu_U-\mu_M>0$
$H_0: \bar{x}_M-\bar{x}_U=0$ ; $H_a: \bar{x}_M-\bar{x}_U<0$
$H_0: \mu_M-\mu_U=0$ ; $H_a: \mu_M-\mu_U<0$ (correct answer)
$H_0: \mu_M=112.3$ ; $H_a: \mu_M<112.3$

Explanation: This question tests setting up hypotheses for comparing mean LDL cholesterol between Mediterranean diet and usual diet groups. The nutritionist claims the Mediterranean diet "reduces" LDL, so we test if the Mediterranean group has lower mean LDL (μ_M - μ_U < 0). The null hypothesis states no difference (μ_M - μ_U = 0). Choice A uses a two-sided alternative when the claim is directional. Choice B reverses the inequality, testing if Mediterranean diet increases LDL. Choice C uses sample means (x̄) instead of population parameters. Choice E tests only the Mediterranean group against a fixed value rather than comparing two groups. For health outcomes where lower values are better (like cholesterol), "reduces" or "lowers" translates to testing if the treatment group mean is less than the control group mean.

Question 3

A sports scientist wants to compare the mean vertical jump height (in cm) of students who completed a 6-week plyometrics program versus students who followed their usual training. A random sample of $n=30$ plyometrics students had a mean jump height of $\bar{x}=52.4$ cm, and a random sample of $n=28$ usual-training students had a mean jump height of $\bar{x}=48.9$ cm. The scientist's research claim is that the plyometrics program increases the population mean jump height. Which hypotheses are appropriate?

$H_0: \bar{x}_P-\bar{x}_U=0$ ; $H_a: \bar{x}_P-\bar{x}_U>0$
$H_0: \mu_P-\mu_U=0$ ; $H_a: \mu_P-\mu_U>0$ (correct answer)
$H_0: \mu_U-\mu_P=0$ ; $H_a: \mu_U-\mu_P>0$
$H_0: \mu_P=52.4$ ; $H_a: \mu_P>52.4$
$H_0: \mu_P-\mu_U=0$ ; $H_a: \mu_P-\mu_U\ne 0$

Explanation: This question tests setting up hypotheses for a two-sample t-test comparing mean vertical jump heights between plyometrics and usual training groups. The null hypothesis always states no difference between population means (μ_P - μ_U = 0), while the alternative reflects the research claim that plyometrics increases jump height, making it one-sided (μ_P - μ_U > 0). Choice A incorrectly uses sample means (x̄) instead of population parameters (μ) in the hypotheses. Choice C reverses the order of subtraction, which would test if usual training is better. Choice D tests only one group against a fixed value rather than comparing two groups. Choice E uses a two-sided alternative when the claim specifically states "increases," requiring a one-sided test. When setting up two-mean hypotheses, always use population parameters (μ), maintain consistent order of subtraction, and match the alternative hypothesis direction to the research claim.

Question 4

A botanist compares the mean height (in cm) of plants grown under LED light versus plants grown under fluorescent light. A random sample of $n=15$ LED-grown plants had a mean height of $\bar{x}=27.4$ cm, and a random sample of $n=16$ fluorescent-grown plants had a mean height of $\bar{x}=26.1$ cm. The botanist claims the population mean plant height differs between the two lighting types. Which hypotheses are appropriate?

$H_0: \mu_{LED}-\mu_{Fl}=0$ ; $H_a: \mu_{LED}-\mu_{Fl}\ne 0$ (correct answer)
$H_0: \mu_{LED}-\mu_{Fl}=0$ ; $H_a: \mu_{LED}-\mu_{Fl}>0$
$H_0: \bar{x}_{LED}-\bar{x}_{Fl}=0$ ; $H_a: \bar{x}_{LED}-\bar{x}_{Fl}\ne 0$
$H_0: \mu_{Fl}-\mu_{LED}=0$ ; $H_a: \mu_{Fl}-\mu_{LED}>0$
$H_0: \mu_{LED}=27.4$ ; $H_a: \mu_{LED}\ne 27.4$

