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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: $H_{0}$.
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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: \(\boxed{\text{We fail to reject the null hypothesis } H_0: p = 0.75}\)