Question

A random sample of 16 students selected from the student body of a large university had an average age of 25 years and a population standard deviation of 2 years. We want to determine if the average age of all the students at the university is significantly more than 24. Assume the distribution of the population of ages is normal.

What is the alternative and null hypotheses?
What is the test statistic?
What is the p-value?
What is your conclusion about the stated hypotheses at a 95% confidence level?

Answer 1

Answer:

$t=\frac{25-24}{\frac{2}{\sqrt{16}}}=2$      

$p_v =P(t_{15}>2)=0.0320$  

If we compare the p value and the significance level given for example $\alpha=0.05$ we see that $p_v<\alpha$ so we can conclude that we reject the null hypothesis, and the true mean is significant higher than 24 years.  

Step-by-step explanation:

1) Data given and notation      

$\bar X=25$ represent the sample mean      

$s=2$ represent the standard deviation for the sample      

$n=16$ sample size      

$\mu_o =24$ 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)  

Confidence =0.95 or 95%

$\alpha=0.05$

State the null and alternative hypotheses.      

We need to conduct a hypothesis in order to determine if the mean is higher than 24, the system of hypothesis would be:      

Null hypothesis:$\mu \leq 24$      

Alternative hypothesis:$\mu > 24$      

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".  

Calculate the statistic      

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

$t=\frac{25-24}{\frac{2}{\sqrt{16}}}=2$      

Calculate the P-value      

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

$df=n-1=16-1=15$  

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

$p_v =P(t_{15}>2)=0.0320$  

Conclusion      

If we compare the p value and the significance level given for example $\alpha=0.05$ we see that $p_v<\alpha$ so we can conclude that we reject the null hypothesis, and the true mean is significant higher than 24 years.