t-Test for Difference of Means - College Statistics

1. Two independent-sample t-tests are performed at the same alpha level ($\alpha = 0.05$, two-tailed). * Test 1: Compares a group of size 10 to a group of size 12. * Test 2: Compares a group of size 50 to a group of size 60. Let $t_1^*$ be the critical value for Test 1 and $t_2^*$ be the critical value for Test 2. Which of the following statements is true?
   Answer: $t_1^* > t_2^*$ When you encounter questions about critical values in t-tests, focus on how degrees of freedom affect the t-distribution. The key insight is that as sample sizes increase, the t-distribution approaches the standard normal distribution, making critical values smaller. For independent-sample t-tests, the degrees of freedom equal $df = n_1 + n_2 - 2$. Test 1 has $df = 10 + 12 - 2 = 20$, while Test 2 has $df = 50 + 60 - 2 = 108$. Since both tests use $\alpha = 0.05$ (two-tailed), you're looking for the critical values that cut off 2.5% in each tail. The t-distribution has heavier tails than the normal distribution, but these tails get lighter as degrees of freedom increase. With fewer degrees of freedom (Test 1), you need a larger critical value to capture the same area in the tails. With more degrees of freedom (Test 2), the distribution is closer to normal, so the critical value is smaller. Therefore, $t_1^* > t_2^*$, making choice A correct. Choice B is wrong because critical values definitely change with different degrees of freedom. Choice C reverses the relationship—smaller sample sizes actually require larger critical values, not smaller ones. Choice D incorrectly suggests that sample variances affect critical values, but critical values depend only on the significance level and degrees of freedom, not on the actual data variances. Remember: smaller samples mean fewer degrees of freedom, which means larger critical values needed for significance. This makes it harder to reject the null hypothesis with small samples.
2. A researcher compares the effectiveness of two different fertilizers on crop yield. They test Fertilizer A on 10 plots and Fertilizer B on 12 plots. Boxplots of the yields for each group show significant skewness and several outliers. The sample standard deviations are 25.4 and 28.1, respectively.
   Answer: The presence of significant skewness and outliers with small sample sizes. The validity of a two-sample t-test relies on three main assumptions: (1) independent samples, (2) normal populations or large enough sample sizes for the Central Limit Theorem to apply, and (3) equal population variances (for the pooled test). The most significant issue here is the violation of the normality assumption. The presence of significant skewness and outliers in small samples (n < 30) means the sampling distribution of the difference in means may not be well-approximated by a t-distribution, making the calculated p-value unreliable.
3. A company claims its new machine can process widgets at a rate that is, on average, more than 5 widgets per hour faster than the old machine. Let $\mu_{new}$ and $\mu_{old}$ be the mean processing rates. Summary data from trials are: * New Machine: $n_1 = 20, \bar{x}_1 = 58, s_1 = 4$ * Old Machine: $n_2 = 20, \bar{x}_2 = 51, s_2 = 3$ To test the company's claim, what are the appropriate null and alternative hypotheses and the value of the test statistic?
   Answer: $H_0: \mu_{new} - \mu_{old} = 5$; $H_a: \mu_{new} - \mu_{old} > 5$; t ≈ 1.79 The company's claim is that the new machine is more than 5 widgets per hour faster, so the alternative hypothesis is $H_a: \mu_{new} - \mu_{old} > 5$. The corresponding null hypothesis is $H_0: \mu_{new} - \mu_{old} = 5$. The test statistic for a non-zero hypothesized difference is $t = \frac{(\bar{x}_1 - \bar{x}_2) - (\mu_1 - \mu_2)_0}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}$. Plugging in the values: $t = \frac{(58 - 51) - 5}{\sqrt{\frac{4^2}{20} + \frac{3^2}{20}}} = \frac{7-5}{\sqrt{\frac{16}{20} + \frac{9}{20}}} = \frac{2}{\sqrt{\frac{25}{20}}} = \frac{2}{\sqrt{1.25}} \approx 1.79$.
4. An analyst wants to test the hypothesis that the treatment reduces the mean task completion time ($H_a: \mu_{Treatment} < \mu_{Control}$). Based on the output provided, what is the p-value for this specific one-tailed test, and what is the correct conclusion at $\alpha = 0.05$?
