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

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

Step-by-step explanation:

Data given

[tex]\bar X=33.9[/tex] represent the sample mean

[tex]s=4.1[/tex] represent the sample standard deviation

[tex]n=22[/tex] sample size

[tex]\mu_o =26[/tex] represent the value that we want to test

[tex]\alpha=0.025[/tex] represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

[tex]p_v[/tex] 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:[tex]\mu \geq 36[/tex]

Alternative hypothesis:[tex]\mu < 36[/tex]

The statistic is given by:

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)

And replacing we got:

[tex]t=\frac{33.9-36}{\frac{4.1}{\sqrt{22}}}=-2.402[/tex]

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

[tex] df = n-1= 22-1=21[/tex]

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:

[tex] t_{\alpha/2}= -2.08[/tex]

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