Difference of Two Means (Setup)

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AP Statistics › Difference of Two Means (Setup)

Questions 1 - 10
1

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?

$H_0: \mu_{\text{caf}}-\mu_{\text{decaf}}=0$; $H_a: \mu_{\text{caf}}-\mu_{\text{decaf}}<0$

$H_0: \bar{x}{\text{caf}}-\bar{x}{\text{decaf}}=0$; $H_a: \bar{x}{\text{caf}}-\bar{x}{\text{decaf}}<0$

$H_0: \mu_{\text{caf}}-\mu_{\text{decaf}}=0$; $H_a: \mu_{\text{caf}}-\mu_{\text{decaf}}>0$

$H_0: \mu_{\text{caf}}=240$; $H_a: \mu_{\text{caf}}<240$

$H_0: \mu_{\text{decaf}}-\mu_{\text{caf}}=0$; $H_a: \mu_{\text{decaf}}-\mu_{\text{caf}}<0$

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.

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?

$H_0: \bar{x}_M-\bar{x}_U=0$; $H_a: \bar{x}_M-\bar{x}_U<0$

$H_0: \mu_M-\mu_U=0$; $H_a: \mu_M-\mu_U<0$

$H_0: \mu_M-\mu_U=0$; $H_a: \mu_M-\mu_U\ne 0$

$H_0: \mu_U-\mu_M=0$; $H_a: \mu_U-\mu_M>0$

$H_0: \mu_M=112.3$; $H_a: \mu_M<112.3$

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.

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?

$H_0: \mu_P-\mu_U=0$; $H_a: \mu_P-\mu_U\ne 0$

$H_0: \mu_P-\mu_U=0$; $H_a: \mu_P-\mu_U>0$

$H_0: \bar{x}_P-\bar{x}_U=0$; $H_a: \bar{x}_P-\bar{x}_U>0$

$H_0: \mu_U-\mu_P=0$; $H_a: \mu_U-\mu_P>0$

$H_0: \mu_P=52.4$; $H_a: \mu_P>52.4$

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.

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?

$H_0: \mu_{LED}-\mu_{Fl}=0$; $H_a: \mu_{LED}-\mu_{Fl}\ne 0$

$H_0: \mu_{Fl}-\mu_{LED}=0$; $H_a: \mu_{Fl}-\mu_{LED}>0$

$H_0: \mu_{LED}=27.4$; $H_a: \mu_{LED}\ne 27.4$

$H_0: \bar{x}{LED}-\bar{x}{Fl}=0$; $H_a: \bar{x}{LED}-\bar{x}{Fl}\ne 0$

$H_0: \mu_{LED}-\mu_{Fl}=0$; $H_a: \mu_{LED}-\mu_{Fl}>0$

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.

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?

$H_0: \mu_E-\mu_R=0$; $H_a: \mu_E-\mu_R>0$

$H_0: \mu_R-\mu_E=0$; $H_a: \mu_R-\mu_E<0$

$H_0: \mu_E=31.6$; $H_a: \mu_E<31.6$

$H_0: \mu_E-\mu_R=0$; $H_a: \mu_E-\mu_R<0$

$H_0: \bar{x}_E-\bar{x}_R=0$; $H_a: \bar{x}_E-\bar{x}_R<0$

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.

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?

$H_0: \mu_W-\mu_Y=0$; $H_a: \mu_W-\mu_Y<0$

$H_0: \mu_Y=62.7$; $H_a: \mu_Y<62.7$

$H_0: \bar{x}_Y-\bar{x}_W=0$; $H_a: \bar{x}_Y-\bar{x}_W<0$

$H_0: \mu_Y-\mu_W=0$; $H_a: \mu_Y-\mu_W>0$

$H_0: \mu_Y-\mu_W=0$; $H_a: \mu_Y-\mu_W<0$

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.

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?

$H_0: \mu_{9}-\mu_{8}=0$; $H_a: \mu_{9}-\mu_{8}>0$

$H_0: \mu_{8}-\mu_{9}=0$; $H_a: \mu_{8}-\mu_{9}\ne 0$

$H_0: \bar{x}{8}-\bar{x}{9}=0$; $H_a: \bar{x}{8}-\bar{x}{9}\ne 0$

$H_0: \mu_{9}=5.4$; $H_a: \mu_{9}\ne 5.4$

$H_0: \mu_{8}-\mu_{9}=0$; $H_a: \mu_{8}-\mu_{9}>0$

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.

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?

$H_0: \mu_L-\mu_N=0$; $H_a: \mu_L-\mu_N>0$

$H_0: \bar{x}_L-\bar{x}_N=0$; $H_a: \bar{x}_L-\bar{x}_N<0$

$H_0: \mu_N-\mu_L=0$; $H_a: \mu_N-\mu_L<0$

$H_0: \mu_L-\mu_N=0$; $H_a: \mu_L-\mu_N<0$

$H_0: \mu_L=126.3$; $H_a: \mu_L<126.3$

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.

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?

$H_0: \mu_{LED}-\mu_F=0$; $H_a: \mu_{LED}-\mu_F\ne 0$

$H_0: \mu_{LED}=34.5$; $H_a: \mu_{LED}>34.5$

$H_0: \bar{x}_{LED}-\bar{x}F=0$; $H_a: \bar{x}{LED}-\bar{x}_F>0$

$H_0: \mu_{LED}-\mu_F=0$; $H_a: \mu_{LED}-\mu_F>0$

$H_0: \mu_F-\mu_{LED}=0$; $H_a: \mu_F-\mu_{LED}<0$

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.

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?

$H_0: \mu_{caf}-\mu_{decaf}=0$; $H_a: \mu_{caf}-\mu_{decaf}>0$

$H_0: \bar{x}{caf}-\bar{x}{decaf}=0$; $H_a: \bar{x}{caf}-\bar{x}{decaf}<0$

$H_0: \mu_{caf}-\mu_{decaf}=0$; $H_a: \mu_{caf}-\mu_{decaf}<0$

$H_0: \mu_{decaf}-\mu_{caf}=0$; $H_a: \mu_{decaf}-\mu_{caf}<0$

$H_0: \mu_{caf}=241.8$; $H_a: \mu_{caf}<241.8$

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.

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