Difference of Two Means (Test) - AP Statistics

A retailer compares mean amount spent per customer for shoppers who received a coupon versus those who did not. Two independent random samples were collected (Coupon: $n=60$ ; No coupon: $n=58$ ). A two-sample $t$ test for $\mu_{\text{coup}}-\mu_{\text{noc}}$ was conducted with $H_0:\mu_{\text{coup}}=\mu_{\text{noc}}$ and $H_a:\mu_{\text{coup}}>\mu_{\text{noc}}$ . The p-value was 0.0007. At $\alpha=0.05$ , what conclusion is appropriate?
Answer: Reject $H_0$ ; there is convincing evidence that the population mean amount spent is higher for customers who receive a coupon than for those who do not. This question examines whether customers with coupons spend more on average ( $H_a$ : μ_coup > μ_noc). The p-value (0.0007) is much smaller than α = 0.05, providing very strong evidence to reject the null hypothesis. This gives convincing evidence that the population mean amount spent is higher for customers who receive a coupon than for those who do not. Choice C overstates by claiming the effect applies to "every customer." Choice E misunderstands the purpose of hypothesis testing, which is specifically to make inferences about populations from samples. When we have such a small p-value, it indicates the observed difference would be extremely unlikely if there were truly no difference in the populations.
A company tests whether a new training program changes mean weekly sales. A random sample of 42 employees used the new program and a separate random sample of 40 employees used the old program. A two-sample $t$ test for $\mu_{\text{new}}-\mu_{\text{old}}$ was conducted with hypotheses $H_0:\mu_{\text{new}}=\mu_{\text{old}}$ and $H_a:\mu_{\text{new}}\ne\mu_{\text{old}}$ . The test produced a p-value of 0.018. Using $\alpha=0.05$ , what conclusion is appropriate?
Answer: Reject $H_0$ ; there is convincing evidence that the mean weekly sales differ for employees using the new versus old program. This question tests understanding of hypothesis test conclusions for comparing two population means. Since the p-value (0.018) is less than the significance level α = 0.05, we reject the null hypothesis. The correct interpretation is that there is convincing evidence of a difference between the mean weekly sales for the two training programs. Choice C incorrectly claims causation and applies to "all employees," which overstates the conclusion. Choice D confuses sample means with the hypothesis test conclusion. When we reject H₀ in a two-tailed test, we conclude there's evidence of a difference, not a specific direction unless we examine the sample statistics.
A company tests whether a redesigned website changes mean time (in seconds) that visitors spend on a page. A random sample of 60 visitors saw the redesigned page (R) and a separate random sample of 55 visitors saw the original page (O). A two-sample $t$ test for $\mu_R-\mu_O$ is conducted with $H_0: \mu_R-\mu_O=0$ and $H_a: \mu_R-\mu_O\neq 0$ . The p-value is 0.041. Using $\alpha=0.05$ , what conclusion is appropriate?
Answer: Reject $H_0$ ; there is convincing evidence that the redesigned page and original page differ in population mean time on page. This question presents a two-tailed test for website redesign effectiveness. The p-value (0.041) is less than α (0.05), so we reject the null hypothesis. With a two-tailed alternative (Ha: μR - μO ≠ 0), rejecting H0 means we have evidence of a difference in either direction between the population means. Choice B correctly states this conclusion. Choice C is incorrect because it specifies a direction (larger) when our two-tailed test only establishes that a difference exists. Choice E incorrectly implies causation, which cannot be established from this observational comparison. In two-tailed tests, we conclude there's a difference but don't specify which group has the higher mean unless we examine the actual sample statistics.
A manufacturer compares mean battery life (hours) for two brands of AA batteries: Brand X and Brand Y. Independent random samples of batteries are tested, and a two-sample $t$ test for $\mu_X-\mu_Y$ is performed with $H_0: \mu_X-\mu_Y=0$ and $H_a: \mu_X-\mu_Y>0$ . The p-value is 0.049. Using $\alpha=0.05$ , what conclusion is appropriate?
Answer: Reject $H_0$ ; there is convincing evidence that Brand X has a higher population mean battery life than Brand Y. This battery life comparison uses a right-tailed test. The p-value (0.049) is just barely less than α (0.05), so we reject the null hypothesis. The alternative Ha: μX - μY > 0 tests whether Brand X has higher mean battery life than Brand Y. Rejecting H0 provides convincing evidence that Brand X has a higher population mean battery life. Choice A incorrectly suggests failing to reject when p < α requires rejection. Choice D misinterprets the conclusion by claiming every Brand X battery lasts longer - we can only make claims about population means, not individual batteries. With p-value = 0.049, we have just enough evidence to reject H0 at the 0.05 level.
