Question

In the past, a company found that the mean number of overtime hours for its employees was 10.5 hours per month. However, a recent sample of 40 employees showed that the mean number of overtime hours in a month was 12.3, with s = 4.4 hours. Test the claim, at a significance level of 0.01, that the mean number of overtime hours per month exceeds 10.5 for all employees of this company:

a) Write the null and alternative hypotheses:

b) Find the value of the test statistic (rounded properly):

c) Find the critical value, and sketch the rejection region

d) We will reject / fail to reject H0.

e) Make a statement with your conclusion in the context of the problem

Answer 1

a) Null hypothesis:
`\mu \leq 10.5`

Alternative hypothesis:
`\mu > =10.5`

b)
`t=(12.3-10.5)/((4.4)/(√(40)))=2.587`

c) We need to find the degrees of freedom

`df = n-1= 40-1 = 39`

And we need to find in the t distribution with 39 degrees of freedom a critical value who accumulates 0.01 of the area in the right and we got

`t_(cric)= 2.426`

d) The calculated value is higher than the critical value so then we can reject the null hypothesis

e) For this case we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly higher than 10.5 hours per month at the significance level of 0.01

Explanation:

Data given

`\bar X=12.3` represent the sample mean

`s=4.4` represent the sample standard deviation

`n=40` sample size

`\mu_o =10.5` represent the value that we want to test

`\alpha=0.01` represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

`p_v` represent the p value for the test (variable of interest)

Part a: System of hypothesis

We need to conduct a hypothesis in order to check if the true mean is higher than 10.5 hours per month, the system of hypothesis would be:

Null hypothesis:
`\mu \leq 10.5`

Alternative hypothesis:
`\mu > =10.5`

The statistic is given by:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

Part b: Calculate the statistic

We can replace in formula (1) the info given like this:

`t=(12.3-10.5)/((4.4)/(√(40)))=2.587`

Part c

We need to find the degrees of freedom

`df = n-1= 40-1 = 39`

And we need to find in the t distribution with 39 degrees of freedom a critical value who accumulate 0.01 of the area in the right and we got

`t_(cric)= 2.426`

Part d

The calculated value is higher than the critical value so then we can reject the null hypothesis

Part e

For this case we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly higher than 10.5 hours per month at the significance level of 0.01