Difference of Two Means (Setup) - AP Statistics

1. A university tests whether students in a flipped classroom have a higher mean final exam score than students in a traditional classroom. A random sample of 45 flipped-class students had a mean score of 81.6, and a random sample of 47 traditional-class students had a mean score of 79.8. Let $\mu_F$ and $\mu_T$ be the true mean final exam scores for flipped and traditional classes, respectively. Which hypotheses are appropriate?
   Answer: $H_0: \mu_F-\mu_T=0$; $H_a: \mu_F-\mu_T>0$ The university wants to test if flipped classroom students have HIGHER mean exam scores than traditional classroom students. This requires $H_a$: μ_F > μ_T, which can be written as μ_F - μ_T > 0. Choice A incorrectly reverses the order (μ_T - μ_F > 0), which would test if traditional students score higher. Choice C uses sample means (x̄) instead of population means (μ), violating the principle that hypotheses concern parameters. Choice D tests for any difference rather than specifically higher scores for flipped classrooms. When group F is hypothesized to have higher values than group T, write $H_a$: μ_F - μ_T > 0.
2. 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?
   Answer: $H_0: \mu_I-\mu_S=0$; $H_a: \mu_I-\mu_S\ne 0$ 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.
3. A fitness coach compares mean change in resting heart rate (in beats per minute) after 8 weeks for two workout plans: Plan X and Plan Y. A random sample of 20 clients on Plan X had a mean change of −4.1 bpm, and a random sample of 22 clients on Plan Y had a mean change of −2.7 bpm. The research claim is that the mean change differs between the two plans. Which hypotheses are appropriate?
   Answer: $H_0: \mu_X-\mu_Y=0$; $H_a: \mu_X-\mu_Y\ne 0$ This AP Statistics question involves hypotheses for differing mean changes in two groups, requiring a two-sided test: Ha: μ_X - μ_Y ≠ 0, as in choice A. It properly uses population parameters. A common distractor, choice C, substitutes sample means for μ. Choice E tests one mean against a constant, not the comparison. Two-mean setup mini-lesson: define μ_X and μ_Y, H0: μ_X - μ_Y = 0, Ha ≠ 0 for claims of difference without specified direction.
4. 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?
   Answer: $H_0: \mu_{8}-\mu_{9}=0$; $H_a: \mu_{8}-\mu_{9}\ne 0$ 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.
5. 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?
   Answer: $H_0: \mu_E-\mu_R=0$; $H_a: \mu_E-\mu_R<0$ 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.
6. 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?
   Answer: $H_0: \mu_{caf}-\mu_{decaf}=0$; $H_a: \mu_{caf}-\mu_{decaf}<0$ 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.
7. A city planner wants to test whether average monthly electricity use differs between apartments with smart thermostats and apartments without smart thermostats. A random sample of 35 smart-thermostat apartments had a mean of 410 kWh/month, and a random sample of 40 non-smart apartments had a mean of 435 kWh/month. Let $\mu_S$ be the true mean monthly electricity use for all smart-thermostat apartments in the city and $\mu_N$ for all non-smart apartments. Which hypotheses are appropriate?
   Answer: $H_0: \mu_S-\mu_N=0$; $H_a: \mu_S-\mu_N\ne 0$ The city planner wants to test whether electricity use DIFFERS between smart and non-smart thermostat apartments, with no specified direction. This calls for a two-sided test with alternative hypothesis μ_S - μ_N ≠ 0. Choice C incorrectly uses sample means (x̄) rather than population means (μ), which is a common conceptual error. Choice B would be appropriate only if we specifically hypothesized that smart thermostats use less electricity. The key insight: when testing for any difference (not a specific direction), use a two-tailed alternative hypothesis with ≠. Also note that while choices A and D are algebraically equivalent (just different orders of subtraction), choice A follows the conventional format of listing the first group mentioned first.
8. A manager compares the mean time (in minutes) to complete a customer service call for two training programs. A random sample of 35 employees using Program A had a mean call time of 11.8 minutes, and a random sample of 33 employees using Program B had a mean call time of 12.6 minutes. The manager's research claim is that Program A leads to a shorter mean call time than Program B. Which hypotheses are appropriate for a two-sample $t$ test comparing population means?