A nutritionist compares mean sodium intake (mg/day) for adults who follow Diet A versus Diet B. Independent random samples were taken from each diet group. A two-sample $t$ test was conducted for $H_0: \mu_A-\mu_B=0$ versus $H_a: \mu_A-\mu_B\ne 0$ . The p-value was 0.41. At the 0.05 significance level, what conclusion is appropriate?
Answer: Fail to reject $H_0$ ; there is not convincing evidence that mean sodium intake differs between Diet A and Diet B. This problem involves a two-sample t-test with a two-tailed alternative hypothesis (μ_A - μ_B ≠ 0). The p-value of 0.41 is much larger than the significance level of 0.05, so we fail to reject the null hypothesis. This means there is not convincing evidence that mean sodium intake differs between Diet A and Diet B. Choice D incorrectly suggests that similar sample means guarantee equal population means, while Choice E wrongly claims that failing to reject H₀ proves the means are exactly equal. When we fail to reject H₀, we simply lack sufficient evidence to conclude the means differ; we cannot prove they are equal. Remember that hypothesis tests can only provide evidence against the null hypothesis, never proof that it's true.
A manufacturer compares the mean lifetime (hours) of bulbs from Machine X and Machine Y. Random samples of bulbs were selected from each machine's production. A two-sample $t$ test was performed for $H_0: \mu_X-\mu_Y=0$ versus $H_a: \mu_X-\mu_Y>0$ . The p-value was 0.055. Using $\alpha=0.05$ , what conclusion is appropriate?
Answer: Fail to reject $H_0$ ; there is not convincing evidence that Machine X produces bulbs with a greater mean lifetime. This question tests whether Machine X produces bulbs with greater mean lifetime than Machine Y, with $H_a$ : μ_X - μ_Y > 0. The p-value of 0.055 is slightly greater than α = 0.05, so we fail to reject the null hypothesis. This means there is not convincing evidence that Machine X produces bulbs with a greater mean lifetime. Choice D contains a logical error, claiming we must reject H₀ when we actually fail to reject it. Choice E incorrectly states that failing to reject H₀ means the machines are identical. In hypothesis testing, we use a strict cutoff: if p-value ≥ α, we fail to reject H₀, even if the p-value is very close to α.
A public health researcher compares mean systolic blood pressure for adults in two towns: Town A and Town B. Independent random samples are taken, and a two-sample $t$ test for $\mu_A-\mu_B$ is conducted with $H_0: \mu_A-\mu_B=0$ and $H_a: \mu_A-\mu_B<0$ . The p-value is 0.095. Using $\alpha=0.10$ , what conclusion is appropriate?
Answer: Reject $H_0$ ; there is convincing evidence that Town A has a lower population mean systolic blood pressure than Town B. This public health study uses a left-tailed test for blood pressure differences. The p-value (0.095) is less than α (0.10), so we reject the null hypothesis. The alternative Ha: μA - μB < 0 tests whether Town A has lower mean systolic blood pressure than Town B. When we reject H0, we have convincing evidence supporting this alternative - that Town A has a lower population mean systolic blood pressure. Choice B incorrectly suggests failing to reject, but with p < α, we must reject. Choice E inappropriately claims causation from living in Town A. Note that using α = 0.10 instead of the more common 0.05 makes it easier to reject H0, and our p-value of 0.095 leads to rejection at this level.
A company tests whether a new training program reduces the mean time (in minutes) to complete a standard task. A random sample of employees used the new program (group T) and a separate random sample used the old program (group C). A two-sample $t$ test was performed for $H_0: \mu_T-\mu_C=0$ versus $H_a: \mu_T-\mu_C<0$ . The p-value was 0.08. Using $\alpha=0.10$ , what conclusion is appropriate?
Answer: Reject $H_0$ ; there is convincing evidence that the new program reduces the mean completion time. In this two-sample t-test, we're testing whether the new training program reduces mean completion time, with $H_a$ : μ_T - μ_C < 0 (new minus old is negative). The p-value of 0.08 is less than the significance level α = 0.10, so we reject the null hypothesis. This provides convincing evidence that the new program reduces the mean completion time. Choice D incorrectly makes a claim about every individual employee rather than the population mean. Choice E wrongly states that failing to reject H₀ would prove no difference exists. When interpreting hypothesis tests, focus on whether the p-value is less than α, and remember that conclusions apply to population parameters, not individual observations.
