Question

suppose 7% of student veterans at a college are involved in sports. a random sample of 134 student veterans is taken. what is the mean of the sampling distribution for the proportion of veterans in sports at this college? when working with samples of size 134, what is the standard error of the sampling distribution for the proportion of veterans in sports at this college? round answer to 3 decimal places. is it unusual that no more than 20 veterans in the sample are involved in sports? (hint: 20 of 134 is about 0.149) round answer to 4 decimal places. is this result unusual? yes, there is a less than 5% chance of this happening by random variation. no, there is at least a 5% chance of this happening by random variation.

Explanation:

Step1: Find the mean of the sampling distribution

The mean of the sampling distribution of the sample - proportion $\mu_{\hat{p}}$ is equal to the population proportion $p$. Given that $p = 0.07$. So, $\mu_{\hat{p}}=p = 0.07$.

Step2: Calculate the standard error of the sampling distribution

The formula for the standard error of the sampling distribution of the sample - proportion is $\sigma_{\hat{p}}=\sqrt{\frac{p(1 - p)}{n}}$, where $n = 134$ and $p = 0.07$. First, calculate $1-p=1 - 0.07 = 0.93$. Then, $\sigma_{\hat{p}}=\sqrt{\frac{0.07\times0.93}{134}}=\sqrt{\frac{0.0651}{134}}\approx\sqrt{0.000485821}\approx0.022$.

Step3: Calculate the z - score for the sample proportion

The sample proportion $\hat{p}=\frac{20}{134}\approx0.149$. The z - score is calculated using the formula $z=\frac{\hat{p}-\mu_{\hat{p}}}{\sigma_{\hat{p}}}$. Substitute $\hat{p}=0.149$, $\mu_{\hat{p}} = 0.07$, and $\sigma_{\hat{p}}\approx0.022$ into the formula: $z=\frac{0.149 - 0.07}{0.022}=\frac{0.079}{0.022}\approx3.591$.

Step4: Find the probability associated with the z - score

Using a standard normal distribution table or calculator, the probability $P(Z\leq3.591)\approx0.9998$. The probability of having no more than 20 veterans involved in sports is approximately $0.9998$.

Answer:

Mean of sampling distribution: 0.07
Standard error: 0.022
Probability: 0.9998
Is this result unusual? No, there is at least a 5% chance of this happening by random variation.