View Policies Show Attempt History Current Attempt in Progress Use the normal distribution and the given sample results to complete the test of the given hypotheses. Assume the results come from a random sample and use a $5 \%$ significance level. Test $H_{0}: p=0.75$ vs $H_{a}: p \neq 0.75$ using the sample results $\hat{p}=0.70$ with $n=122$ Round your answer for the test statistic to two decimal places, and your answer for the $p$-value to three decimal places. test statistic $=$ $p$-value $=\mathbf{i}$ Conclusion: $H_{0}$. Save for Later Attempts: 1 of 4 used Submit Answer

Step 1 ：Calculate the test statistic using the formula: \(Z = \frac{{\hat{p} - p_0}}{{\sqrt{\frac{{p_0(1 - p_0)}}{n}}}}\)

Step 2 ：Substitute the given values into the formula: \(Z = \frac{{0.70 - 0.75}}{{\sqrt{\frac{{0.75(1 - 0.75)}}{122}}}}\)

Step 3 ：Simplify the equation to get: \(Z = \frac{{-0.05}}{{\sqrt{0.1875 / 122}}}\)

Step 4 ：Further simplify to get the test statistic: \(Z = -1.29\)

Step 5 ：Find the p-value corresponding to \(Z = -1.29\) in a standard normal distribution table or a calculator. The p-value is 0.197

Step 6 ：Compare the p-value to the significance level. If the p-value is less than the significance level, reject the null hypothesis. If the p-value is greater than the significance level, fail to reject the null hypothesis

Step 7 ：Since the p-value (0.197) is greater than the significance level (0.05), we fail to reject the null hypothesis

Step 8 ：Therefore, the conclusion is: \(\boxed{\text{We fail to reject the null hypothesis } H_0: p = 0.75}\)