   Answer: The p-value is 0.029. Since 0.029 < 0.05, there is sufficient evidence to conclude the treatment reduces mean time. The output provides a two-tailed p-value of 0.058 for the test of difference = 0 (vs not =). The analyst is interested in a one-tailed test, $H_a: \mu_{Treatment} < \mu_{Control}$, which is equivalent to $H_a: \mu_{Treatment} - \mu_{Control} < 0$. The observed T-Value is -1.95, which is in the direction specified by this alternative hypothesis. Therefore, the p-value for the one-tailed test is half of the two-tailed p-value: $0.058 / 2 = 0.029$. Since this p-value (0.029) is less than the significance level ($\alpha = 0.05$), we reject the null hypothesis and conclude there is sufficient evidence that the treatment reduces mean task completion time.
5. A researcher tests if a new fertilizer (F) increases plant height compared to an old fertilizer (O). The hypotheses are $H_0: \mu_F - \mu_O = 0$ and $H_a: \mu_F - \mu_O > 0$. The data yields $n_F = 30, \bar{x}_F = 55$ cm and $n_O = 30, \bar{x}_O = 57$ cm. The resulting test statistic is $t = -2.50$.
   Answer: The sample data are in the opposite direction of the alternative hypothesis, so the p-value will be large. The alternative hypothesis $H_a: \mu_F - \mu_O > 0$ suggests we are looking for evidence that the new fertilizer produces taller plants. However, the sample data show the opposite: the mean height for the new fertilizer (55 cm) is less than the mean height for the old fertilizer (57 cm). This leads to a negative test statistic (t = -2.50). For a right-tailed test, a negative t-statistic provides no evidence in favor of the alternative hypothesis. The p-value is the probability of observing a result as extreme or more extreme in the direction of the alternative hypothesis. Thus, the p-value would be the area to the right of t = -2.50, which is very large (close to 1).
6. A pharmaceutical company develops a new drug to reduce blood pressure. A clinical trial is conducted where one group receives the new drug and a control group receives a placebo. Let $\mu_{drug}$ be the true mean reduction in systolic blood pressure for the drug group and $\mu_{placebo}$ be the true mean reduction for the placebo group. The company will only market the drug if it is shown to be more effective than the placebo. The research team conducts a hypothesis test and obtains a p-value of 0.03.
   Answer: $H_0: \mu_{drug} - \mu_{placebo} = 0$; $H_a: \mu_{drug} - \mu_{placebo} > 0$. The result is significant at $\alpha = 0.05$, providing evidence the drug is more effective. The research question is whether the drug is more effective than the placebo, which translates to a one-tailed alternative hypothesis: $H_a: \mu_{drug} > \mu_{placebo}$, or $H_a: \mu_{drug} - \mu_{placebo} > 0$. The null hypothesis is that there is no difference: $H_0: \mu_{drug} - \mu_{placebo} = 0$. The obtained p-value is 0.03. Since 0.03 is less than the standard significance level of $\alpha = 0.05$, the null hypothesis is rejected in favor of the alternative. This provides statistically significant evidence that the drug is more effective than the placebo.
7. A study compared the mean recovery times for two surgical procedures. The resulting p-value for a two-sample t-test was 0.25. The lead surgeon concluded, 'This p-value proves that the two procedures have the same mean recovery time.'
   Answer: The conclusion is flawed because failing to find evidence of a difference is not proof of no difference. This is a common misinterpretation of hypothesis testing. A large p-value (typically > 0.05) leads to a failure to reject the null hypothesis. This does not mean the null hypothesis is true or has been proven. It simply means that the sample data did not provide sufficient evidence to conclude that a difference exists. This is analogous to a 'not guilty' verdict in a trial: it doesn't prove innocence, only that there was not enough evidence for a conviction. The phrase 'absence of evidence is not evidence of absence' applies here.
