Question

for the following situation, find the mean and standard deviation of the population. list all samples (with replacement) of the given size from that population and find the mean of each. find the mean and standard deviation of the sampling distribution and compare them with the mean and standard deviation of the population. the scores of three students in a study group on a test are 91, 90, 92. use a sample size of 3. the mean of the sampling distribution is 91.00 (round to two decimal places as needed.) the standard deviation of the sampling distribution is 0.47 (round to two decimal places as needed.) choose the correct comparison of the population and sampling distribution below. a. the means are equal but the standard deviation of the sampling distribution is larger. b. the means are not equal and the standard deviation of the sampling distribution is larger. c. the means are equal but the standard deviation of the sampling distribution is smaller. d. the means are not equal and the standard deviation of the sampling distribution is smaller. e. the means and standard deviations are equal.

Explanation:

Step1: Calculate population mean

The population data is \(91,90,92\). The population mean \(\mu=\frac{91 + 90+92}{3}=\frac{273}{3}=91\).

Step2: Calculate population standard - deviation

First, find the squared - differences from the mean: \((91 - 91)^2=0\), \((90 - 91)^2 = 1\), \((92 - 91)^2=1\). The population variance \(\sigma^{2}=\frac{0 + 1+1}{3}=\frac{2}{3}\approx0.67\), and the population standard deviation \(\sigma=\sqrt{\frac{2}{3}}\approx0.82\).

Step3: Recall properties of sampling distribution

The mean of the sampling distribution of the sample mean \(\mu_{\bar{x}}\) is equal to the population mean \(\mu\). Given \(\mu_{\bar{x}} = 91.00\) (from the problem) and \(\mu = 91\), they are equal. The standard deviation of the sampling distribution of the sample mean (also known as the standard error) \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\), where \(n = 3\). We know \(\sigma_{\bar{x}}=0.47\) and \(\sigma\approx0.82\), so \(\sigma_{\bar{x}}<\sigma\).

Answer:

C. The means are equal but the standard deviation of the sampling distribution is smaller.