Explanation: This problem tests understanding of two-sided hypothesis tests when comparing plant heights under different lighting. The botanist claims the population mean heights "differ" without specifying which is taller, requiring a two-sided alternative (μ_LED - μ_Fl ≠ 0). The null hypothesis states no difference (μ_LED - μ_Fl = 0). Choice B incorrectly uses a one-sided alternative (>) when no direction is specified. Choice C uses sample means (x̄) instead of population parameters. Choice D reverses subtraction and uses one-sided test. Choice E tests only LED plants against a fixed value. When researchers claim two groups "differ" or are "different" without stating which is larger, always use a two-sided alternative hypothesis with the not-equal symbol.

Question 5

A city planner compares mean commute time (in minutes) for commuters who use a new express bus route versus commuters who use the regular route. A random sample of $n=50$ express-route commuters had a mean commute of $\bar{x}=31.6$ minutes, and a random sample of $n=47$ regular-route commuters had a mean commute of $\bar{x}=34.2$ minutes. The planner's claim is that the express route decreases the population mean commute time. Which hypotheses are appropriate?

$H_0: \mu_E-\mu_R=0$ ; $H_a: \mu_E-\mu_R<0$ (correct answer)
$H_0: \mu_E-\mu_R=0$ ; $H_a: \mu_E-\mu_R>0$
$H_0: \mu_R-\mu_E=0$ ; $H_a: \mu_R-\mu_E<0$
$H_0: \bar{x}_E-\bar{x}_R=0$ ; $H_a: \bar{x}_E-\bar{x}_R<0$
$H_0: \mu_E=31.6$ ; $H_a: \mu_E<31.6$

Explanation: This problem requires testing whether the express bus route decreases commute time compared to the regular route. The null hypothesis states no difference (μ_E - μ_R = 0), and since the claim is that express "decreases" time, we test if express has lower mean time (μ_E - μ_R < 0). Choice B incorrectly uses > which would test if express takes longer. Choice C reverses subtraction and uses the wrong inequality. Choice D uses sample means (x̄) instead of population parameters (μ). Choice E tests only the express route against a specific value. When testing if one treatment "decreases" or "reduces" a time-based outcome, the alternative hypothesis should reflect that the treatment group has a smaller population mean.

Question 6

A fitness coach compares mean resting heart rate (in beats per minute) for clients who follow a yoga program versus clients who follow a weight-training program. A random sample of 25 yoga clients had a mean resting heart rate of 62.7 bpm, and a random sample of 27 weight-training clients had a mean of 65.1 bpm. The coach's claim is that yoga leads to a lower mean resting heart rate than weight training. Which hypotheses are appropriate?

$H_0: \mu_Y-\mu_W=0$ ; $H_a: \mu_Y-\mu_W>0$
$H_0: \mu_W-\mu_Y=0$ ; $H_a: \mu_W-\mu_Y<0$
$H_0: \mu_Y-\mu_W=0$ ; $H_a: \mu_Y-\mu_W<0$ (correct answer)
$H_0: \bar{x}_Y-\bar{x}_W=0$ ; $H_a: \bar{x}_Y-\bar{x}_W<0$
$H_0: \mu_Y=62.7$ ; $H_a: \mu_Y<62.7$

Explanation: This question tests setting up hypotheses for the claim that yoga leads to lower resting heart rate than weight training. The claim is μ_Y < μ_W, which can be written as μ_Y - μ_W < 0. The null hypothesis assumes no difference: μ_Y - μ_W = 0. The alternative hypothesis supports the claim: μ_Y - μ_W < 0. Choice C correctly uses population parameters with the proper subtraction order to yield a < symbol. Choice D incorrectly uses sample statistics, and choice E incorrectly tests against a specific value. When testing if one treatment produces lower values, ensure the subtraction order in the alternative hypothesis produces the < symbol.

