Question

The CEO of a large electric utility claims that more than 75 percent of his customers are very satisfied with the service they receive. To test this claim, the local newspaper surveyed 100 customers, using simple random sampling. Among the sampled customers, 77% say they are very satisfied. We want to decide if the 77% is enough evidence to lead us to accept or reject the CEO’s claim?

Answer 1

Answer:

$z=\frac{0.77-0.75}{\sqrt{\frac{0.75(1-0.75)}{100}}}=0.462$  

Now we can find the p value using the alternative hypothesis with this probability:

$p_v =P(z>0.462)=0.322$  

Since the p value is large enough, we have evidence to conclude that the true proportion for this case is NOT significanctly higher than 0.75 since we FAIL to reject the null hypothesis at any significance level lower than 30%

Step-by-step explanation:

Information provided

n=100 represent the random sample selected

$\hat p=0.77$ estimated proportion of students that are satisfied

is the value that we want to test

z would represent the statistic

$p_v$ represent the p value

System of hypothesis

We want to verify if more than 75 percent of his customers are very satisfied with the service they receive, then the system of hypothesis is.:  

Null hypothesis:$p\leq 0.75$  

Alternative hypothesis:$p > 0.75$  

The statistic is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)  

Replacing the info given we got:

$z=\frac{0.77-0.75}{\sqrt{\frac{0.75(1-0.75)}{100}}}=0.462$  

Now we can find the p value using the alternative hypothesis with this probability:

$p_v =P(z>0.462)=0.322$  

Since the p value is large enough we have evidence to conclude that the true proportion for this case is NOT significanctly higher than 0.75 since we FAIL to reject the null hypothesis at any significance level lower than 30%