Trials in an experiment with a polygraph include 97 results that include 24 cases of wrong results and 73 cases of correct results. Use a 005 significance level to test the claim that such polygraph results are correct less than $80 \%$ of the time. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. Use the P-value method. Use the normal distribution as an approximation of the binomial distribution. Let $\mathrm{p}$ be the population proportion of correct polygraph results. Identify the null and alternative hypotheses Choose the correct answer below. A. \[ \begin{array}{l} H_{0}-p=0.20 \\ H_{1} . p \neq 0.20 \end{array} \] C. \[ \begin{array}{l} H_{0}: p=0.80 \\ H_{1}: p<0.80 \end{array} \] E \[ \begin{array}{l} H_{0}: p=0.80 \\ H_{1}: p \neq 0.80 \end{array} \] B. \[ \begin{array}{l} H_{0} p=0.80 \\ H_{1} p>0.80 \end{array} \] D. \[ \begin{array}{l} H_{0} p=0.20 \\ H_{1} p>0.20 \end{array} \] Fit. \[ \begin{array}{l} H_{0}, p=0.20 \\ H_{1} . p<020 \end{array} \] The test statistic is $z=\square$ (Rourh to two decimal places as needed)
Solution
Step 1 ：The null hypothesis (H0) is that the proportion of correct polygraph results is 80% or more, and the alternative hypothesis (H1) is that the proportion of correct polygraph results is less than 80%. So, the hypotheses are: \[H_{0}: p=0.80\] \[H_{1}: p<0.80\]
Step 2 ：Next, we calculate the test statistic, which is a z-score (z). The formula for the z-score is: \[z = \frac{{\hat{p} - p_{0}}}{{\sqrt{\frac{{p_{0} * (1 - p_{0})}}{n}}}}\] where \(\hat{p}\) is the sample proportion, \(p_{0}\) is the proportion in the null hypothesis, and n is the sample size.
Step 3 ：In this case, \(\hat{p} = \frac{73}{97} = 0.7526\), \(p_{0} = 0.80\), and n = 97.
Step 4 ：Substituting these values into the z-score formula, we get: \[z = \frac{0.7526 - 0.80}{\sqrt{\frac{0.80 * (1 - 0.80)}{97}}} = \frac{-0.0474}{\sqrt{0.0041}} = \frac{-0.0474}{0.0640} = -0.74\] (rounded to two decimal places).
Step 5 ：So, the test statistic is \(\boxed{z = -0.74}\).