Question 7

A school district compares mean number of absences in a semester for students who start school at 8:00 a.m. versus students who start at 9:00 a.m. A random sample of 40 students with an 8:00 start had a mean of 6.1 absences, and a random sample of 42 students with a 9:00 start had a mean of 5.4 absences. The district's claim is that the mean number of absences differs between the two start times. Which hypotheses are appropriate?

$H_0: \mu_{8}-\mu_{9}=0$ ; $H_a: \mu_{8}-\mu_{9}>0$
$H_0: \mu_{8}-\mu_{9}=0$ ; $H_a: \mu_{8}-\mu_{9}\ne 0$ (correct answer)
$H_0: \bar{x}_{8}-\bar{x}_{9}=0$ ; $H_a: \bar{x}_{8}-\bar{x}_{9}\ne 0$
$H_0: \mu_{9}-\mu_{8}=0$ ; $H_a: \mu_{9}-\mu_{8}>0$
$H_0: \mu_{9}=5.4$ ; $H_a: \mu_{9}\ne 5.4$

Explanation: This question requires setting up a two-sided hypothesis test since the district claims the mean number of absences differs between start times without specifying which is higher. The null hypothesis states no difference: μ_8 - μ_9 = 0. The alternative hypothesis uses ≠ to indicate any difference: μ_8 - μ_9 ≠ 0. Choice B correctly uses population parameters with a two-sided alternative. Choice C incorrectly uses sample means instead of population means, and choice E incorrectly tests against a specific value. When a claim mentions a difference without specifying direction, always use a two-tailed test with ≠ in the alternative hypothesis.

Question 8

A medical clinic compares mean systolic blood pressure (mmHg) between patients who follow a low-sodium diet and patients who do not. A random sample of $n=40$ low-sodium patients had mean $\bar{x}=126.3$ mmHg, and a random sample of $n=42$ patients with no diet change had mean $\bar{x}=131.7$ mmHg. The claim is that the low-sodium diet results in a lower population mean systolic blood pressure. Which hypotheses are appropriate?

$H_0: \mu_L-\mu_N=0$ ; $H_a: \mu_L-\mu_N<0$ (correct answer)
$H_0: \mu_N-\mu_L=0$ ; $H_a: \mu_N-\mu_L<0$
$H_0: \mu_L=126.3$ ; $H_a: \mu_L<126.3$
$H_0: \bar{x}_L-\bar{x}_N=0$ ; $H_a: \bar{x}_L-\bar{x}_N<0$
$H_0: \mu_L-\mu_N=0$ ; $H_a: \mu_L-\mu_N>0$

Explanation: This question evaluates hypotheses for two population means in AP Statistics, comparing blood pressure for low-sodium and no-change diets. The claim is low-sodium lowers mean BP (μ_L < μ_N), so H0: μ_L - μ_N = 0 and Ha: μ_L - μ_N < 0. Distractor choice B reverses order with <0, testing no-change lower, and choice D uses sample means. Choice E has >0, opposing the claim. Mini-lesson: Subscript clearly; null equality, <0 alternative for 'lower.' Ensure order matches inequality. Samples (126.3 vs. 131.7) guide tests, not statements.

Question 9

A botanist compares mean plant height (cm) after 6 weeks for plants grown under LED light versus fluorescent light. A random sample of $n=22$ LED-grown plants had mean $\bar{x}=34.5$ cm, and a random sample of $n=21$ fluorescent-grown plants had mean $\bar{x}=33.8$ cm. The claim is that LED lighting leads to a higher population mean height than fluorescent lighting. Which hypotheses are appropriate?

