Trials in an experiment with a polygraph include 97 results that include 22 cases of wrong results and 75 cases of correct results. Use a 0.05 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 p be the population proportion of correct polygraph results. Identify the null and alternative hypotheses. Choose the correct answer below.О В. Но: р3D0.20О А. Но: р3D0.80H;: p+0.80H;: p#0.20О с. Но: р3D0.20H;: p> 0.20O D. Ho: p= 0.80H;: p>0.80O F. Ho: p= 0.20Е. Но: р30.80H;: p<0.80H;: p<0.20The test statistic is z=|. (Round to two decimal places as needed.)The P-value is(Round to four decimal places as needed.)Identify the conclusion about the null hypothesis and the final conclusion that addresses the original claim.Ho. Theresufficient evidence to support the claim that the polygraph results are correct less than 80% of the time.Fail to rejectReject - Theresufficient evidence to support the claim that the polygraph results are correct less than 80% of the time.is notis

Given,n=97x=75p^=xnp^=7597=0.7732α=0.05

Answered: Trials in an experiment with a…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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To evaluate if auditors might be helped in determining the chances of fraud if they carefully measure cash flow, samples of midlevel auditors from CPA firms were asked to indicate the chance of material fraud on a scale from 0 to 100 for a case. A random sample of 36 auditors used the cash-flow information. Their mean assessment was 35.51, and the sample standard deviation was 24.14. For an independent random sample of 36 auditors not using the cash-flow information, the sample mean and standard deviation were, respectively, 40.58 and 29.18. Assuming that the two population distributions are normal with equal variances, test against a two-sided alternative the null hypothesis that the population means are equal. (Use α = 0.05.) Click the icon to view a table of critical values for the Student's t-distribution. Let μx be the mean assessment for auditors who measured cash flow and μy be the mean assessment for auditors not using the cash-flow information. Determine the null and…

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