Question

Two random samples, A and B, were selected from the same population to estimate the population mean. For each sample, the mean, standard deviation, and margin of error for a 95 percent confidence interval for the population mean are shown in the table. Mean Standard Deviation Margin of Error Sample A 45 6.45 1.02Sample B 43 7.84 0.72Which of the following could explain why the margin of error of sample A is greater than the margin of error sample B? (A) The sample size of A is greater than the sample size of B. (B) The sample size of A is less than the sample size of B. (C) The sample size of A is equal to the sample size of B. (D) The mean of sample A is greater than the mean of sample B. (E) The standard deviation of sample A is less than the standard deviation of sample B.

Answer 1

Answer:

$n_A = \frac{6.45^2}{(\frac{1.02}{1.96})^2}=153.61 \approx 154$

$n_B = \frac{7.84^2}{(\frac{0.72}{1.96})^2}=455.49 \approx 456$

For this case as we can see we have a larger sample size for sample B, so then the best option for this case would be:

(B) The sample size of A is less than the sample size of B.

Step-by-step explanation:

For this case we have the following data given:

$\bar X_A= 45$ represent the sample mean for A

$s_A= 6.45$ represent the sample deviation for A

$ME_A = 1.02$ represent the margin of error for A

$\bar X_B= 43$ represent the sample mean for B

$s_B= 7.84$ represent the sample deviation for B

$ME_B= 0.72$ represent the margin of error for B

And for this case we are assuming that we have the same confidence level of 95%

For this case we an use the fact that the sample deviation is an unbiased estimator for the population deviation $\hat \sigma = \hat s$ and we can use the following formula for the margin of error of the sample mean the following formula:

$ME= z_{\alpha/2} \frac{\hat s}{\sqrt{n}}$

For this case the value of the significance is given by $\alpha =1-0.95 =0.05$ and the value for $\alpha/2 =0.025$ , so then the value for $z_{\alpha/2}$ represent a quantile of the normal standard distribution that accumulates 0.025 of the area on each tail of the normal standard distribution and for this case is $z_{\alpha/2}=\pm 1.96$.

So then since we have the value for z if we solve for n from the margin of error formula we got:

$n = \frac{\hat s^2}{(\frac{ME}{z})^2}$

And for the case A we can find the sample size and we got:

$n_A = \frac{6.45^2}{(\frac{1.02}{1.96})^2}=153.61 \approx 154$

And for the case B we can find the sample size and we got:

$n_B = \frac{7.84^2}{(\frac{0.72}{1.96})^2}=455.49 \approx 456$

For this case as we can see we have a larger sample size for sample B, so then the best option for this case would be:

(B) The sample size of A is less than the sample size of B.