CI for Difference of Means - College Statistics

Let Interval 1 be a 95% confidence interval for $\mu_A - \mu_B$ based on sample sizes $n_A=30, n_B=30$ . Let Interval 2 be a 95% confidence interval for $\mu_C - \mu_D$ from a separate study with sample sizes $n_C=120, n_D=120$ . Assume the sample standard deviations are identical across all four groups ( $s_A=s_B=s_C=s_D$ ). What is the relationship between the width of Interval 1 (W1) and the width of Interval 2 (W2)?
Answer: W1 is approximately twice W2. When comparing confidence interval widths, you need to understand how sample size affects the standard error. The width of a confidence interval for the difference between two means depends on the standard error of that difference. For a confidence interval comparing two means, the standard error is $SE = s\sqrt{\frac{1}{n_1} + \frac{1}{n_2}}$ , where s is the pooled standard deviation. Since all sample standard deviations are identical here, we can focus on how the sample sizes affect this formula. For Interval 1 (n = 30 each): $SE_1 = s\sqrt{\frac{1}{30} + \frac{1}{30}} = s\sqrt{\frac{2}{30}} = s\sqrt{\frac{1}{15}}$ For Interval 2 (n = 120 each): $SE_2 = s\sqrt{\frac{1}{120} + \frac{1}{120}} = s\sqrt{\frac{2}{120}} = s\sqrt{\frac{1}{60}}$ The ratio of standard errors is: $\frac{SE_1}{SE_2} = \frac{\sqrt{\frac{1}{15}}}{\sqrt{\frac{1}{60}}} = \sqrt{\frac{60}{15}} = \sqrt{4} = 2$ Since confidence interval width is proportional to standard error, W1 is approximately twice W2, making A correct. B is wrong because it reverses the relationship—larger samples create narrower intervals, not wider ones. C incorrectly assumes sample size doesn't matter for interval width. D gives the wrong multiplier; $\sqrt{2}$ would apply if one sample were twice the size of the other, but here one is four times larger. Remember: When sample sizes increase by a factor of 4, standard errors (and confidence interval widths) decrease by a factor of 2. The square root relationship is key.
Two 95% confidence intervals for $\mu_1 - \mu_2$ are created from the same set of sample data: $n_1=10, s_1=5$ and $n_2=15, s_2=12$ . Interval A is constructed using the conservative approach for degrees of freedom. Interval B is constructed using the Satterthwaite approximation for degrees of freedom. Which statement accurately compares the two intervals?
Answer: Interval A will be wider than Interval B. The conservative degrees of freedom is $df_A = \min(n_1-1, n_2-1) = \min(9, 14) = 9$ . The Satterthwaite approximation results in a higher degrees of freedom, which will be between 9 and $n_1+n_2-2 = 23$ . A lower degrees of freedom corresponds to a larger t-critical value ( $t^*$ ). Since Interval A uses a smaller df, its $t^*$ will be larger, resulting in a larger margin of error and a wider interval compared to Interval B. Both intervals will have the same center, $\bar{x}_1 - \bar{x}_2$ . - A is incorrect because the different methods for df lead to different t-critical values. - C is incorrect because the conservative method always produces a df less than or equal to the Satterthwaite df, leading to a wider or equal-width interval, never narrower. - D is incorrect because while the center is the same, we can determine which is wider based on the properties of the df calculation methods.
A biologist compares the mean weight of a species of fish from two independent lakes. A sample from Lake 1 has $n_1 = 15$ fish, and a histogram of their weights is strongly skewed to the right. A sample from Lake 2 has $n_2 = 18$ fish, and a histogram of their weights is approximately symmetric. Why should the biologist be cautious about interpreting a two-sample t-interval for the difference in mean weights?
Answer: The strong skewness in one of the small samples suggests the normality condition for the procedure is not met. The two-sample t-procedure relies on the assumption that both samples are drawn from normally distributed populations or that the sample sizes are large enough for the Central Limit Theorem to apply (typically $n \ge 30$ ). In this case, both sample sizes are small (15 and 18), and the data from Lake 1 show strong skewness. This combination means the normality condition is likely violated, making the t-interval potentially unreliable. - A is incorrect because the t-procedure for the difference of means does not require equal sample sizes. - B is incorrect because this is only an assumption for the pooled t-procedure. The unpooled (Welch's) t-procedure, which is more common, does not require equal population standard deviations. - D is incorrect because the problem states the lakes are independent, meaning the samples are independent even if the fish are the same species.