A psychologist compares mean reaction time (milliseconds) for participants who drank caffeinated coffee versus decaf. Two independent random samples were taken (Caffeine: $n=22$ ; Decaf: $n=24$ ). A two-sample $t$ test for $\mu_{\text{caf}}-\mu_{\text{dec}}$ used $H_0:\mu_{\text{caf}}=\mu_{\text{dec}}$ and $H_a:\mu_{\text{caf}}<\mu_{\text{dec}}$ . The p-value was 0.20. Using $\alpha=0.05$ , what conclusion is appropriate?
Answer: Fail to reject $H_0$ ; there is not convincing evidence that mean reaction time is lower after caffeinated coffee than after decaf. This problem tests whether caffeinated coffee leads to lower mean reaction times than decaf ( $H_a$ : μ_caf < μ_dec). The p-value (0.20) greatly exceeds α = 0.05, so we fail to reject the null hypothesis. This means there is not convincing evidence that mean reaction time is lower after caffeinated coffee than after decaf. Choice C makes an inappropriate universal causal claim. Choice D incorrectly states that the results "prove" equality—we never prove the null hypothesis true. In hypothesis testing, a large p-value indicates the observed difference could easily occur by chance under the null hypothesis, so we don't have sufficient evidence to support the alternative hypothesis.
A researcher compares mean resting heart rate (beats per minute) between runners and non-runners. Independent random samples were taken from each group. A two-sample $t$ test was conducted for $H_0: \mu_R-\mu_N=0$ versus $H_a: \mu_R-\mu_N<0$ . The p-value was 0.001. At $\alpha=0.05$ , what conclusion is appropriate?
Answer: Reject $H_0$ ; there is convincing evidence that runners have a lower mean resting heart rate than non-runners. This question tests a one-tailed hypothesis where $H_a$ : μ_R - μ_N < 0, meaning we're testing if runners have a lower mean resting heart rate than non-runners. With a p-value of 0.001, which is much less than α = 0.05, we reject the null hypothesis. This provides convincing evidence that runners have a lower mean resting heart rate than non-runners. Choice D incorrectly implies causation from an observational study, while Choice E makes an incorrect claim about the exact difference between populations. When conducting hypothesis tests, we can only conclude about the direction of the difference (higher or lower), not establish causation or determine exact values. Statistical significance indicates evidence of a difference, not the magnitude of that difference.
A botanist compares mean plant height (cm) for seedlings grown under LED light versus fluorescent light. Independent random samples were used (LED: $n=16$ ; Fluorescent: $n=16$ ). A two-sample $t$ test for $\mu_{\text{LED}}-\mu_{\text{Fl}}$ was performed with $H_0:\mu_{\text{LED}}=\mu_{\text{Fl}}$ and $H_a:\mu_{\text{LED}}\ne\mu_{\text{Fl}}$ . The p-value was 0.083. Using $\alpha=0.10$ , what conclusion is appropriate?
Answer: Reject $H_0$ ; there is convincing evidence that the population mean heights differ under LED and fluorescent light. This question tests a two-tailed hypothesis comparing plant heights under different light types. The p-value (0.083) is less than the significance level α = 0.10, so we reject the null hypothesis. This provides convincing evidence that the population mean heights differ under LED and fluorescent light. Choice C incorrectly makes a directional causal claim when we only tested for any difference. Choice E misunderstands the relationship between sample and population means. When using α = 0.10 instead of the more common 0.05, we're accepting a higher risk of Type I error but increasing our power to detect differences—here, the evidence is sufficient at this less stringent level.
An engineer compares mean battery life (hours) for two brands, Brand X and Brand Y. Independent random samples were tested (Brand X: $n=18$ ; Brand Y: $n=20$ ). A two-sample $t$ test for $\mu_X-\mu_Y$ used $H_0:\mu_X=\mu_Y$ and $H_a:\mu_X\ne\mu_Y$ . The p-value was 0.62. At $\alpha=0.05$ , what conclusion is appropriate?
Answer: Fail to reject $H_0$ ; there is not convincing evidence of a difference in population mean battery life between Brand X and Brand Y. This problem involves a two-tailed test comparing mean battery life between two brands. With a p-value of 0.62, which greatly exceeds α = 0.05, we fail to reject the null hypothesis. This means there is not convincing evidence of a difference in population mean battery life between Brand X and Brand Y. Choice C incorrectly states that the brands have "exactly the same mean"—failing to reject H₀ doesn't prove equality, it only indicates insufficient evidence for a difference. Choice E makes an inappropriate causal claim. In hypothesis testing, a large p-value suggests the observed difference could easily occur by chance if the null hypothesis were true, so we maintain our assumption of no difference.