   Answer: $H_0: \mu_A - \mu_B = 0$; $H_a: \mu_A - \mu_B < 0$ This question requires setting up hypotheses for a two-sample t-test where the research claim is that Program A leads to shorter mean call times than Program B. Since we want to test if A < B, we can write this as μ_A - μ_B < 0 in the alternative hypothesis. The null hypothesis states there is no difference between the population means (μ_A - μ_B = 0). Choice A incorrectly tests a single mean against a specific value rather than comparing two populations. Choice C has the wrong direction in the alternative hypothesis (> 0 would mean A takes longer than B). Choice D incorrectly uses sample means (x̄) instead of population means (μ). Choice E reverses the order of subtraction, which would test if B > A, but this is mathematically equivalent to testing if A < B. When setting up hypotheses for comparing two means, ensure the alternative hypothesis direction matches the research claim and always use population parameters.
9. A school district compares the mean number of minutes of homework completed on a weeknight by students in two programs. A random sample of $n=38$ students in the tutoring program had a mean of $\bar{x}=52.1$ minutes, and a random sample of $n=41$ students not in tutoring had a mean of $\bar{x}=47.6$ minutes. The district's research claim is that students in tutoring complete more homework on average. Which hypotheses are appropriate for testing the claim about the population means?
   Answer: $H_0: \mu_T-\mu_N=0$; $H_a: \mu_T-\mu_N>0$ This question assesses the skill of setting up hypotheses for the difference of two population means in AP Statistics, specifically for comparing homework completion times between tutored and non-tutored students. The claim states that students in tutoring complete more homework on average, which translates to testing if the population mean for tutored students (μ_T) is greater than that for non-tutored students (μ_N), so the hypotheses are H0: μ_T - μ_N = 0 and Ha: μ_T - μ_N > 0. A common distractor here is choice A, which incorrectly uses sample means (\bar{x}) instead of population means (μ), as hypothesis tests are always about population parameters, not sample statistics. Another distractor is choice D, which reverses the order of subtraction, effectively testing the opposite claim. In a mini-lesson on two-mean setup: For independent samples, define the parameters clearly with subscripts matching the groups; the null hypothesis always assumes no difference (μ1 - μ2 = 0), while the alternative reflects the claim's direction—one-sided for 'more' or 'less,' two-sided for 'different.' Ensure the order of subtraction aligns with the claim, such as group of interest minus comparison group for a positive difference if claiming higher. Remember, sample means like 52.1 and 47.6 inform the test but do not appear in the hypotheses.
10. 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?
    Answer: $H_0: \mu_M-\mu_U=0$; $H_a: \mu_M-\mu_U<0$ 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.
11. A school district is comparing average nightly sleep (in hours) for students who participate in a later school start pilot program versus students on the traditional schedule. A random sample of 42 pilot students had a mean of 7.6 hours, and a random sample of 38 traditional-schedule students had a mean of 7.1 hours. The district's research claim is that students in the pilot program sleep more on average. Which hypotheses are appropriate for a two-sample test of the difference in population means?
    Answer: $H_0: \mu_{\text{pilot}}-\mu_{\text{trad}}=0$; $H_a: \mu_{\text{pilot}}-\mu_{\text{trad}}>0$ This question assesses the skill of setting up hypotheses for a two-sample t-test on the difference of two population means in AP Statistics. The research claim is that pilot students sleep more on average, so the alternative hypothesis should reflect $\mu_{\text{pilot}} > \mu_{\text{trad}}$, or $\mu_{\text{pilot}} - \mu_{\text{trad}} > 0$, with the null being equality. Choice B correctly uses population means ($\mu$) and matches the directional claim. A common distractor, like choice A, uses sample means ($\bar{x}$) instead of population parameters, which is incorrect because hypotheses test populations, not samples. Another distractor, choice E, tests only one mean against a sample value, ignoring the two-sample comparison. In a two-mean setup mini-lesson: define $\mu_1$ and $\mu_2$ for each group, set $H_0: \mu_1 - \mu_2 = 0$, choose $ $H_a$ $ as >, <, or $\ne$, based on the claim, and ensure the subtraction order aligns with which group is hypothesized to be larger.
12. 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?
    Answer: $H_0: \mu_M-\mu_T=0$; $H_a: \mu_M-\mu_T<0$ 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.
13. 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?
    Answer: $H_0: \mu_Y-\mu_W=0$; $H_a: \mu_Y-\mu_W<0$ 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.
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?
    Answer: $H_0: \mu_{\text{caff}} - \mu_{\text{decaf}} = 0$; $H_a: \mu_{\text{caff}} - \mu_{\text{decaf}} < 0$ 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.
15. 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?
    Answer: $H_0: \mu_{app}-\mu_{non}=0$; $H_a: \mu_{app}-\mu_{non}>0$ 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.
