Question

A Cell Phone company sells cellular phones and airtime in a State. At a recent meeting, the marketing manager states that the average age of the customers is 40 years. Before actually completing the advertising plan, it was decided to select a random sample of customers. Among the questions asked in the survey of 50 customers was the customer’s ages. The mean and the standard deviation of the data based on the survey are 38 years and 7 years. a. Formulate a hypothesis to test the marketing manager’s claim. b. Does the sample support manager’s claim. Test at 0.05 level of significance.

Answer 1

Answer:

The null hypothesis is rejected.

There is enough evidence to support the claim that the average age of the customers differs from 40 years.

The sample does not support the manager claim (the average age seems to differ from 40 years).

Step-by-step explanation:

This is a hypothesis test for the population mean.

The manager claims that the average age of customers is 40 years. As this is an equality, we will test if the average age differs from 40. If the null hypothesis failed to be rejected, the claim of the manager is right.

Then, the claim is that the average age of the customers differs from 40 years.

Then, the null and alternative hypothesis are:

$H_0: \mu=40\\\\H_a:\mu\neq 40$

The significance level is 0.05.

The sample has a size n=50.

The sample mean is M=38.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=7.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{7}{\sqrt{50}}=0.99$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{38-40}{0.99}=\dfrac{-2}{0.99}=-2.02$

The degrees of freedom for this sample size are:

$df=n-1=50-1=49$

This test is a two-tailed test, with 49 degrees of freedom and t=-2.02, so the P-value for this test is calculated as (using a t-table):

$P-value=2\cdot P(t$

As the P-value (0.049) is smaller than the significance level (0.05), the effect is  significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the average age of the customers differs from 40 years.