Suppose that in a random selection of 100 colored candies, $29 \%$ of them are blue. The candy company claims that the percentage of blue candies is equal to $26 \%$. Use a 0.01 significance level to test that claim. A. \[ \begin{array}{l} H_{0}: p=0.26 \\ H_{1}: p \neq 0.26 \end{array} \] B. \[ \begin{array}{l} H_{0}: p=0.26 \\ H_{1}: p>0.26 \end{array} \] c. \[ \begin{array}{l} H_{0}: p=0.26 \\ H_{1}: p<0.26 \end{array} \] D. \[ \begin{array}{l} H_{0}: p \neq 0.26 \\ H_{1}: p=0.26 \end{array} \] Identify the test statistic for this hypothesis test. The test statistic for this hypothesis test is $\square$. (Round to two decimal places as needed.) Clear all Check answer an example Get more help - Search

Step 1 ：Identify the null hypothesis \(H_{0}\) and the alternative hypothesis \(H_{1}\). The null hypothesis is usually a statement of no effect or status quo, in this case, it would be that the proportion of blue candies is equal to 26%. The alternative hypothesis is what we are testing against the null hypothesis. Since the question does not specify whether we are testing if the proportion is greater or less than 26%, we should assume it's a two-tailed test, meaning the proportion is not equal to 26%. So, the correct hypotheses are: \[H_{0}: p=0.26\] \[H_{1}: p \neq 0.26\]

Step 2 ：Calculate the test statistic using the formula for the test statistic of a proportion, which is \((p_{hat} - p_{0}) / \sqrt{(p_{0}*(1-p_{0}))/n}\), where \(p_{hat}\) is the sample proportion, \(p_{0}\) is the proportion under the null hypothesis, and \(n\) is the sample size. Given that \(p_{hat} = 0.29\), \(p_{0} = 0.26\), and \(n = 100\), the test statistic is calculated as follows: \(test\_statistic = (0.29 - 0.26) / \sqrt{(0.26*(1-0.26))/100} = 0.68\)

Step 3 ：Final Answer: The correct hypotheses are \[H_{0}: p=0.26\] \[H_{1}: p \neq 0.26\] and the test statistic for this hypothesis test is \(\boxed{0.68}\)