A student is asked to calculate the margin of error for a 95% confidence interval for $\mu_1 - \mu_2$ given summary statistics: $\bar{x}_1 = 50, s_1=10, n_1=40$ and $\bar{x}_2=45, s_2=12, n_2=40$ . The student writes the following expression for the margin of error: $ME = 1.96 \sqrt{\frac{10^2+12^2}{40}}$ . What is the most significant conceptual error in the student's calculation?
Answer: The formula for the standard error is incorrect; the variances of the two groups should not have been added together before division. When calculating confidence intervals for the difference between two means, you need to properly construct the standard error that accounts for variability from both samples. The correct standard error for $\mu_1 - \mu_2$ should be $SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}$ . Notice that each sample's variance is divided by its own sample size before adding. The student's formula $\sqrt{\frac{s_1^2 + s_2^2}{n}}$ incorrectly adds the variances first, then divides by the sample size. This violates the fundamental principle that when you're dealing with the difference of two independent random variables, you add their individual variances, not pool them before standardizing. Answer A correctly identifies this error. The student treated this like a single-sample problem instead of recognizing that variability comes from two distinct sources. Answer B is wrong because pooling is optional when sample sizes are equal, not required. You can use either pooled or unpooled methods. Answer C misses the main issue. While using a t-critical value instead of 1.96 would be more precise given unknown population standard deviations, this is a minor correction compared to the fundamental formula error. Answer D reflects a misunderstanding of confidence interval components. The margin of error is the "plus or minus" part that gets added to and subtracted from the point estimate—it doesn't include the point estimate itself. Study tip: For two-sample problems, always write out the standard error formula carefully, ensuring each sample's variance is divided by its own sample size before combining terms.
A medical researcher conducts a study on a new drug to reduce blood pressure. A 95% confidence interval for the difference in mean systolic blood pressure reduction ( $\mu_{\text{new drug}} - \mu_{\text{control}}$ ) is found to be ( $-0.5, 3.5$ ) mmHg. Based only on this interval, which of the following is the most appropriate conclusion?
Answer: At the $\alpha = 0.05$ significance level, there is not sufficient evidence to conclude a difference in mean effectiveness between the new drug and the control. A confidence interval for a difference in means that contains zero indicates that a value of zero for the true difference is plausible. Therefore, we cannot reject the null hypothesis that the means are equal ( $H_0: \mu_{\text{new drug}} - \mu_{\text{control}} = 0$ ) at the corresponding alpha level ( $1 - 0.95 = 0.05$ ). - A is incorrect because it misinterprets the confidence level as a probability about a specific interval. The 95% confidence level refers to the long-run success rate of the method, not the probability that this particular interval captures the true parameter. - B is incorrect because it confuses the sampling distribution of a statistic (the sample mean difference) with the confidence interval for a parameter (the population mean difference). - D is incorrect because the interval also contains positive values, meaning it is plausible that the new drug is more effective. We cannot conclude it is less effective.
A research team constructs a 90% confidence interval for the difference in mean test scores between students taught with Method A and Method B. They plan to repeat the study next year. Which of the following proposed changes would, holding all other factors constant, result in a narrower confidence interval?
Answer: Increasing the number of students sampled in both groups. The width of the confidence interval is determined by the margin of error, which is calculated as $t^* \times \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}$ . Increasing the sample sizes ( $n_1$ and $n_2$ ) will decrease the standard error, which in turn decreases the margin of error and makes the interval narrower. - A is incorrect because a higher confidence level requires a larger critical value ( $t^*$ ), which would make the interval wider. - B is incorrect because the difference in sample means determines the center of the interval, not its width. - C is incorrect because larger sample standard deviations ( $s_1$ and $s_2$ ) would increase the standard error, making the interval wider.
A 95% confidence interval for the difference between two population means, $\mu_1 - \mu_2$ , is calculated to be (12, 38). The sample sizes used were $n_1 = 30$ and $n_2 = 30$ . What was the observed difference in the sample means, $\bar{x}_1 - \bar{x}_2$ ?
