Question

Zappos is an online retailer based in Nevada and employs 1,300 employees. One of their competitors, Amazon, would like to test the hypothesis that the average age of a Zappos employee is less than 36 years old. A random sample of 22 Zappos employees was found to have an average age of 33.9 years. The standard deviation for this sample was 4.1 years. Amazon would like to set α = 0.025. The conclusion for this hypothesis test would be that because the test statistic is

Answer 1

`t=(33.9-36)/((4.1)/(√(22)))=-2.402`

`df = n-1= 22-1=21`

`t_(\alpha/2)= -2.08`

Since the calculated values is lower than the critical value we have enough evidence to reject the null hypothesis at the significance level of 2.5% and we can say that the true mean is lower than 36 years old

Explanation:

Data given

`\bar X=33.9` represent the sample mean

`s=4.1` represent the sample standard deviation

`n=22` sample size

`\mu_o =26` represent the value that we want to test

`\alpha=0.025` 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)

System of hypothesis

We need to conduct a hypothesis in order to check if the true mean is less than 36 years old, the system of hypothesis would be:

Null hypothesis:
`\mu \geq 36`

Alternative hypothesis:
`\mu < 36`

The statistic is given by:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

And replacing we got:

`t=(33.9-36)/((4.1)/(√(22)))=-2.402`

Now we can calculate the critical value but first we need to find the degreed of freedom:

`df = n-1= 22-1=21`

So we need to find a critical value in the t distribution with df =21 who accumulates 0.025 of the area in the left and we got:

`t_(\alpha/2)= -2.08`

Since the calculated values is lower than the critical value we have enough evidence to reject the null hypothesis at the significance level of 2.5% and we can say that the true mean is lower than 36 years old