A business school has a goal that the average number of years of work experience of its MBA applicants is more than three years. Historical data suggest that the variance has been constant at around six months, and thus, the population variance can be assumed to be known. Based on last year’s applicants, it was found that among a sample of 47, the average number of years of work experience is 3.1. Can the school state emphatically that it is meeting its goal? Formulate the appropriate hypothesis test and conduct the test.

If we compare the p value and the significance level assumed [tex]\alpha=0.01[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the height of men actually its NOT significant higher than 0.3 at 1% of signficance.

Step-by-step explanation:

Data given and notation

[tex]\bar X=3.1[/tex] represent the sample mean

[tex]\sigma=\sqrt{6}[/tex] represent the sample standard deviation for the sample

[tex]n=47[/tex] sample size

[tex]\mu_o =3[/tex] represent the value that we want to test

[tex]\alpha[/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)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is higher than 3, the system of hypothesis would be:

Null hypothesis:[tex]\mu \leq 3[/tex]

Alternative hypothesis:[tex]\mu > 3[/tex]

If we analyze the size for the sample is > 30 and we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

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

z-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:

[tex]z=\frac{3.1-3}{\frac{2.449}{\sqrt{47}}}=0.280[/tex]

P-value

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

[tex]p_v =P(z>0.280)=0.390[/tex]

Conclusion

If we compare the p value and the significance level assumed [tex]\alpha=0.01[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the height of men actually its NOT significant higher than 0.3 at 1% of signficance.