Answer: 25 The point estimate for the difference in population means ( $\mu_1 - \mu_2$ ) is the difference in sample means ( $\bar{x}_1 - \bar{x}_2$ ). This point estimate is always the center of the confidence interval. The center of the interval (12, 38) is the average of its endpoints: $\frac{12 + 38}{2} = \frac{50}{2} = 25$ . - A is incorrect because it represents half the width of the interval (the margin of error), $\frac{38 - 12}{2} = 13$ , not the center. - C is incorrect because it is the full width of the interval (38 - 12 = 26). - D is incorrect because the point estimate can be found directly from the interval's endpoints without needing the standard deviations or critical value.
A statistician is comparing the mean results of two independent samples from populations with unknown standard deviations. A statistical test for the equality of the two population variances yields a p-value of 0.02. The sample sizes are $n_1=20$ and $n_2=25$ . Which procedure is most appropriate for constructing a confidence interval for the difference between the two population means?
Answer: A two-sample t-interval without pooling the variances. The choice between a pooled and unpooled (Welch's) t-interval depends on whether it is reasonable to assume the population variances are equal. The test for equality of variances yielded a p-value of 0.02. Since this p-value is small (e.g., less than $\alpha=0.05$ ), we reject the null hypothesis of equal variances. Therefore, the assumption for pooling variances is not met, and the unpooled t-interval is the appropriate procedure. - A is incorrect because population standard deviations are unknown. - B is incorrect because the samples are independent, not paired. - C is incorrect because the evidence suggests the population variances are not equal, making the pooled procedure inappropriate.
A marketing analyst compares the mean spending of customers at two different store locations. They collect data from the first 100 customers who enter Store A on a Monday morning and the first 100 customers who enter Store B on a Saturday afternoon. After finding a significant difference, they construct a 95% confidence interval. What is the most significant threat to the validity of this confidence interval for generalizing to all customers of the two stores?
Answer: The samples are convenience samples, not random samples from the populations of interest. Statistical inference, including the construction of confidence intervals, relies on the assumption of random sampling from the population of interest. By using the first 100 customers on specific days and times, the analyst has collected convenience samples. These samples are likely not representative of all customers (e.g., weekday shoppers may differ from weekend shoppers), introducing bias and making it inappropriate to generalize the findings to the entire customer population of each store. - A is incorrect because the unpooled t-procedure is designed to handle unequal variances; it's a statistical issue that can be addressed. - C is incorrect because sample sizes of 100 are generally considered large enough for the Central Limit Theorem to apply. - D is incorrect because the t-distribution is specifically used when population standard deviations are unknown.
A medical researcher wants to estimate the difference in mean recovery time for two surgical procedures. They want to be 95% confident that their estimate is within 3 days of the true mean difference. Based on pilot data, they estimate the population standard deviation for recovery time is about 8 days for both procedures. Assuming equal sample sizes ( $n_1 = n_2 = n$ ), what is the minimum sample size $n$ required for each group to achieve the desired margin of error?
Answer: 55 The formula for the margin of error (ME) for the difference of two means is $ME = z^* \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}$ . For planning purposes, we use the z-critical value for 95% confidence, which is 1.96. We are given $ME=3$ , $\sigma_1=\sigma_2=8$ , and $n_1=n_2=n$ . Plugging in the values: $3 = 1.96 \sqrt{\frac{8^2}{n} + \frac{8^2}{n}} = 1.96 \sqrt{\frac{64+64}{n}} = 1.96 \sqrt{\frac{128}{n}}$ . Solving for $n$ : $\frac{3}{1.96} = \sqrt{\frac{128}{n}} \implies (\frac{3}{1.96})^2 = \frac{128}{n} \implies 2.341 = \frac{128}{n} \implies n = \frac{128}{2.341} \approx 54.67$ . Since sample size must be an integer, we round up to 55. - A is incorrect because this is approximately the sample size needed for a one-sample CI with the same parameters: $3 = 1.96 \frac{8}{\sqrt{n}} \implies n \approx 27.3$ which rounds up to 28. - B is incorrect and likely results from a calculation error. - D is incorrect because this is approximately the total sample size ( $2n$ ), not the size for each group.
