A publisher reports that 46% of their readers own a personal computer. A marketing executive wants to test the claim that the percentage is actually over the reported percentage. A random sample of 380 found that 49% of the readers owned a personal computer. Is there sufficient evidence at the 0.05 level to support the executive’s claim?

Answer:

Given Data:
Hypothesized Population Proportion $ (p) = 0.46 $
The Sample Proportion $ (\hat{p}) = 0.49 $

The Level of significance $ (\alpha) = 0.05 $

Solution:
The null and alternative hypothesis:

$ H_0: p = 0.46 $
$ H_1: p > 0.46 $

The test statistic $ (z) : $

$ Z = \frac{\hat{p} – p}{\sqrt{\frac{p(1-p)}{n}}} $
$ = \frac{0.49 – 0.46}{\sqrt{\frac{0.46(1-0.46)}{380}}} $

$ = 1.173 $

The p-value:
$ P(z > 1.173) = 0.1203 $

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 46%.

Final Answer:
The null and alternative hypothesis:
$ H_0: p = 0.46 $
$ H_1: p > 0.46 $

The test statistic $ (z) = 1.173 $
The p-value $= 0.1203 $
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 46%.