Question

An online retailer employs​ 1,300 employees. One of their competitors would like to test the hypothesis that the average age of these​ 1,300 employees is less than 36 years old. A random sample of 22 employees was found to have an average age of 33.9 years. The standard deviation for this sample was 4.1 years. The competitor would like to set alphaαequals=0.025. What is the correct hypothesis statement for this hypothesis​ test?

Answer 1

Answer:

Null hypothesis:$\mu \geq 36$  

Alternative hypothesis:$\mu < 36$  

$p_v =P(t_{21}$  

Since the $p_v< \alpha$ we can reject the null hypothesis at 2.5 % of significance.  

Step-by-step explanation:

Data given and notation  

$\bar X=33.9$ represent the sample mean  

$s=4.1$ represent the standard deviation for the sample  

$n=22$ sample size  

$\mu_o =36$ represent the value that we want to test  

$\alpha$ 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)  

State the null and alternative hypotheses to be tested  

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

Null hypothesis:$\mu \geq 36$  

Alternative hypothesis:$\mu < 36$  

Compute the test statistic  

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

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

$t=\frac{33.9-36}{\frac{4.1}{\sqrt{22}}}=-2.402$  

Now we need to find the degrees of freedom for the t distirbution given by:

$df=n-1=22-1=21$  

What do you conclude? Use the p-value approach  

Since is a one left tailed test the p value would be:  

$p_v =P(t_{21}$  

Since the $p_v< \alpha$ we can reject the null hypothesis at 2.5 % of significance.