Question

1. a doctor would like to estimate the average number of hours per week patients over 50 spend exercising. which of the following sample sizes would yield the smallest margin of error for estimating the true mean number of hours per week patients over 50 spend exercising? a) 10 b) 20 c) 50 d) 100

Explanation:

Step1: Recall margin - of - error formula

The formula for the margin of error $E$ for a confidence interval for the population mean (when the population standard deviation $\sigma$ is known) is $E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$, where $z_{\alpha/2}$ is the z - score corresponding to the level of confidence, $\sigma$ is the population standard deviation, and $n$ is the sample size.

Step2: Analyze the relationship between $n$ and $E$

From the formula $E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$, we can see that the margin of error $E$ is inversely proportional to the square - root of the sample size $n$. That is, as $n$ increases, $E$ decreases.

Step3: Compare given sample sizes

We are given sample sizes $n_1 = 10$, $n_2=20$, $n_3 = 50$, and $n_4=100$. Since $100>50 > 20>10$, the largest sample size will give the smallest margin of error.

Answer:

D. 100