Question

A data set includes data from student evaluations of courses. The summary statistics are nequals99​, x overbarequals3.58​, sequals0.55. Use a 0.01 significance level to test the claim that the population of student course evaluations has a mean equal to 3.75. Assume that a simple random sample has been selected. Identify the null and alternative​ hypotheses, test​ statistic, P-value, and state the final conclusion that addresses the original claim.

Answer 1

Null hypothesis:
`\mu = 3.75`

Alternative hypothesis:
`\mu \\eq 3.75`

`t=(3.58-3.75)/((0.55)/(√(99)))=-3.075`

`df=n-1=99-1=98`

`p_v =2*P(t_((98))<-3.075)=0.0027`

For this case since the p value is lower than the significance level we have enough evidence to reject the null hypothesis at the significance level os 0.01 and the the true mean is different from 3.75

Explanation:

Data given

`\bar X=3.58` represent the sample mean

`s=0.55` represent the sample standard deviation

`n=99` sample size

`\mu_o =3.75` represent the value that we want to test

`\alpha=0.01` 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)

System of hypothesis

We need to conduct a hypothesis in order to check if the true mean is equal to 3.75 or not, the system of hypothesis would be:

Null hypothesis:
`\mu = 3.75`

Alternative hypothesis:
`\mu \\eq 3.75`

The statistic is given by:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

Calculate the statistic

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

`t=(3.58-3.75)/((0.55)/(√(99)))=-3.075`

P-value

The degrees of freedom are:

`df=n-1=99-1=98`

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

`p_v =2*P(t_((98))<-3.075)=0.0027`

Conclusion

For this case since the p value is lower than the significance level we have enough evidence to reject the null hypothesis at the significance level os 0.01 and the the true mean is different from 3.75