$H_0: \mu_F-\mu_{LED}=0$ ; $H_a: \mu_F-\mu_{LED}<0$
$H_0: \mu_{LED}-\mu_F=0$ ; $H_a: \mu_{LED}-\mu_F>0$ (correct answer)
$H_0: \bar{x}_{LED}-\bar{x}_F=0$ ; $H_a: \bar{x}_{LED}-\bar{x}_F>0$
$H_0: \mu_{LED}=34.5$ ; $H_a: \mu_{LED}>34.5$
$H_0: \mu_{LED}-\mu_F=0$ ; $H_a: \mu_{LED}-\mu_F\ne 0$

Explanation: This question probes setup for difference of means in AP Statistics, comparing plant heights under LED and fluorescent lights. The claim is LED leads to higher height (μ_LED > μ_F), so H0: μ_LED - μ_F = 0 and Ha: μ_LED - μ_F > 0. Distractor choice A reverses groups with <0, testing fluorescent higher, and choice C uses sample means. Choice E is two-sided, not matching directional claim. Mini-lesson: Use specific subscripts; null zero difference, >0 for 'higher.' Align subtraction to inequality. Sample means (34.5 vs. 33.8) inform but aren't in hypotheses.

Question 10

A psychologist compares mean reaction time (in milliseconds) for participants who drank caffeinated coffee versus participants who drank decaffeinated coffee before a computer task. A sample of $n=22$ caffeinated participants had a mean reaction time of $\bar{x}=241.8$ ms, and a sample of $n=24$ decaffeinated participants had a mean reaction time of $\bar{x}=255.6$ ms. The psychologist claims caffeine leads to faster reactions (lower population mean reaction time). Which hypotheses are appropriate?

$H_0: \mu_{caf}-\mu_{decaf}=0$ ; $H_a: \mu_{caf}-\mu_{decaf}<0$ (correct answer)
$H_0: \mu_{caf}-\mu_{decaf}=0$ ; $H_a: \mu_{caf}-\mu_{decaf}>0$
$H_0: \mu_{decaf}-\mu_{caf}=0$ ; $H_a: \mu_{decaf}-\mu_{caf}<0$
$H_0: \bar{x}_{caf}-\bar{x}_{decaf}=0$ ; $H_a: \bar{x}_{caf}-\bar{x}_{decaf}<0$
$H_0: \mu_{caf}=241.8$ ; $H_a: \mu_{caf}<241.8$

Explanation: This question involves testing whether caffeine leads to faster reactions (lower reaction times) compared to decaf. The null hypothesis states no difference (μ_caf - μ_decaf = 0), and since "faster reactions" means lower reaction time, the alternative is μ_caf - μ_decaf < 0. Choice B uses > which would test if caffeine makes reactions slower. Choice C reverses the subtraction order and uses the wrong inequality direction. Choice D incorrectly uses sample means (x̄) in the hypotheses. Choice E tests only the caffeinated group against a fixed value. For reaction time studies, "faster" or "quicker" responses correspond to lower numerical values, so improvement is tested with a less-than alternative hypothesis.

Question 11

A public health researcher compares the mean number of minutes of moderate-to-vigorous physical activity per day for adults who use a fitness-tracking app versus adults who do not. A sample of $n=45$ app users had a mean of $\bar{x}=38.2$ minutes, while a sample of $n=40$ non-users had a mean of $\bar{x}=34.6$ minutes. The researcher claims that app users have a higher population mean activity time. Which hypotheses are appropriate?

$H_0: \mu_{non}-\mu_{app}=0$ ; $H_a: \mu_{non}-\mu_{app}>0$
$H_0: \bar{x}_{app}-\bar{x}_{non}=0$ ; $H_a: \bar{x}_{app}-\bar{x}_{non}>0$
$H_0: \mu_{app}-\mu_{non}=0$ ; $H_a: \mu_{app}-\mu_{non}>0$ (correct answer)
$H_0: \mu_{app}=38.2$ ; $H_a: \mu_{app}>38.2$
$H_0: \mu_{app}-\mu_{non}=0$ ; $H_a: \mu_{app}-\mu_{non}\ne 0$

Explanation: This problem requires setting up hypotheses to test whether fitness app users have higher mean physical activity than non-users. The null hypothesis states no difference between population means (μ_app - μ_non = 0), and since the claim is that app users have "higher" activity, we need a one-sided alternative (μ_app - μ_non > 0). Choice A reverses the subtraction order, which would test if non-users have higher activity. Choice B incorrectly uses sample means (x̄) rather than population parameters (μ) in the hypotheses. Choice D tests only the app group against a specific value instead of comparing two groups. Choice E uses a two-sided alternative (≠) when the directional claim requires a one-sided test. Remember that hypothesis tests about population means always use μ notation, and the alternative hypothesis must match the directionality of the research claim.