8. A researcher is comparing two groups and is deciding whether to use a pooled or unpooled (Welch's) two-sample t-test. The sample sizes are $n_1 = 15$ and $n_2 = 18$. The sample standard deviations are $s_1 = 10.5$ and $s_2 = 11.2$. A common rule of thumb suggests that if the larger sample standard deviation is not more than twice the smaller one, one can assume equal variances.
   Answer: The unpooled (Welch's) t-test is generally preferred as it is more robust to violations of the equal variance assumption. While the rule of thumb (11.2 / 10.5 ≈ 1.07 < 2) is satisfied, modern statistical practice recommends using the unpooled (Welch's) t-test as the default choice. It performs nearly as well as the pooled test when the population variances are equal and performs much better (i.e., maintains the correct Type I error rate) when they are not. Since one rarely knows if the population variances are truly equal, the safer and more robust option is the unpooled test. Thus, it is generally preferred.
9. A researcher conducts a two-tailed test for the difference in mean scores between Group 1 and Group 2. The hypotheses are $H_0: \mu_1 - \mu_2 = 0$ and $H_a: \mu_1 - \mu_2 \neq 0$. The test yields a statistic of $t = 2.10$ and a p-value of 0.04. If the researcher had instead defined the difference as $\mu_2 - \mu_1$ and tested the hypotheses $H_0: \mu_2 - \mu_1 = 0$ and $H_a: \mu_2 - \mu_1 \neq 0$, what would the new test statistic and p-value be?
   Answer: t = –2.10, p-value = 0.04 The test statistic is calculated as $t = \frac{(\text{sample difference}) - (\text{hypothesized difference})}{SE}$. When the order of subtraction is reversed from (Group 1 - Group 2) to (Group 2 - Group 1), the sample difference ($\bar{x}_1 - \bar{x}_2$) becomes ($\bar{x}_2 - \bar{x}_1$), which simply negates its value. Therefore, the new test statistic will be the negative of the original: t = –2.10. For a two-tailed test, the p-value is the probability of observing a test statistic as extreme or more extreme than the one calculated, in either direction (i.e., $P(|T| \ge |t|)$). Since $|-2.10| = |2.10|$, the p-value remains exactly the same.
10. A large-scale educational study involving 10,000 students in Group A and 10,000 students in Group B compared two online learning platforms. The study found that the mean final exam score for Group A was 85.2 and for Group B was 85.6. A two-sample t-test for the difference in means yielded a p-value of 0.001.
    Answer: The difference in mean scores is statistically significant, but it may not be practically significant. This question tests the distinction between statistical significance and practical significance. A very small p-value (like 0.001) indicates that the observed difference is unlikely to be due to random chance, so the result is statistically significant. However, the magnitude of the difference in means (85.6 - 85.2 = 0.4 points) is very small. In a practical context, a 0.4-point difference on an exam may be considered negligible. Large sample sizes provide high power to detect even very small true differences, which may not be meaningful in a real-world context.
11. A researcher conducts a two-sample t-test to compare the mean effectiveness of two therapies and obtains a p-value of 0.08. Since the result is not significant at $\alpha = 0.05$, they fail to reject the null hypothesis of no difference. The researcher suspects that a true difference exists, but the study lacked sufficient power. Which of the following actions, if taken in a new study, would be most likely to increase the power of the test to detect a true difference of the same magnitude?
    Answer: Increase the sample sizes for both groups. Statistical power is the probability of correctly rejecting a false null hypothesis. Power is increased by four main factors: increasing the sample size, increasing the significance level ($\alpha$), increasing the true effect size, and decreasing the population variance. Of the choices given, increasing the sample sizes is a standard and effective method for a researcher to increase the power of a new study. Decreasing $\alpha$ (A) would decrease power. Using a two-tailed test instead of a correctly specified one-tailed test (B) would decrease power. Equalizing sample sizes (D) can be beneficial, but increasing the total sample size has a much larger and more certain impact on power.
12. A 95% confidence interval for the difference in mean test scores between two teaching methods (Method A minus Method B) is calculated to be (–0.5, 4.5). A researcher wants to use this information to conduct a two-tailed hypothesis test of $H_0: \mu_A - \mu_B = 0$ versus $H_a: \mu_A - \mu_B \neq 0$.
