In the past the average age of employees of a large corporation has been 40 years. Recently, the company has been hiring older individuals. In order to determine whether there has been an increase in the average age of all the employees, a sample of 25 employees was selected. The average age in the sample was 45 years with a standard deviation of 5 years. Assume the distribution of the population is normal let α= 0.5 A, state the null and the alterative hypothesis B, test to determine whether or not the mean age of all the employees is significantly more than 40 years.

If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that mean age of all employees is significantly more than 40 years at 5% of signficance.

Step-by-step explanation:

1) Data given and notation

[tex]\bar X=45[/tex] represent the sample mean

[tex]s=15[/tex] represent the sample standard deviation

[tex]n=25[/tex] sample size

[tex]\mu_o =40[/tex] represent the value that we want to test

[tex]\alpha=0.05[/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)

2) State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean age of all the employees is significantly more than 40 years, the system of hypothesis are :

Null hypothesis:[tex]\mu \leq 40[/tex]

Alternative hypothesis:[tex]\mu > 40[/tex]

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

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

3) Calculate the statistic

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

[tex]t=\frac{45-40}{\frac{5}{\sqrt{25}}}=5[/tex]

4) P-value

First we need to calculate the degrees of freedom given by:

[tex]df=n-1=25-1=24[/tex]

Since is a one-side upper test the p value would be:

[tex]p_v =P(t_{(24)}>5)=2.08x10^{-5}[/tex]

5) Conclusion

If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean age of all employees it's significantly more than 40 years at 5% of signficance.