Question 12

An educator compares mean quiz scores (out of 20 points) for students taught with a new interactive method versus a traditional lecture method. A random sample of $n=26$ interactive-method students had a mean score of $\bar{x}=16.8$ , and a random sample of $n=24$ lecture-method students had a mean score of $\bar{x}=15.9$ . The educator's claim is that the interactive method changes the population mean quiz score (could be higher or lower). Which hypotheses are appropriate?

$H_0: \mu_I-\mu_L=0$ ; $H_a: \mu_I-\mu_L\ne 0$ (correct answer)
$H_0: \mu_I-\mu_L=0$ ; $H_a: \mu_I-\mu_L>0$
$H_0: \bar{x}_I-\bar{x}_L=0$ ; $H_a: \bar{x}_I-\bar{x}_L\ne 0$
$H_0: \mu_L-\mu_I=0$ ; $H_a: \mu_L-\mu_I\ne 0$
$H_0: \mu_I=16.8$ ; $H_a: \mu_I\ne 16.8$

Explanation: This question tests understanding of two-sided hypothesis tests for comparing two population means. The educator claims the interactive method "changes" the mean score, which could be higher or lower, requiring a two-sided alternative hypothesis (μ_I - μ_L ≠ 0). The null hypothesis always states no difference (μ_I - μ_L = 0). Choice B incorrectly uses a one-sided alternative (>) when the claim doesn't specify direction. Choice C uses sample means (x̄) instead of population parameters (μ). Choice D has the same hypotheses as A but with reversed subtraction order, which is mathematically equivalent for a two-sided test. Choice E tests only one group against a fixed value rather than comparing two groups. When a research claim uses words like "changes," "differs," or "is different," without specifying direction, always use a two-sided alternative hypothesis.

Question 13

A nutrition researcher tests whether adults who follow a Mediterranean-style diet have a lower mean LDL cholesterol level than adults who follow a typical diet. A random sample of 48 Mediterranean-diet adults had a mean LDL of 112 mg/dL, and a random sample of 52 typical-diet adults had a mean LDL of 119 mg/dL. Let $\mu_M$ and $\mu_T$ be the true mean LDL levels for the two diet groups. Which hypotheses are appropriate?

$H_0: \mu_M-\mu_T=0$ ; $H_a: \mu_M-\mu_T>0$
$H_0: \mu_T-\mu_M=0$ ; $H_a: \mu_T-\mu_M<0$
$H_0: \mu_M-\mu_T=0$ ; $H_a: \mu_M-\mu_T<0$ (correct answer)
$H_0: \bar{x}_M-\bar{x}_T=0$ ; $H_a: \bar{x}_M-\bar{x}_T<0$
$H_0: \mu_M=112$ ; $H_a: \mu_M<112$

Explanation: The researcher wants to test if Mediterranean diet followers have LOWER LDL cholesterol than typical diet followers. This requires testing if μ_M < μ_T, which translates to μ_M - μ_T < 0 in the alternative hypothesis. Choice A incorrectly has μ_M - μ_T > 0, which would test if Mediterranean dieters have higher LDL (the opposite of what's claimed). Choice D makes the error of using sample means (x̄) instead of population means (μ) in the hypotheses. When the research question asks if group A has lower values than group B, the alternative hypothesis should be H_a: μ_A - μ_B < 0. Always double-check that your inequality direction matches the research question.

Question 14

A psychologist compares the mean reaction time (in milliseconds) for two groups: participants who drank caffeinated coffee and participants who drank decaffeinated coffee. A random sample of 40 caffeinated participants had a mean reaction time of 255 ms, and a random sample of 41 decaffeinated participants had a mean reaction time of 268 ms. The research claim is that caffeine reduces mean reaction time. Which hypotheses are appropriate for a two-sample $t$ test comparing population means?