16. A school district wants to know whether students who participate in an after-school tutoring program spend more time on homework per week, on average, than students who do not participate. A random sample of 42 tutoring participants had a mean of 6.8 hours/week, and a random sample of 50 nonparticipants had a mean of 6.1 hours/week. Let $\mu_T$ be the true mean weekly homework time for all tutoring participants in the district and $\mu_N$ be the true mean weekly homework time for all nonparticipants. Which hypotheses are appropriate for testing the district's claim?
    Answer: $H_0: \mu_T-\mu_N=0$; $H_a: \mu_T-\mu_N>0$ This question tests the ability to set up hypotheses for comparing two population means. The district wants to test if tutoring participants spend MORE time on homework than non-participants, making this a one-sided test with the alternative hypothesis showing μ_T - μ_N > 0. The common mistake in choice A is using sample means (x̄) instead of population means (μ) in the hypotheses - remember that hypotheses are always about population parameters, not sample statistics. Choice D reverses the order of subtraction, which would test if non-participants spend more time. When testing if group 1 has a larger mean than group 2, write the difference as μ₁ - μ₂ > 0 in the alternative hypothesis.
17. A city compares mean monthly electricity use (in kWh) between homes with smart thermostats and homes without smart thermostats. A random sample of 50 homes with smart thermostats had a mean of 910 kWh, and a random sample of 47 homes without smart thermostats had a mean of 980 kWh. The research claim is that homes with smart thermostats use less electricity on average. Which hypotheses are appropriate?
    Answer: $H_0: \mu_{\text{smart}}-\mu_{\text{no smart}}=0$; $H_a: \mu_{\text{smart}}-\mu_{\text{no smart}}<0$ The skill tested is hypothesis setup for two-sample mean comparisons in AP Statistics, claiming smart thermostats reduce usage, so Ha: μ_smart - μ_no smart < 0. Choice A correctly applies this with μ. Distractors like choice C use sample statistics. Choice E tests only smart homes against their mean. Mini-lesson: assign μ_smart and μ_no smart, null difference = 0, alternative < 0 to reflect lower usage claim for smart group.
18. A public health analyst compares the mean number of days per month that adults exercise for at least 30 minutes in two cities. A random sample of 60 adults from City A had a mean of 12.4 days, and a random sample of 55 adults from City B had a mean of 11.1 days. The research claim is that adults in City A exercise more days per month on average than adults in City B. Which hypotheses are appropriate for a two-sample $t$ test comparing population means?
    Answer: $H_0: \mu_A - \mu_B = 0$; $H_a: \mu_A - \mu_B > 0$ This question involves setting up hypotheses where City A adults are claimed to exercise more days per month than City B adults. The alternative hypothesis should be μ_A - μ_B > 0, indicating City A has a higher mean. The null hypothesis states no difference (μ_A - μ_B = 0). Choice B incorrectly has the alternative as < 0, which would mean City A exercises less. Choice C uses sample means (x̄) instead of population means (μ). Choice D reverses the order, testing if City B exercises more. Choice E tests only City A against a specific value rather than comparing the two cities. Remember that hypothesis tests for comparing two populations always involve differences between population parameters, and the direction of the alternative must match the research claim.
19. 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?
    Answer: $H_0: \mu_{LED}-\mu_F=0$; $H_a: \mu_{LED}-\mu_F>0$ 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.
20. An environmental scientist measures mean nitrate concentration (mg/L) in water samples from two rivers. A random sample of $n=18$ samples from River A had mean $\bar{x}=6.4$ mg/L, and a random sample of $n=20$ samples from River B had mean $\bar{x}=5.7$ mg/L. The claim is that River A has a higher population mean nitrate concentration than River B. Which hypotheses are appropriate?
    Answer: $H_0: \mu_A-\mu_B=0$; $H_a: \mu_A-\mu_B>0$ This question examines setting up hypotheses for the difference of two means in AP Statistics, for nitrate concentrations in Rivers A and B. The claim is River A has higher mean than B (μ_A > μ_B), so H0: μ_A - μ_B = 0 and Ha: μ_A - μ_B > 0. Distractor choice C reverses order, testing if B is higher, and choice D uses sample means instead of population. Choice B has a two-sided alternative, but the claim specifies direction ('higher'). Mini-lesson: Use clear subscripts for each population; null is equality, one-sided >0 for 'higher' claim. Ensure subtraction order supports the inequality. Sample means (6.4 vs. 5.7) are for testing, not hypotheses.

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