Question

A counselor at a community college claims that the mean GPA for students who have transferred to State University is more than 3.1. A random sample of 36 such students had a mean GPA of 3.25 with a standard deviation of 0.3. Test the counselor's claim at the 99% level of confidence. a) Yes. The counselor's claim is valid. b) No. The counselor's claim is not valid. c) both a and b are correct, depending on the total number of students transferred to State Universty d) None of the above.

Answer 1

Answer:

$t=\frac{3.25-3.1}{\frac{0.3}{\sqrt{36}}}=3$  

If we compare the p value and the significance level for example $\alpha=1-0.99=0.01$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis.

And the best conclusion would be:

a) Yes. The counselor's claim is valid

Because we reject the null hypothesis in favor to the alternative hypothesis.

Step-by-step explanation:

Data given and notation  

$\bar X=3.25$ represent the sample mean

$s=0.3$ represent the sample standard deviation

$n=36$ sample size  

$\mu_o =3.1$ 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)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is more than 3.1, the system of hypothesis would be:  

Null hypothesis:$\mu \leq 3.1$  

Alternative hypothesis:$\mu > 3.1$  

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:  

$t=\frac{3.25-3.1}{\frac{0.3}{\sqrt{36}}}=3$  

P-value  

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

$df=n-1=36-1=35$  

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

$p_v =P(t_{(35)}>3)=0.00247$  

Conclusion  

If we compare the p value and the significance level for example $\alpha=1-0.99=0.01$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis.

And the best conclusion would be:

a) Yes. The counselor's claim is valid

Because we reject the null hypothesis in favor to the alternative hypothesis.