Question

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 64 employees was selected. The average age in the sample was 45 years with a standard deviation of 16 years. Let α = .05

Answer 1

Answer:

$p_v =P(t_{(63)}>2.5)=0.0075$  

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can conclude that the mean age is significantly higher than 45 years at 5% of significance.  

Step-by-step explanation:

1) Data given and notation  

$\bar X=45$ represent the mean height for the sample  

$s=16$ represent the sample standard deviation for the sample  

$n=64$ sample size  

$\mu_o =40$ represent the value that we want to test

$\alpha=0.05$ 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.  

We need to conduct a hypothesis in order to check if the mean age is higher than 40 years, the system of hypothesis would be:  

Null hypothesis:$\mu \leq 40$  

Alternative hypothesis:$\mu > 40$  

If we analyze the size for the sample is < 30 and we don't know the population deviation so 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".  

Calculate the statistic

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

$t=\frac{45-40}{\frac{16}{\sqrt{64}}}=2.5$    

P-value

The first step is calculate the degrees of freedom, on this case:  

$df=n-1=64-1=63$  

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

$p_v =P(t_{(63)}>2.5)=0.0075$  

Conclusion  

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can conclude that the mean age is significantly higher than 45 years at 5% of significance.