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

Question

Trials in an experiment with a polygraph include 99 results that include 22 cases of wrong results and 77 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. A. H_0: p = 0.80 H_1: p > 0.80 B. H_0: p = 0.20 H_1: p notequalto 0.20 C. H_0: p = 0.80 H_1: p 0.20 E. H_0: p = 0.80 H_1: p notequalto 0.80 F. H_0: p = 0.20 H_1: p

Explanation / Answer

Given that,
possibile chances (x)=77
sample size(n)=99
success rate ( p )= x/n = 0.7778
success probability,( po )=0.8
failure probability,( qo) = 0.2
null, Ho:p=0.8
alternate, H1: p<0.8
level of significance, = 0.05
from standard normal table,left tailed z /2 =1.64
since our test is left-tailed
reject Ho, if zo < -1.64
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.77778-0.8/(sqrt(0.16)/99)
zo =-0.5528
| zo | =0.5528
critical value
the value of |z | at los 0.05% is 1.64
we got |zo| =0.553 & | z | =1.64
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value: left tail - Ha : ( p < -0.55277 ) = 0.29021
hence value of p0.05 < 0.29021,here we do not reject Ho
ANSWERS
---------------
null, Ho:p=0.8
alternate, H1: p<0.8
test statistic: -0.5528
critical value: -1.64
decision: do not reject Ho
p-value: 0.29021

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