A researcher is analyzing data from two independent samples. They report that the standard error for the first sample mean ( $\bar{x}_1$ ) is $SE_1 = 3$ , and the standard error for the second sample mean ( $\bar{x}_2$ ) is $SE_2 = 4$ . What is the standard error for the difference between the two sample means, $\bar{x}_1 - \bar{x}_2$ ?
Answer: 5 The standard error of the difference of two independent means is calculated from the variances of the individual sample means. The formula is $SE_{\text{diff}} = \sqrt{(SE_1)^2 + (SE_2)^2}$ . Given $SE_1 = 3$ and $SE_2 = 4$ , we have $SE_{\text{diff}} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$ . This formula comes from the property that the variance of the difference of two independent random variables is the sum of their variances. - A is incorrect as it results from subtracting the standard errors (4 - 3 = 1). - C is incorrect as it results from adding the standard errors (3 + 4 = 7), a common mistake. Variances add, not standard deviations. - D is incorrect because the standard errors of the individual means ( $SE_1 = s_1/\sqrt{n_1}$ and $SE_2 = s_2/\sqrt{n_2}$ ) contain all the necessary information.
A researcher computes a 95% confidence interval for the difference in means between two small, independent groups ( $n_1=10, n_2=12$ ). After the calculation, it is discovered that one data point in the first group was entered incorrectly and is a severe outlier. If this outlier is removed, which of the following changes to the confidence interval is most likely to occur, assuming the sample mean of the first group also changes?
Answer: The interval will become narrower because the sample standard deviation of the first group will likely decrease substantially. A severe outlier dramatically inflates the standard deviation of a sample. Removing it will cause the sample standard deviation ( $s_1$ ) to decrease substantially. The width of the interval is proportional to the standard error, $\sqrt{s_1^2/n_1 + s_2^2/n_2}$ . A large decrease in $s_1$ will cause a large decrease in the standard error and thus a narrower interval. While decreasing $n_1$ from 10 to 9 will slightly increase the standard error and the t-critical value, the effect of removing the outlier's influence on $s_1$ will be much more pronounced. - A is incorrect because the effect of the reduced standard deviation will far outweigh the effect of the slightly smaller sample size. - B is incorrect because the width is highly dependent on the standard deviations, which will change. - D is incorrect because the confidence level is a choice made by the researcher; it does not change based on data values.
Two manufacturing processes, A and B, produce steel rods. A random sample of 50 rods from process A has a mean length of 105.2 cm with a standard deviation of 1.8 cm. A random sample of 40 rods from process B has a mean length of 104.5 cm with a standard deviation of 2.1 cm. Using the conservative approach for degrees of freedom, what is the margin of error for a 95% confidence interval for the difference in mean lengths ( $\mu_A - \mu_B$ )?
Answer: 0.85 cm The margin of error (ME) is $t^* \times SE$ , where $SE = \sqrt{\frac{s_A^2}{n_A} + \frac{s_B^2}{n_B}}$ . First, calculate the standard error: $SE = \sqrt{\frac{1.8^2}{50} + \frac{2.1^2}{40}} = \sqrt{0.0648 + 0.11025} = \sqrt{0.17505} \approx 0.4184$ cm. Next, find the conservative degrees of freedom: $df = \min(n_A-1, n_B-1) = \min(49, 39) = 39$ . The critical value $t^*$ for 95% confidence with df=39 is approximately 2.023. Finally, calculate the margin of error: $ME = 2.023 \times 0.4184 \approx 0.8464$ cm, which rounds to 0.85 cm. - A is incorrect because it uses the z-critical value of 1.96 instead of the t-critical value: $1.96 \times 0.4184 \approx 0.82$ . - C is incorrect because it results from incorrectly adding the standard errors of the means instead of their variances: $t^* (\frac{s_A}{\sqrt{n_A}} + \frac{s_B}{\sqrt{n_B}}) = 2.023 (\frac{1.8}{\sqrt{50}} + \frac{2.1}{\sqrt{40}}) \approx 2.023(0.255 + 0.332) \approx 1.19$ . - D is incorrect because it is only the standard error (0.42 cm), not the full margin of error.
A study compared the mean number of sick days taken annually by employees at two companies. Company A has a wellness program and Company B does not. A 95% confidence interval for the difference in means ( $\mu_A - \mu_B$ ) was found to be ( $-2.1, -0.3$ ). Which of the following statements is a valid conclusion from this result?
