Question

A city manager wants to know whether less than 25 secretaries are being used per day. If in a random sample of 60 days, the mean number of temporary secretaries employed is 23.5, with a standard deviation s=6.1, is there at the .05 level of significance sufficient evidence to reject the null hypothesis that M(mule)=25?

1. State Null Hypothesis and Alternative Hypothesis

2. State level significance.

3. State criterion and test statistic

4.Calculate.

5. Decision.

Answer 1

Let denotes the true mean number of temporary secretaries employed.

To test against

Here

sample mean

sample standard deviation

and sample size

The test statistic can be written as

which under H₀ follows a t distribution with n-1 df.

We reject H₀ at 5% level of significance if P-value < 0.05

Now,

The value of the test statistic

P-value =

Since P-value < 0.05, so we reject H₀ at 5% level of significance and we can conclude that the true mean number of temporary secretaries employed is significantly less than 25 secretaries.