Question

a baseball players career batting average before this season was.280. (note that even though this is called a batting average, its actually a proportion of at - bats in which he gets a hit.) this season, he started a new training regimen, and wants to know if the new regimen has improved his batting. out of 500 at - bats this season, his batting average was.310. the player wants to test $h_0:p =.280$, $h_a:p >.280$. a. in the sample of 500 at - bats this season, what was $hat{p}$?.310 b. if we took a new random sample of 500 at - bats, what values of $hat{p}$ would be equally good, or better, evidence against $h_0$? a. $hat{p}geq.280$ b. $hat{p}geq.310$ or $hat{p}leq.250$ c. $hat{p}geq.310$ d. $hat{p}leq.310$ c. if we took many samples of 500 at - bats, what would we expect the mean of $hat{p}$ to be? assume the null hypothesis is true. 155 d. if we took many samples of 500 at - bats, what would we expect the standard deviation of $hat{p}$ to be? assume the null hypothesis is true. 0.02068

Explanation:

Step1: Recall sample - proportion concept

The sample proportion $\hat{p}$ is the proportion of successes in the sample. In part (a), it is given that out of 500 at - bats, the batting average (proportion of hits) is 0.310, so $\hat{p}=0.310$.

Step2: Analyze evidence against $H_0$

The null hypothesis is $H_0:p = 0.280$ and the alternative hypothesis is $H_a:p>0.280$. Values of $\hat{p}$ that are at least as extreme as the observed $\hat{p}$ value in the direction of the alternative hypothesis are evidence against $H_0$. Since the observed $\hat{p}=0.310$ and the alternative is $p > 0.280$, values of $\hat{p}\geq0.310$ are equally good or better evidence against $H_0$.

Step3: Find the mean of $\hat{p}$ under $H_0$

If the null hypothesis $H_0:p = p_0$ is true, the mean of the sampling distribution of $\hat{p}$ is $\mu_{\hat{p}}=p_0$. Here, $p_0 = 0.280$, so $\mu_{\hat{p}}=0.280$.

Step4: Calculate the standard deviation of $\hat{p}$ under $H_0$

The formula for the standard deviation of the sampling distribution of $\hat{p}$ is $\sigma_{\hat{p}}=\sqrt{\frac{p_0(1 - p_0)}{n}}$, where $p_0$ is the proportion under the null hypothesis and $n$ is the sample size. Given $p_0 = 0.280$, $1-p_0=1 - 0.280 = 0.720$, and $n = 500$. Then $\sigma_{\hat{p}}=\sqrt{\frac{0.280\times0.720}{500}}=\sqrt{\frac{0.2016}{500}}=\sqrt{0.0004032}\approx0.0201$.

Answer:

a. 0.310
b. C. $\hat{p}\geq0.310$
c. 0.280
d. Approximately 0.0201