Answer to Question #181807 in Statistics and Probability for Nicole Cubillas

Question

A telecommunications company claims that households receive an average of 45 telephone calls per month.To test the claim at a= 0.10,a researcher surveyed 35 households and found out that the average number of calls was 40.8.Is there enough evidence to say that the company's claim is incorrect if the population standard deviation is known to be 3.2 calls?

Answer 1

$\mu = 45 \\ \sigma = 3.2 \\ n = 35 \\ \bar{x} = 40.8 \\ α = 0.10 \\ H_0 : \mu = 45 \\ H_1 : \mu ≠ 45$

Test concerning averages

$z = \frac{\bar{x}- \mu}{\sigma / \sqrt{n}} \\ z = \frac{40.8-45}{3.2/ \sqrt{35}} = -0.776$

Critical value at 0.10 significance level z = ±1.645

Critical regions: Two-tailed test. Reject H₀ if z ≤ -1.645 or z≥1.645

Conclusion: Since z = -0.776 is greater than -1.645, the null hypothesis can not be rejected. There is enough evidence to say that the company's claim is correct.