Random samples of size 49 are taken from a population that has 200 elements, a mean of 180, and a variance of 196. The distribution of the population is unknown. The mean and the standard error of the distribution of sample means are _____.
A) 180 and 24.39
B) 180 and 28
C) 180 and 1.74
D) 180 and 2

Question

Quizplus · Accepted Answer

The mean of the distribution of sample means is equal to the population mean, which is 180. The standard error of the distribution of sample means is calculated using the formula $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$, where $\sigma$ is the population standard deviation and $n$ is the sample size. Given a population variance of 196, the standard deviation ($\sigma$) is the square root of 196, which is 14. Therefore, the standard error is $\sigma_{\bar{x}} = \frac{14}{\sqrt{49}} = \frac{14}{7} = 2$. However, because the population size is finite and the sample size is more than 5% of the population (49 out of 200), we should apply the finite population correction (FPC) formula: $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \sqrt{\frac{N - n}{N - 1}}$, where $N$ is the population size. Applying the FPC, we get $\sigma_{\bar{x}} = 2 \times \sqrt{\frac{200 - 49}{200 - 1}}$, which slightly adjusts the standard error. However, the options provided do not account for this correction, and the closest correct calculation without considering the FPC is 2. Therefore, the correct choice based on the options provided and without applying the FPC is 180 for the mean and 2 for the standard error, making option C the best answer given the options.

Random Samples of Size 49 Are Taken from a Population