$H_0: \mu_{\text{caff}} - \mu_{\text{decaf}} = 0$ ; $H_a: \mu_{\text{caff}} - \mu_{\text{decaf}} < 0$ (correct answer)
$H_0: \mu_{\text{caff}} - \mu_{\text{decaf}} = 0$ ; $H_a: \mu_{\text{caff}} - \mu_{\text{decaf}} > 0$
$H_0: \mu_{\text{decaf}} - \mu_{\text{caff}} = 0$ ; $H_a: \mu_{\text{decaf}} - \mu_{\text{caff}} < 0$
$H_0: \bar{x}_{\text{caff}} - \bar{x}_{\text{decaf}} = 0$ ; $H_a: \bar{x}_{\text{caff}} - \bar{x}_{\text{decaf}} < 0$
$H_0: \mu_{\text{caff}} = 255$ ; $H_a: \mu_{\text{caff}} < 255$

Explanation: This question requires setting up hypotheses where caffeine is claimed to reduce reaction time. Since lower reaction time is better (faster), and caffeinated participants had mean 255 ms while decaffeinated had 268 ms, the claim is that μ_caff < μ_decaf, which can be written as μ_caff - μ_decaf < 0. The null hypothesis states no difference (μ_caff - μ_decaf = 0). Choice B incorrectly has > 0, which would mean caffeine increases reaction time. Choice C reverses the order but maintains the correct relationship. Choice D uses sample means instead of population means. Choice E tests only the caffeinated group against a specific value. When dealing with "reduces" or "decreases," ensure you understand whether smaller or larger values are desirable for the variable being measured.

Question 15

A teacher compares mean quiz scores (out of 20) for students who used paper flashcards versus a digital flashcard app. A random sample of $n=33$ paper users had mean $\bar{x}=15.2$ , and a random sample of $n=35$ app users had mean $\bar{x}=16.1$ . The claim is that app users have a higher population mean quiz score than paper users. Which hypotheses are appropriate?

$H_0: \mu_{paper}-\mu_{app}=0$ ; $H_a: \mu_{paper}-\mu_{app}<0$
$H_0: \mu_{app}-\mu_{paper}=0$ ; $H_a: \mu_{app}-\mu_{paper}>0$ (correct answer)
$H_0: \mu_{app}=16.1$ ; $H_a: \mu_{app}>16.1$
$H_0: \bar{x}_{app}-\bar{x}_{paper}=0$ ; $H_a: \bar{x}_{app}-\bar{x}_{paper}>0$
$H_0: \mu_{app}-\mu_{paper}=0$ ; $H_a: \mu_{app}-\mu_{paper}\ne 0$

Explanation: This question tests hypothesis setup for two means in AP Statistics, comparing quiz scores for paper and app flashcard users. The claim is app users score higher (μ_app > μ_paper), so H0: μ_app - μ_paper = 0 and Ha: μ_app - μ_paper > 0. Distractor choice A reverses groups, testing if paper is lower, while choice D uses sample means. Choice E is two-sided, but the claim is directional. Mini-lesson: Subscript parameters by group; null assumes no difference, >0 alternative for 'higher.' Match subtraction to claim direction. Samples (15.2 vs. 16.1) inform but don't define hypotheses.

Question 16

A fitness app company wants to know whether the mean resting heart rate differs between users who follow an "interval training" plan and users who follow a "steady cardio" plan. A random sample of $n=30$ interval users had mean $\bar{x}=64.8$ bpm, and a random sample of $n=28$ steady-cardio users had mean $\bar{x}=67.2$ bpm. The research claim is that the population mean resting heart rates are different for the two plans. Which hypotheses are appropriate?