    Answer: The null hypothesis would not be rejected, and the p-value for the test would be greater than 0.05. There is a direct correspondence between a two-sided hypothesis test and a confidence interval. The null hypothesis $H_0: \mu_A - \mu_B = 0$ is rejected at significance level $\alpha$ if and only if the value 0 is not contained within the $(1-\alpha)100\%$ confidence interval for the difference. Since the 95% confidence interval (–0.5, 4.5) contains 0, the null hypothesis would not be rejected at the $\alpha = 0.05$ level. This implies that the corresponding p-value must be greater than 0.05.
13. A study compares the salaries of 20 randomly selected employees from a tech company and 20 from a manufacturing company. A two-sample t-test is performed. Later, it is discovered that the highest salary in the tech company sample, which was an extreme outlier, was recorded incorrectly and should be removed. The sample mean for the tech company was much higher than for the manufacturing company.
    Answer: The sample mean and standard deviation for the tech company will both decrease, likely leading to a smaller t-statistic. Removing an extreme high outlier from a sample will have two primary effects: it will decrease the sample mean ($\bar{x}$) and it will decrease the sample standard deviation ($s$), as the data will be less spread out. The t-statistic is calculated as $(\bar{x}_1 - \bar{x}_2) / SE$. Both the numerator (the difference in means) and the denominator (which depends on the standard deviations) will likely decrease. However, the outlier's effect on the mean is typically more pronounced. The reduction in the difference of means (numerator) will generally be larger than the reduction in the standard error (denominator), leading to a smaller t-statistic and, consequently, a larger p-value.
14. An ecologist measures the shell diameter of a specific snail species in two different coastal areas, Area A (sandy) and Area B (rocky). They collect 15 snails from Area A and 40 snails from Area B. The data from Area A show a roughly symmetric, mound-shaped distribution. The data from Area B are moderately right-skewed. The ecologist is concerned about the different sample sizes and the skewness in the Area B data.
    Answer: Proceed with an unpooled (Welch's) t-test, as the test is robust due to the large sample size in the skewed group. When you encounter questions about comparing means between two groups with different sample sizes and distributions, focus on the conditions required for two-sample t-tests and how robust these tests are to violations. The key insight here is understanding when t-tests remain valid despite imperfect conditions. With Area B having 40 snails, this large sample size makes the t-test robust to moderate skewness due to the Central Limit Theorem. The sampling distribution of the mean becomes approximately normal even when the population isn't perfectly normal. Since the sample sizes are unequal (15 vs 40), an unpooled (Welch's) t-test is appropriate because it doesn't assume equal variances. Answer A correctly identifies this approach. Answer B is too restrictive—it assumes any deviation from normality invalidates the t-test. While normality is an assumption, the test is robust to violations when sample sizes are reasonably large (typically n ≥ 30). Answer C suggests a pooled t-test, but this assumes equal variances between groups, which is questionable given the different environments and distributions. The unequal sample sizes also make pooling less appropriate. Answer D reflects a common misconception—t-tests don't require equal sample sizes. Welch's t-test was specifically designed to handle unequal sample sizes and variances. Remember this pattern: For two-sample comparisons, large sample sizes (especially n ≥ 30) allow t-tests to work well even with moderate skewness. Always choose unpooled methods when sample sizes or variances differ significantly between groups.
15. Two different studies are conducted to compare the mean blood pressure reduction from two drugs, Drug A and Drug B. Summary statistics are as follows: * Study 1: $n_A = 25, n_B = 25$; sample variances $s_A^2 = 100, s_B^2 = 120$. * Study 2: $n_A = 50, n_B = 50$; sample variances $s_A^2 = 50, s_B^2 = 60$. Assuming the difference in sample means ($\bar{x}_A - \bar{x}_B$) is the same for both studies, how would the width of a 95% confidence interval for $\mu_A - \mu_B$ from Study 2 compare to the one from Study 1?
