Answer:
Given Data:
Hypothesized Population Proportion $ (p) = 0.49 $
The Sample Proportion $ (\hat{p}) = 0.52 $
The sample size $(n) = 390 $
The Level of significance $ (\alpha) = 0.05 $
Solution:
The null and alternative hypothesis:
$ H_0: p = 0.49 $
$ H_1: p > 0.49 $
The test statistic $ (z) : $
$ Z = \frac{\hat{p} – p}{\sqrt{\frac{p(1-p)}{n}}} $
$ = \frac{0.52 – 0.49}{\sqrt{\frac{0.49(1-0.49)}{390}}} $
$ = 1.185 $
The p-value:
$ P(z > 1.185) = 0.1180 $
The conclusion:
The p-value is greater than the significance level. Therefore, we fail to reject the null hypothesis. There is not sufficient evidence to support the claim that the percentage is over the reported percentage of 49%.
Final Answer:
The null and alternative hypothesis:
$ H_0: p = 0.49 $
$ H_1: p > 0.49 $
The test statistic $ (z) = 1.185 $
The p-value $= 0.1180 $
The conclusion:
The p-value is greater than the significance level. Therefore, we fail to reject the null hypothesis. There is not sufficient evidence to support the claim that the percentage is over the reported percentage of 49%.
