Answer to Question #130252 in Statistics and Probability for Gemechu

Question

Telephone company claims that the average length of local calls is 4.5 minutes.randome sample of 50 calls is observed and it was mean call length is 3.9 minutes with standard deviation of 2.8 minutes.Is sample results significantly different than claims if alpha is 0.010

Answer 1

Given that the population mean $\mu$ = 4.5 minutes as claimed by the company.

On observing a sample set of N = 50

sample mean is found to be $\bar x$ = 3.9 and

standard deviation $\sigma$ = 2.8

Let us assume that the null hypothesis H₀ : $\mu$ = 4.5 (company's claim is correct)

Alternate hypothesis will be H_{a :} $\mu \not = 4.5$

For the above given data we can calculate the value of Z_cal = $\frac{\bar x - \mu}{\frac{\sigma}{\sqrt N}}$

Z_cal = $\frac {3.9 - 4.5}{\frac{2.8}{\sqrt50}} = -1.515$

At the given level of significance $\alpha$ = 0.01 we can find the value of Z₍$\frac {\alpha}{2}$ ₎ from the standard normal table.

Since this is a two tailed test we consider $\alpha$/2 = 0.01/2 = 0.005

Therefore Z₍$\frac {\alpha}{2}$ _{) =} Z_0.005 = 2.5758

Since, |Z_cal| = |-1.515| = 1.515 is found to be less than < Z₍$\frac {\alpha}{2}$ ₎ = 2.5758,

we do not have sufficient evidence to reject the company's claim.

Hence we can say that the sample results are not significantly different than the company's claim.