QUESTION IMAGE
Question
trials in an experiment with a polygraph include 98 results that include 23 cases of wrong results and 75 cases of correct results. use a 0.01 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. e. ( h_{0}:p = 0.20 ) ( h_{1}:plt0.20 ) f. ( h_{0}:p = 0.20 ) ( h_{1}:p
eq0.20 ) the test statistic is ( z=-0.86 ). (round to two decimal places as needed.) the p - value is.195. (round to three decimal places as needed.) identify the conclusion about the null hypothesis and the final conclusion that addresses the original claim. ( h_{0} ). there sufficient evidence to support the claim that the polygraph results are correct less than 80% of the time.
Step1: State the hypotheses
The claim is that polygraph results are correct less than \(80\%\) of the time. So the null hypothesis \(H_{0}:p = 0.8\) (since \(1 - 0.2=0.8\) if we consider the proportion of correct results) and the alternative hypothesis \(H_{1}:p<0.8\) (left - tailed test as we are testing for less than).
Step2: Calculate the sample proportion
The sample proportion \(\hat{p}=\frac{75}{98}\approx0.765\)
Step3: Calculate the test statistic
The formula for the test statistic \(z=\frac{\hat{p}-p}{\sqrt{\frac{p(1 - p)}{n}}}\)
Substitute \(p = 0.8\), \(\hat{p}=0.765\), and \(n = 98\)
Step4: Find the P - value
For a left - tailed test with \(z=-0.86\), using the standard normal distribution table or a calculator, the P - value is \(P(Z < - 0.86)=0.195\)
Step5: Make a decision about the null hypothesis
Since the significance level \(\alpha = 0.01\) and \(P - value=0.195>\alpha\)
We fail to reject \(H_{0}\)
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We fail to reject \(H_{0}\). There is not sufficient evidence to support the claim that the polygraph results are correct less than \(80\%\) of the time.