Trials in an experiment with a polygraph include 96 results that include 22 cases of wrong results and 74 cases of correct resul
ts. Use a 0.01 significance level to test the claim that such polygraph results are correct less than 80% of the time Based on the results should polygraph test results be prohibited as evidence in trials? Identify the null hypothesis, alternative hypothesis, test statistics, p value, conclusion about null hypotheses and final conclusion that address the original claim. Use the p value methos. Use the normal distribution as an approximation of the binomial distribution. A. I dentify the Null and alternative hypothesis B. The test staistic is z= (round to two decimals) C. The P value is=( rounds to four decimals) D. Identify the conclusion about the null hypotheses and the final conclusion that address the original claim. Choose (fail to reject, reject) h0. There choose(is, is not) sufficient evidence to support the claim that the polygraph result are correct less than 80% of time. answer H0: P ≥ 0.80 Ha: P < 0.80 Estimated p = 74 / 98 = 0.7551 Variance of proportion = p*(1-p)/n = 0.8(0.2)/98 =0.0016327 S.D. of p is sqrt[0.001633] = 0.0404 z = ( 0.7551 - 0.8 ) / 0.0404 = -1.1112 P-value = P( z < -1.1112) = 0.1335 Since the p-value is greater than 0.05, we do not reject the null hypothesis. Based on the results there is no evidence that polygraph test results should be prohibited as evidence in trials.
The rule of exponential multiplication is (a^x)*(a^y)=a^(x+y) using this with your expression, we can do (10^1)*(10^3)*(10^2)=10^(1+3+2)=10^6 Final answer: 10^6 or 1 million Hope I helped :)
