Question

type the correct answer in the box. use numerals instead of words. if necessary, use / for the fraction bar. the difference of two sample means is 22, and the standard deviation of the difference of the sample means is 10. the difference of the means of the two populations at a 95% confidence interval is ± .

Explanation:

Step1: Recall 95% confidence interval multiplier

For a 95% confidence interval, the critical value (multiplier) is approximately 1.96 (from the standard normal distribution, \( z_{\alpha/2} \) for 95% confidence is about 1.96).

Step2: Calculate the margin of error

The margin of error (ME) for the difference of means is calculated as \( ME = z_{\alpha/2} \times \text{standard deviation of difference of sample means} \). Here, \( z_{\alpha/2} \approx 1.96 \) and the standard deviation of the difference of sample means is 10. So, \( ME = 1.96 \times 10 = 19.6 \). But wait, maybe we are using a simpler approximation? Wait, sometimes for large samples, people use 2 as an approximation for 1.96. Wait, but let's check. Wait, the problem says "the difference of the means of the two populations at a 95% confidence interval is ± [something]". Wait, actually, the confidence interval for the difference of population means is \( (\bar{x}_1 - \bar{x}_2) \pm z_{\alpha/2} \times SE \), where \( SE \) is the standard error (standard deviation of the difference of sample means). Here, the difference of sample means is 22, but the question is about the margin of error? Wait, no, the question says "the difference of the means of the two populations at a 95% confidence interval is ± [ ]". Wait, maybe it's the margin of error. Wait, let's re-read. "The difference of the means of the two populations at a 95% confidence interval is ± [ ]". Wait, the formula for the confidence interval for the difference of two population means (when using z - test) is \( (\bar{x}_1 - \bar{x}_2) \pm z_{\alpha/2} \times \sigma_{\bar{x}_1 - \bar{x}_2} \). Here, \( \bar{x}_1 - \bar{x}_2 = 22 \), \( \sigma_{\bar{x}_1 - \bar{x}_2} = 10 \), and for 95% confidence, \( z_{\alpha/2} \approx 1.96 \), but sometimes in basic stats, they use 2 as an approximation. Wait, but 1.96*10 = 19.6, which is approximately 20? Wait, no, maybe the problem expects using 2 as the multiplier. Wait, let's check: 1.96 is approximately 2, so 2*10 = 20. But wait, maybe the problem is using the empirical rule? Wait, no, 95% confidence interval for normal distribution is about 1.96 standard deviations. But maybe in the context of the problem, they want the margin of error, which is \( z_{\alpha/2} \times SE \). So if we use 1.96, it's 19.6, but maybe rounded to 20? Wait, but let's see. Wait, the question is "the difference of the means of the two populations at a 95% confidence interval is ± [ ]". Wait, maybe it's the margin of error. So let's calculate: \( 1.96 \times 10 = 19.6 \), which is approximately 20? Wait, no, 1.96*10 is 19.6. But maybe the problem is using a t - test? No, for large samples, z - test is used. Wait, maybe the problem expects the use of 2 as the multiplier. Let's check: 2*10 = 20. Alternatively, maybe the question has a typo, but given that it's a problem for maybe a basic stats course, they might use 2 as an approximation for 1.96. So the margin of error is 1.96*10 = 19.6, which is approximately 20, but maybe the exact value with 1.96 is 19.6. Wait, but let's do the calculation properly. \( z_{\alpha/2} \) for 95% confidence is 1.96. So \( 1.96\times10 = 19.6 \). But maybe the problem wants 20? Wait, no, 1.96 is more accurate. Wait, but let's see the problem statement: "Use numerals instead of words. If necessary, use / for the fraction bar." So 19.6 or 20? Wait, maybe the problem is using the approximation of 2 for 1.96. Let's check: 1.96 is approximately 2, so 2*10 = 20. So the answer is 19.6 (or 20). Wait, but let's confirm. The formula for the margin of error (E) in a 95% confidence interval for the difference of two means (when the standard deviation of the difference of sample means is known) is \( E = z_{\alpha/2} \times \sigma_{\bar{x}_1 - \bar{x}_2} \). For 95% confidence, \( z_{\alpha/2} \approx 1.96 \), so \( E = 1.96\times10 = 19.6 \). So the difference of the means of the two populations at a 95% confidence interval is \( \pm 19.6 \) (or approximately \( \pm 20 \) if using the approximation). But maybe the problem expects 19.6 or 20? Wait, let's see, maybe in the problem's context, they use 2 as the multiplier. So 2*10 = 20. So I think the answer is 19.6 (or 20, but more accurately 19.6). Wait, but let's check with 1.96: 1.96*10 = 19.6. So the margin of error is 19.6, so the confidence interval is \( 22 \pm 19.6 \), but the question is asking for the ± part, which is the margin of error. So the answer is 19.6 (or 20 if approximated). But maybe the problem uses 2, so 20. Wait, maybe the problem is from a basic stats course where they use 2 for 95% confidence. So I think the answer is 19.6 (or 20). But let's go with 19.6 as the more accurate value, or 20 as the approximate. Wait, let's calculate 1.96*10: 19.6. So the answer is 19.6.

Answer:

19.6