Trials in an experiment with a polygraph induce 99 results that include 22 cases

Question

Trials in an experiment with a polygraph induce 99 results that include 22 cases of wrong recurs and 77 cases of comment results use a 0.01signfcance lessthanorequalto to test the claim that such polygraph results are commend less than of the density is, iterative hypotenuses test statistic, P-value, conclusion about the null hypothesis and final confusion that addresses or gland am use the P value method use the normal of the binomial distribution Let p be the population proportion of correct polygraph results identity the null and hypotheses Choose the concert answer below H_0 = = 0.80 H_1 p

Explanation / Answer

Since the population param is .80 we are going to test this.
The claim that such that polygraph are corrected less than 80% of the time.
So, since the claim goes as the alternative hypothesis , therefore,

D is the right option p < .80.

The test statistic is z = (pcap-p0)/(sqrt(p0*p0'/n)

pcap = 77/99 = .78

z = (.78-.80)/(sqrt(.80*.20/99))= -.50

z = -.50

The p value is 1-.3085 = .5915

Conclusion:
This value of p is greater than .01 and therefore we can't reject null hypothesis that p = 0.80

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