Question

- The CEO of a large electric utility claims that more than 80% 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, 81% said that they were very satisfied. Do these results provide sufficient evidence to accept or reject the CEO's claimAssume significance level 0.05. State the null and alternative hypothesis, the sample proportion, test statistic. Using StatKey state the p-vale and if we reject the null or not.

Answer 1

Solution:

Here, we have to use one sample z test for the population proportion.

The null and alternative hypotheses for this test are given as below:

Null hypothesis: H0: Not more than 80% of his customers are very satisfied with the service they receive.

Alternative hypothesis: Ha: More than 80% of his customers are very satisfied with the service they receive.

H0: p ≤ 0.80 versus Ha: p > 0.80

This is an upper tailed test.

We are given

Level of significance = α = 0.05

Test statistic formula for this test is given as below:

Z = (p̂ - p)/sqrt(pq/n)

Where, p̂ = Sample proportion, p is population proportion, q = 1 - p, and n is sample size

n = sample size = 100

p̂ = x/n = 0.81

p = 0.80

q = 1 - p = 0.20

Z = (p̂ - p)/sqrt(pq/n)

Z = (0.81 - 0.80)/sqrt(0.80*0.20/100)

Z = 0.25

Test statistic = 0.25

P-value = 0.4013

(by using z-table)

P-value > α = 0.05

So, we do not reject the null hypothesis

There is not sufficient evidence to conclude that More than 80% of his customers are very satisfied with the service they receive.