Question

A data set includes data from student evaluations of courses. The summary statistics are nequals83​, x overbarequals3.32​, sequals0.56. Use a 0.05 significance level to test the claim that the population of student course evaluations has a mean equal to 3.50. 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

Hypothesis

Null hypothesis:
`\mu =3.50`

Alternative hypothesis:
`\mu \\eq 3.50`

Statistic

`t=(3.32-3.50)/((0.56)/(√(83)))=-2.928`

P value

The degrees of freedom are given by:

`df=n-1=83-1=82`

And the p value would be given by:

`p_v =2*P(t_((82))<-2.928)=0.0044`

Conclusion

Since the p value is lower than the significance level of 0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly different from 3.5

Explanation:

Information given

`\bar X=3.32` represent the sample mean

`s=0.56` represent the sample deviation

`n=83` sample size

`\mu_o =3.5` represent the value to test

t would represent the statistic

`p_v` represent the p value

System of hypothesis

We want to verify if the true mean is equal to 3.5, the system of hypothesis are:

Null hypothesis:
`\mu =3.50`

Alternative hypothesis:
`\mu \\eq 3.50`

Statistic

The statistic would be given by:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

Replacing we got:

`t=(3.32-3.50)/((0.56)/(√(83)))=-2.928`

P value

The degrees of freedom are given by:

`df=n-1=83-1=82`

And the p value would be given by:

`p_v =2*P(t_((82))<-2.928)=0.0044`

Conclusion

Since the p value is lower than the significance level of 0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly different from 3.5