A school district compares mean math test scores for students taught with Method 1 versus Method 2. Two independent random samples of students were selected (Method 1: $n=28$ ; Method 2: $n=30$ ). A two-sample $t$ test for $\mu_1-\mu_2$ was performed with $H_0:\mu_1=\mu_2$ and $H_a:\mu_1>\mu_2$ . The p-value was 0.004. Using $\alpha=0.01$ , what conclusion is appropriate?
Answer: Reject $H_0$ ; there is convincing evidence that Method 1 has a higher population mean score than Method 2. This question tests a one-tailed hypothesis where we're examining if Method 1 produces higher mean scores than Method 2 ( $H_a$ : μ₁ > μ₂). The p-value (0.004) is less than α = 0.01, so we reject the null hypothesis. This provides convincing evidence that Method 1 has a higher population mean score than Method 2. Choice C incorrectly claims the effect applies to "every student," which is an overstatement—hypothesis tests make conclusions about population means, not individual outcomes. Choice D reverses the direction of the conclusion. When conducting hypothesis tests, we make inferences about population parameters based on sample data, not claims about every individual in the population.
A sports scientist investigates whether mean vertical jump height differs between athletes who follow Program 1 and those who follow Program 2. Two independent random samples were taken (Program 1: $n=12$ ; Program 2: $n=14$ ). A two-sample $t$ test for $\mu_1-\mu_2$ was conducted with $H_0:\mu_1-\mu_2=0$ and $H_a:\mu_1-\mu_2\neq 0$ . The p-value was 0.049. Using $\alpha=0.05$ , what conclusion is appropriate?
Answer: Reject $H_0$ ; there is sufficient evidence of a difference in population mean jump height between the two programs. This question assesses interpreting a two-sample t-test for mean jump height differences between two programs. The two-sided test yields p=0.049 just under α=0.05, so we reject H0, indicating sufficient evidence of a population mean difference. A common distractor is choice D, which confuses sample mean differences with the hypothesis test outcome. In concluding two-mean tests, emphasize population-level inferences and avoid absolutes like 'for all athletes' in choice E, as results are probabilistic. Rejection supports the alternative but doesn't prove causation or direction without additional context.
A city compares mean commute time (minutes) for residents who take the bus versus residents who drive. Two independent random samples were selected (bus: $n=55$ ; drive: $n=50$ ). A two-sample $t$ test for $\mu_{bus}-\mu_{drive}$ used hypotheses $H_0:\mu_{bus}-\mu_{drive}=0$ and $H_a:\mu_{bus}-\mu_{drive}\neq 0$ . The p-value was 0.004. Using $\alpha=0.01$ , what conclusion is appropriate?
Answer: Reject $H_0$ ; there is sufficient evidence that the population mean commute times differ for bus riders and drivers. This question examines a two-sample t-test for differences in mean commute times between bus riders and drivers. The two-sided alternative allows for any difference, and with p=0.004 less than α=0.01, we reject H0, providing evidence of a population mean difference. Choice D is a distractor, incorrectly assuming different sample means automatically mean different population means without considering significance. For two-mean tests, conclusions should avoid causal claims like in choice E and focus on evidence for or against the null in the population context. Note that even with rejection, we don't specify direction unless the test is one-sided or further analysis is done.
A company tests whether a new training program changes mean weekly sales compared with the old program. A random sample of 30 employees used the new program and a separate random sample of 28 employees used the old program. A two-sample $t$ test for $\mu_{new}-\mu_{old}$ was conducted with hypotheses $H_0: \mu_{new}-\mu_{old}=0$ and $H_a: \mu_{new}-\mu_{old}>0$ . The test produced a p-value of 0.018. Using $\alpha=0.05$ , what conclusion is appropriate?
Answer: Reject $H_0$ ; there is sufficient evidence that the new program increases the population mean weekly sales compared with the old program. This question assesses the skill of interpreting a two-sample t-test for the difference of two means in AP Statistics. The hypotheses are set up to test if the new training program increases mean weekly sales compared to the old one, with a one-sided alternative. Given the p-value of 0.018 is less than the significance level $α=0.05$ , we reject the null hypothesis, providing sufficient evidence that the population mean sales are higher with the new program. A common distractor is choice D, which incorrectly states that failing to reject means the sample means are equal and programs are equally effective, but actually, failing to reject doesn't prove equality—it just means insufficient evidence of a difference. In two-sample t-tests, conclusions should focus on population means, not samples, and avoid causal language unless the study design supports it. Remember, rejecting $H_0$ supports the alternative but doesn't prove it for every case, as seen in choice E's overstatement about every employee.