Answer: We are 95% confident that the mean number of sick days at Company A is between 0.3 and 2.1 days lower than at Company B. The interval for $\mu_A - \mu_B$ is ( $-2.1, -0.3$ ). Since all values in the interval are negative, this suggests that $\mu_A - \mu_B < 0$ , which means $\mu_A < \mu_B$ . The magnitude of the difference is between 0.3 and 2.1. Thus, we are 95% confident that the mean for A is between 0.3 and 2.1 days lower than the mean for B. - A is incorrect because it misinterprets a confidence interval for a difference as a confidence interval for a single mean ( $\mu_A$ ). - B is incorrect because this was likely an observational study. A confidence interval can show a statistically significant association, but it cannot establish causation without a properly designed experiment. - C is incorrect because the interval does not contain 0, which indicates there is a statistically significant difference at the $\alpha=0.05$ level. The negative values simply indicate the direction of the difference.
After constructing a 99% confidence interval for the difference of two population means, a student writes, "We can be 99% certain that the true difference between the population means is contained in our calculated interval of (5.2, 8.9)." Which statement best evaluates the student's interpretation?
Answer: This interpretation is incorrect because the confidence level describes the long-run performance of the method, not the probability associated with a single, specific interval. This is a classic misinterpretation of confidence intervals. The 99% confidence level does not assign a probability to a specific, already-calculated interval. Instead, it refers to the process: if we were to take many random samples and construct a confidence interval from each one, about 99% of those intervals would capture the true population parameter. Once an interval is calculated, the true parameter is either in it or not; the probability is 0 or 1. - A is incorrect because this is a common misconception. - B is incorrect because the confidence interval is an estimate for the difference in population means, not sample means. - D is incorrect because while 'confident' is standard terminology, the core issue is the probabilistic claim about a specific interval, not the choice between 'certain' and 'confident'.
An exercise physiologist wants to determine if a new type of running shoe improves 5k race times. They recruit 20 runners and have each runner complete a 5k race on two separate occasions: once with their standard shoes and once with the new shoes. The order of shoe use is randomized for each runner. The physiologist analyzes the data by constructing a 95% two-sample t-interval for the difference in mean race times. What is the primary flaw in this analytical approach?
Answer: The two sets of race times are not independent samples, so a two-sample t-interval is inappropriate. The primary flaw is that the data are paired, not independent. Each runner provides two data points (one for each shoe type), so the measurements are dependent. The correct procedure would be to calculate the difference in time for each runner and then construct a one-sample t-interval for the mean of these differences (a paired t-interval). - A is incorrect because while small samples can be a concern for normality, the more fundamental error is the violation of the independence assumption required for a two-sample procedure. - B is incorrect because the unpooled two-sample t-procedure does not require equal variances. Even if it were a concern, it's secondary to the violation of independence. - C is incorrect because a z-interval is only appropriate when the population standard deviations are known, which is not the case here.
A researcher calculates a 95% confidence interval for the difference in mean effectiveness of two allergy medicines, $\mu_A - \mu_B$ , and obtains ( $-0.2, 3.4$ ). A colleague suggests that 99% confidence should have been used instead. How would switching to a 99% confidence interval affect the interval and the resulting conclusion?
Answer: The 99% interval would be wider and would also contain 0, reinforcing the original conclusion of no significant difference. Increasing the confidence level from 95% to 99% requires a larger critical value ( $t^*$ or $z^*$ ), which increases the margin of error and makes the confidence interval wider. The center of the interval, $\bar{x}_A - \bar{x}_B$ , is unaffected by the confidence level. Since the original 95% interval ( $-0.2, 3.4$ ) already contains 0, the wider 99% interval will necessarily also contain 0. Therefore, the conclusion of 'no statistically significant difference' (at the corresponding alpha level) would be maintained or reinforced. - A and C are incorrect because higher confidence leads to a wider, not narrower, interval. The center does not change. - D is incorrect because while the interval is wider (less precise), the term 'more accurate' is ambiguous. The 99% confidence level means the method has a higher success rate in capturing the true mean, but any single interval is not necessarily more 'accurate'.

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