Mathematics, 24.01.2020 21:31 vhife4901

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.

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Answer from: breella0

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

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

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.

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Answer from: Quest

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Two random samples, a and b, were selected from the same population to estimate the population mean....

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