A public health researcher compares mean systolic blood pressure for adults who exercise regularly versus those who do not. Two independent random samples were taken (Exercise: $n=50$ ; No exercise: $n=48$ ). A two-sample $t$ test for $\mu_{\text{ex}}-\mu_{\text{no}}$ was conducted with $H_0:\mu_{\text{ex}}=\mu_{\text{no}}$ and $H_a:\mu_{\text{ex}}<\mu_{\text{no}}$ . The p-value was 0.031. Using $\alpha=0.05$ , what conclusion is appropriate?
Answer: Reject $H_0$ ; there is convincing evidence that the population mean systolic blood pressure is lower for adults who exercise regularly than for those who do not. This question tests understanding of a one-tailed test where we're examining if exercisers have lower mean systolic blood pressure ( $H_a$ : μ_ex < μ_no). The p-value (0.031) is less than α = 0.05, so we reject the null hypothesis. This provides convincing evidence that the population mean systolic blood pressure is lower for adults who exercise regularly than for those who do not. Choice C incorrectly claims causation for "every adult," which overstates the conclusion. Choice E misinterprets what failing to reject would mean. When we reject H₀ in a one-tailed test with "less than" alternative, we conclude there's evidence supporting the specific directional claim in the alternative hypothesis.
A school compares mean time (in minutes) to complete a standardized math test for students using a paper booklet versus an online version. Two independent random samples were taken (paper: $n=40$ ; online: $n=38$ ). A two-sample $t$ test for $\mu_{paper}-\mu_{online}$ used $H_0:\mu_{paper}-\mu_{online}=0$ and $H_a:\mu_{paper}-\mu_{online}<0$ . The p-value was 0.041. Using $\alpha=0.05$ , what conclusion is appropriate?
Answer: Reject $H_0$ ; there is sufficient evidence that the population mean completion time is lower for paper than for online. This question tests the application of a two-sample t-test for the difference in mean completion times between paper and online math tests. The one-sided alternative hypothesizes that paper is faster, and since the p-value of 0.041 is less than α=0.05, we reject H0, supporting that the population mean time is lower for paper. A frequent distractor is choice B, which mistakenly says 'fail to reject' while claiming sufficient evidence, confusing the decision rule. When concluding two-mean tests, specify the direction if it's a one-sided test and evidence supports it, but don't generalize to causation like in choice E. Focus on population inferences, as sample differences alone (as in choice D) don't guarantee population differences without statistical significance.
A teacher compares mean time (seconds) to solve a puzzle for students who listened to music (group M) versus students who worked in silence (group S). Two independent random samples were taken. A two-sample $t$ test was conducted for $H_0: \mu_M-\mu_S=0$ versus $H_a: \mu_M-\mu_S>0$ . The p-value was 0.62. At the 0.05 level, what conclusion is appropriate?
Answer: Fail to reject $H_0$ ; there is not convincing evidence that listening to music increases the mean puzzle-solving time. This problem tests whether listening to music increases mean puzzle-solving time, with $H_a$ : μ_M - μ_S > 0. The p-value of 0.62 is much larger than α = 0.05, so we fail to reject the null hypothesis. This means there is not convincing evidence that listening to music increases the mean puzzle-solving time. Choice D incorrectly claims that sample results can prove population means are equal, while Choice E wrongly extends the conclusion to every individual student. When we fail to reject H₀, we're simply stating that the data doesn't provide sufficient evidence for the alternative hypothesis. This doesn't prove the null hypothesis is true or that there's no effect for any individual.
A teacher compares mean quiz scores for students using two different study apps. Two independent random samples are taken: App A and App B. A two-sample $t$ test for $\mu_A-\mu_B$ is conducted with $H_0: \mu_A-\mu_B=0$ and $H_a: \mu_A-\mu_B\neq 0$ . The p-value is 0.52. At $\alpha=0.05$ , what conclusion is appropriate?
Answer: Fail to reject $H_0$ ; there is not convincing evidence of a difference in population mean quiz scores between the two apps. This problem presents a two-tailed test comparing study apps. The p-value (0.52) is much larger than α (0.05), so we fail to reject the null hypothesis. With a two-tailed alternative (Ha: μA - μB ≠ 0), failing to reject means we don't have convincing evidence of any difference between the population mean quiz scores. Choice A incorrectly suggests rejecting H0 when the large p-value clearly indicates we should fail to reject. Choice D wrongly claims this "proves" equality - hypothesis tests never prove the null hypothesis. A large p-value simply means our sample data are consistent with the null hypothesis of no difference between population means.

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