    Answer: The interval from Study 2 would be narrower. The width of a confidence interval for the difference of two means is proportional to the standard error of the difference, $SE = \sqrt{\frac{s_A^2}{n_A} + \frac{s_B^2}{n_B}}$. For Study 1, $SE_1 = \sqrt{\frac{100}{25} + \frac{120}{25}} = \sqrt{4+4.8} = \sqrt{8.8} \approx 2.97$. For Study 2, $SE_2 = \sqrt{\frac{50}{50} + \frac{60}{50}} = \sqrt{1+1.2} = \sqrt{2.2} \approx 1.48$. Study 2 has both larger sample sizes and smaller sample variances, both of which contribute to a smaller standard error. A smaller standard error results in a narrower confidence interval.
16. A sociologist is studying the average number of hours of television watched per week by adults in two cities. Summary statistics are collected: * City A: $n_A = 40, \bar{x}_A = 15.5, s_A = 4.0$ * City B: $n_B = 50, \bar{x}_B = 17.5, s_B = 5.0$ Assuming the conditions for inference are met, what is the value of the unpooled test statistic for testing $H_0: \mu_A = \mu_B$?
    Answer: –2.11 The unpooled (Welch's) t-test statistic is calculated as $t = \frac{(\bar{x}_A - \bar{x}_B) - 0}{\sqrt{\frac{s_A^2}{n_A} + \frac{s_B^2}{n_B}}}$. Plugging in the values: $t = \frac{15.5 - 17.5}{\sqrt{\frac{4.0^2}{40} + \frac{5.0^2}{50}}} = \frac{-2}{\sqrt{\frac{16}{40} + \frac{25}{50}}} = \frac{-2}{\sqrt{0.4 + 0.5}} = \frac{-2}{\sqrt{0.9}} \approx -2.11$. Distractor A results from forgetting to square the standard deviations. Distractor B is the result of using the pooled standard error. Distractor C results from incorrectly adding the standard errors of the mean.
17. In a study comparing two groups with $n_1 = 10$ and $n_2 = 10$, the sample standard deviations were $s_1 = 5.0$ and $s_2 = 7.0$. Assuming the population variances are equal, the researcher calculates a pooled standard deviation, $s_p$. Which of the following must be true about the value of $s_p$?
    Answer: Its value will be between 5.0 and 7.0. When you encounter problems involving pooled standard deviation, you're dealing with a way to combine variability measures from two groups under the assumption of equal population variances. The pooled standard deviation creates a weighted average that accounts for both sample sizes and variances. The pooled standard deviation formula is: $s_p = \sqrt{\frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2}}$ Let's calculate: $s_p = \sqrt{\frac{(10-1)(5.0)^2 + (10-1)(7.0)^2}{10+10-2}} = \sqrt{\frac{9(25) + 9(49)}{18}} = \sqrt{\frac{225 + 441}{18}} = \sqrt{37} = 6.08$ Answer A is correct because the pooled standard deviation will always fall between the individual sample standard deviations when sample sizes are equal. This makes intuitive sense—you're combining information from both groups, creating a compromise value. Answer B is wrong because pooling doesn't simply add variability; it creates a weighted average that falls between the original values. Answer C incorrectly assumes the pooled standard deviation equals the simple arithmetic mean of the two standard deviations (6.0). The pooling formula weights by degrees of freedom, not just averages. Answer D is incorrect because with equal sample sizes, neither standard deviation has more influence—they're weighted equally. Remember: pooled standard deviation with equal sample sizes always falls between the individual standard deviations. The exact value depends on the specific numbers, but it won't exceed the bounds set by the original values.
18. A researcher wants to determine if a new fuel additive improves gas mileage. Which of the following experimental designs is most appropriate for analysis using an independent-samples t-test for the difference in means?
    Answer: Randomly select 50 cars, assign 25 to use the additive and 25 to not use it, then compare the two groups' mean mileage. An independent-samples t-test is appropriate when the two groups being compared are independent of each other. In choice B, the 50 cars are randomly assigned to two separate groups (treatment and control), and these groups are independent. Choices A and D describe repeated measures (or crossover) designs, where the same subjects (cars) are measured under both conditions. Choice C describes a matched-pairs design. These three designs (A, C, D) create dependent samples and must be analyzed with a paired-samples t-test, not an independent-samples t-test.

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