Question

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

Answer 1

Given that according to a recent poll 53 percent, p=0.53 of Americans would vote for the incumbent president. If a random sample 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 requirements which are:

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

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

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

To test the claim we need to conduct the test hypothesis which is:

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

Rejection region:

Based on the given significance level the Zc or the Z-critical score is calculated using the excel formula for normal distribution which is =NORM.S.INV(0.985) thus the Z score computed is Zc​=+/-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 normal distribution which is =2*(NORM.S.DIST(-1.603, TRUE)) The P-value computed as 0.109.

Conclusion:

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