$H_0: \mu_I-\mu_S=0$ ; $H_a: \mu_I-\mu_S\ne 0$ (correct answer)
$H_0: \mu_I-\mu_S=0$ ; $H_a: \mu_I-\mu_S<0$
$H_0: \bar{x}_I-\bar{x}_S=0$ ; $H_a: \bar{x}_I-\bar{x}_S\ne 0$
$H_0: \mu_S-\mu_I=0$ ; $H_a: \mu_S-\mu_I<0$
$H_0: \mu_I=64.8$ ; $H_a: \mu_I\ne 64.8$

Explanation: This question evaluates the ability to formulate hypotheses for the difference of two population means in AP Statistics, focusing on resting heart rates between interval and steady-cardio training plans. The research claim is that the population means differ (μ_I ≠ μ_S), leading to a two-sided alternative: H0: μ_I - μ_S = 0 and Ha: μ_I - μ_S ≠ 0. A frequent distractor is choice B, which uses a one-sided alternative (<0), but the claim specifies 'different' without direction, requiring a two-sided test. Choice C is misleading as it uses sample means instead of population parameters, which is incorrect for hypotheses. Mini-lesson on two-mean setup: In comparing two independent population means, use μ with appropriate subscripts; the null is always equality to zero difference, and the alternative matches the claim—two-sided for no specified direction. The sample statistics (like 64.8 and 67.2) are used in calculations but not in stating hypotheses. Always verify the direction or lack thereof in the claim to avoid one-sided errors.

Question 17

A grocery store tests whether a new checkout layout reduces mean customer wait time. A random sample of $n=25$ customers in the new layout had mean wait time $\bar{x}=3.9$ minutes, and a random sample of $n=27$ customers in the old layout had mean wait time $\bar{x}=4.6$ minutes. The store's claim is that the new layout has a lower population mean wait time than the old layout. Which hypotheses are appropriate?

$H_0: \mu_{new}=3.9$ ; $H_a: \mu_{new}<3.9$
$H_0: \mu_{old}-\mu_{new}=0$ ; $H_a: \mu_{old}-\mu_{new}<0$
$H_0: \mu_{new}-\mu_{old}=0$ ; $H_a: \mu_{new}-\mu_{old}<0$ (correct answer)
$H_0: \bar{x}_{new}-\bar{x}_{old}=0$ ; $H_a: \bar{x}_{new}-\bar{x}_{old}<0$
$H_0: \mu_{new}-\mu_{old}=0$ ; $H_a: \mu_{new}-\mu_{old}\ne 0$

Explanation: This question tests hypothesis setup for the difference of two means in AP Statistics, examining wait times between new and old checkout layouts. The claim is that the new layout has a lower population mean wait time (μ_new < μ_old), so hypotheses are H0: μ_new - μ_old = 0 and Ha: μ_new - μ_old < 0. Distractor choice B reverses the subtraction order, testing if old is less than new, which opposes the claim. Choice D uses sample means, but hypotheses must involve population parameters. Mini-lesson: For two-sample mean tests, subscript parameters clearly (e.g., new vs. old); null assumes no difference, alternative direction (<0) reflects 'lower' for the first group. Sample means like 3.9 and 4.6 guide the test statistic but stay out of hypotheses. Check subtraction order to ensure it matches the claimed inequality.

Question 18

A company compares mean daily number of customer support tickets handled by employees working from home versus employees working in the office. A random sample of $n=26$ remote employees had mean $\bar{x}=18.7$ tickets, and a random sample of $n=24$ in-office employees had mean $\bar{x}=17.9$ tickets. The claim is that remote employees handle more tickets per day on average in the population. Which hypotheses are appropriate?

