Question

A data set includes data from student evaluations of courses. The summary statistics are nequals92​, x overbarequals4.09​, sequals0.55. Use a 0.10 significance level to test the claim that the population of student course evaluations has a mean equal to 4.25. 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 = 4.25`

Alternative hypothesis:
`\mu \\eq 4.25`

`t=(4.09-4.25)/((0.55)/(√(92)))=-2.79`

`p_v =2*P(t_((91))<-2.79)=0.00642`

If we compare the p value and the significance level given
`\alpha=0.1` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the population mean is significant different from 4.25 at 10% of significance.

Explanation:

1) Previous concepts and data given

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

`\bar X=4.09` represent the sample mean

s=0.55 represent the sample standard deviation

n=92 represent the sample selected

`\alpha=0.1` significance level

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean for evaluation of courses is equal to 4.25, the system of hypothesis would be:

Null hypothesis:
`\mu = 4.25`

Alternative hypothesis:
`\mu \\eq 4.25`

If we analyze the size for the sample is > 30 but 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=(\bar X-\mu_o)/((s)/(√(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=(4.09-4.25)/((0.55)/(√(92)))=-2.79`

P-value

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

`df=n-1=92-1=91`

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

`p_v =2*P(t_((91))<-2.79)=0.00642`

Conclusion

If we compare the p value and the significance level given
`\alpha=0.1` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the population mean is significant different from 4.25 at 10% of significance.