Question

A random sample of 16 students selected from the student body of a large university had an average age of 25 years. We want to determine if the average age of all the students at the university is significantly different from 24. Assume the distribution of the population of ages is normal with a standard deviation of 2 years. At a .05 level of significance, it can be concluded that the mean age is:

Answer 1

Answer:

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

$p_v =2*P(Z>2)=0.0455$  

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 reject the null hypothesis, and we can conclude that the true mean differs from 24 at 5% of significance

Step-by-step explanation:

Data given and notation  

$\bar X=25$ represent the sample mean

$\sigma=2$ represent the sample population deviation for the sample  

$n=16$ sample size  

$\mu_o =24$ 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 true mean is different from 24, the system of hypothesis would be:  

Null hypothesis:$\mu = 24$  

Alternative hypothesis:$\mu \neq 24$  

If we analyze the size for the sample is < 30 but 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:  

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$  (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:  

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

P-value

Since is a two sided test the p value would be:  

$p_v =2*P(Z>2)=0.0455$  

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 reject the null hypothesis, and we can conclude that the true mean differs from 24 at 5% of significance