$H_0: \mu_{remote}-\mu_{office}=0$ ; $H_a: \mu_{remote}-\mu_{office}>0$ (correct answer)
$H_0: \mu_{remote}-\mu_{office}=0$ ; $H_a: \mu_{remote}-\mu_{office}\ne 0$
$H_0: \mu_{office}-\mu_{remote}=0$ ; $H_a: \mu_{office}-\mu_{remote}>0$
$H_0: \bar{x}_{remote}-\bar{x}_{office}=0$ ; $H_a: \bar{x}_{remote}-\bar{x}_{office}>0$
$H_0: \mu_{remote}=18.7$ ; $H_a: \mu_{remote}>18.7$

Explanation: This question tests two-mean hypothesis formulation in AP Statistics, for tickets handled by remote and office employees. The claim is remote handle more (μ_remote > μ_office), so H0: μ_remote - μ_office = 0 and Ha: μ_remote - μ_office > 0. Distractor choice C reverses with >0, testing office more, and choice D uses sample means. Choice B is two-sided, but claim is directional. Mini-lesson: Define μ by group; null no difference, >0 for 'more.' Subtraction order supports claim. Samples (18.7 vs. 17.9) are for analysis, not hypotheses.

Question 19

A city compares mean commute times for residents who use a new express bus route versus residents who use the regular route. A random sample of $n=52$ express riders had mean $\bar{x}=28.4$ minutes, and a random sample of $n=49$ regular riders had mean $\bar{x}=31.0$ minutes. The claim is that the express route reduces the population mean commute time. Which hypotheses are appropriate?

$H_0: \mu_E-\mu_R=0$ ; $H_a: \mu_E-\mu_R\ne 0$
$H_0: \mu_R-\mu_E=0$ ; $H_a: \mu_R-\mu_E>0$
$H_0: \bar{x}_E-\bar{x}_R=0$ ; $H_a: \bar{x}_E-\bar{x}_R<0$
$H_0: \mu_E-\mu_R=0$ ; $H_a: \mu_E-\mu_R<0$ (correct answer)
$H_0: \mu_E=28.4$ ; $H_a: \mu_E<28.4$

Explanation: This question assesses two-mean hypothesis setup in AP Statistics, for commute times on express and regular bus routes. The claim is express reduces mean time (μ_E < μ_R), so H0: μ_E - μ_R = 0 and Ha: μ_E - μ_R < 0. Distractor choice B reverses order with >0, testing longer express times, and choice C uses sample means. Choice A is two-sided, ignoring the directional claim. Mini-lesson: Use group subscripts; null is zero difference, <0 for 'reduces.' Align subtraction for correct inequality. Sample means (28.4 vs. 31.0) are computational, not hypothetical.

Question 20

A car manufacturer compares the mean fuel efficiency (in miles per gallon) of two tire types. A random sample of 20 cars with Tire Type 1 had a mean of 30.8 mpg, and a random sample of 22 cars with Tire Type 2 had a mean of 29.9 mpg. The research claim is that Tire Type 1 and Tire Type 2 have different population mean fuel efficiencies. Which hypotheses are appropriate for a two-sample $t$ test comparing population means?

$H_0: \mu_1 - \mu_2 = 0$ ; $H_a: \mu_1 - \mu_2 \ne 0$ (correct answer)
$H_0: \mu_1 - \mu_2 = 0$ ; $H_a: \mu_1 - \mu_2 > 0$
$H_0: \mu_2 - \mu_1 = 0$ ; $H_a: \mu_2 - \mu_1 > 0$
$H_0: \bar{x}_1 - \bar{x}_2 = 0$ ; $H_a: \bar{x}_1 - \bar{x}_2 \ne 0$
$H_0: \mu_1 = 30.8$ ; $H_a: \mu_1 \ne 30.8$

Explanation: This question involves a two-sided hypothesis test because the research claim is that the tire types have different population mean fuel efficiencies, without specifying which is better. The null hypothesis states no difference (μ₁ - μ₂ = 0), and the alternative states they differ (μ₁ - μ₂ ≠ 0). Choice B incorrectly uses a one-sided alternative with >, assuming Tire Type 1 is better. Choice C reverses the order and uses a one-sided test. Choice D uses sample means (x̄) instead of population means (μ). Choice E tests only Tire Type 1 against a specific value. When the research question asks whether two groups are "different" without specifying direction, always use a two-sided alternative hypothesis with ≠.