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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 $H_{0}: p=0.5$ vs $H_{a}: p> 0.5$ using the sample results $\hat{p}=0.64$ with $n=50$

Round your answer for the test statistic to two decimal places, and your answer for the $p$-value to three decimal places.
\[
\begin{array}{l}
\text { test statistic }=\mathbf{i} \\
p \text {-value }=\mathbf{i}
\end{array}
\]
Conclusion:
$H_{0}$.
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Hint

Step 1 ：We are given the null hypothesis \(H_{0}: p=0.5\) and the alternative hypothesis \(H_{a}: p>0.5\). The sample results are \(\hat{p}=0.64\) with a sample size of \(n=50\). We are to use a \(5 \%\) significance level.

Step 2 ：The test statistic is calculated using the formula \((\hat{p} - p_{0}) / \sqrt{(p_{0} * (1 - p_{0})) / n}\), where \(\hat{p}\) is the sample proportion, \(p_{0}\) is the hypothesized population proportion, and \(n\) is the sample size.

Step 3 ：Substituting the given values into the formula, we get a test statistic of \(1.98\).

Step 4 ：The p-value is the probability of observing a test statistic as extreme as, or more extreme than, the observed test statistic, under the assumption that the null hypothesis is true. For a one-tailed test where the alternative hypothesis is that \(p\) is greater than \(p_{0}\), the p-value is the probability of observing a test statistic greater than the observed test statistic. This can be found using the cumulative distribution function of the standard normal distribution.

Step 5 ：Calculating the p-value, we get \(0.024\).

Step 6 ：The p-value is less than the significance level of \(0.05\), so we reject the null hypothesis. This means that there is sufficient evidence to support the alternative hypothesis that the population proportion \(p\) is greater than \(0.5\).

Step 7 ：The final answer is: The test statistic is \(\boxed{1.98}\) and the p-value is \(\boxed{0.024}\).