Question

The manager of a local nightclub has recently surveyed a random sample of 280 customers of the club. She would now like to determine whether or not the mean age of her customers is over 35. If so, she plans to alter the entertainment to appeal to an older crowd. If not, no entertainment changes will be made. Suppose she found that the sample mean was 35.6 years and the population standard deviation was 5 years. Use a=0.05

Report the value of the test statistics_______ and the critical value or p-value____
We (select one: reject, fail to reject)_______ the null hypothesis. That means that we (select one: found, did not find)________ evidence to support the alternative.

Answer 1

`z = (35.6-35)/((5)/(√(280)))= 2.007`

`p_v = P(z>2.007) = 0.0224`

Since the p value is lower than the significance level given of 0.05 we have enough evidence to reject the null hypothesis on this case. And the best conclusion for this case is:

We (reject) the null hypothesis. That means that we (found) evidence to support the alternative.

Explanation:

We have the following info given:

`\bar X = 35.6` represent the sampel mean for the age of customers

`\sigma = 5` represent the population standard deviation

`n = 280` represent the sample size selected

We want to test if the mean age of her customers is over 35 so then the system of hypothesis for this case are:

Null hypothesis:
`\mu \leq 35`

Alternative hypothesis
`\mu >35`

The statistic for this case is given by:

`z = (\bar X -\mu)/((\sigma)/(√(n)))`

And replacing the data given we got:

`z = (35.6-35)/((5)/(√(280)))= 2.007`

We can calculate the p value since we are conducting a right tailed test like this:

`p_v = P(z>2.007) = 0.0224`

Since the p value is lower than the significance level given of 0.05 we have enough evidence to reject the null hypothesis on this case. And the best conclusion for this case is:

We (reject) the null hypothesis. That means that we (found) evidence to support the alternative.