In a random sample of 96 people, 9 of them are left-handed. Let \( p \) be the true proportion of lefthanded people in the population. Does the sample indicate that \( p \) is higher than 0.03 ? Use a 0.008 level of significance. A) What is the level of significance (comparison level)? B) State the Null and Alternate hypotheses. \( H_{0}: p=0.03: H_{1}: p>0.03 \) \( H_{0}: p>0.03: H_{1}: p=0.03 \) \( H_{0}: p=0.03 ; H_{1}: p<0.03 \) \( H_{0}: p=0.03 ; H_{1}: p \neq 0.03 \) C) Is this a (left-tail, right-tail, one-tail, two-tail) problem? left-tail right-tail one-tail two-tail D) Using 4 decimal places for \( \sigma_{\hat{P}} \), what's TRUE about \( \widehat{P} 96^{?} \) \( \widehat{P}_{96} \sim N(0.0174,0.03) \) \( \hat{P}_{96} \sim N(0.03,0.0174) \) \( \widehat{P}_{96} \sim N(0.0938,0.0174) \) \( \widehat{P}_{96} \sim N(0.03,0.0298) \) \( \widehat{P}_{96} \sim N(0.0938,0.0298) \) E) Using 4 decimal places for \( \hat{p} \), find the \( z \)-score of \( \hat{p} \) \( z=\square \). Round your answer to 2 decimal places. F) What is the p-value of this test? \( \square \) . Round your answer to 4 decimal places. G) Does the p-value suggest strong evidence to reject \( H_{0} \) ? No. Yes.

**A)** The level of significance is \[ \alpha = 0.008. \] **B)** Since we want to test if the true proportion is higher than 0.03, we have \[ H_{0} : p = 0.03 \quad \text{versus} \quad H_{1} : p > 0.03. \] Thus, the correct option is: \( H_{0}: p=0.03:\, H_{1}: p>0.03 \). **C)** Because the alternative hypothesis is \( p > 0.03 \), the rejection region will be in the right tail of the normal distribution. Hence, it is a **right-tail** test. **D)** Under \( H_{0} \) the sampling distribution of the sample proportion is approximately \[ \hat{p} \sim N\left( p, \sqrt{\frac{p(1-p)}{n}} \right). \] Here, with \( p=0.03 \) and \( n=96 \), we have \[ \sigma_{\hat{p}} = \sqrt{\frac{0.03(0.97)}{96}}. \] Calculating the variance first: \[ 0.03 \times 0.97 = 0.0291, \] \[ \frac{0.0291}{96} \approx 0.000303125, \] \[ \sigma_{\hat{p}} \approx \sqrt{0.000303125} \approx 0.0174. \] Thus, the correct description is: \[ \hat{p}_{96} \sim N(0.03, 0.0174). \] **E)** The sample proportion is \[ \hat{p} = \frac{9}{96} \approx 0.0938. \] The test statistic (z-score) is computed using \[ z=\frac{\hat{p}-p}{\sigma_{\hat{p}}}=\frac{0.0938-0.03}{0.0174}. \] Calculating the numerator: \[ 0.0938-0.03=0.0638. \] Then, \[ z \approx \frac{0.0638}{0.0174} \approx 3.67. \] Thus, the z-score is \[ z \approx 3.67. \] **F)** Since this is a right-tail test, the p-value is \[ P(Z>3.67). \] Looking this up in a standard normal table or using a calculator gives approximately \[ \text{p-value} \approx 0.0001. \] **G)** Because the p-value \( (0.0001) \) is much smaller than the significance level \( \alpha = 0.008 \), there is strong evidence to reject \( H_{0} \). Thus, the answer is: **Yes.**