# According to a recent poll 53

According to a recent poll 53 percent of Americans would votefor the incumbent president. If a random sample of 100 peopleresults in 45 percent who would vote for the incumbent, test theclaim that the actual percentage is 53 percent. Use a 0.03significance level. show all 5 steps.

Given that according to a recent poll 53 percent, p=0.53 ofAmericans would vote for the incumbent president. If a randomsample of n = 100 people results in 45 percent = 0.45 who would vote for the incumbent.

So, to find the correct test we need to check the requirementswhich are:

1) If n*p(1-p) >10 and

2) If the sample is randomly selected then the distribution isnormal.

so, as n*p(1-p) >10 = 100*0.53(1-0.53) = 24.91 which isgreater than 10 and the sample is randomly selected hence thedistribution is normal thus the Z test is applicable for hypothesistesting.

To test the claim we need to conduct the test hypothesis whichis:

Based on the hypothesis it is a two-tailed test.

Rejection region:

Based on the given significance level the Zc or the Z-criticalscore is calculated using the excel formula for normal distributionwhich is =NORM.S.INV(0.985) thus the Z score computed isZc​=+/-2.17.

Thus, Reject Ho if Z >2.17 or Z<-2.17

Test Statistic:

P-value:

The P-value is computed using the excel formula for normaldistribution which is =2*(NORM.S.DIST(-1.603, TRUE)) The P-valuecomputed as 0.109.

Conclusion:

Since the Z score computed is greater than -Zc and P-value isgreater than 0.03 level of significance hence we cannot reject thenull hypothesis thus we conclude that there is insufficientevidence to warrant the claim.

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