Statistics students believe that the mean score on a first statistics test is 65. The instructor thinks that the mean score is higher. She samples 10 statistics students and obtains the scores: Grades 63.9 68.4 96 85.5 96 62.7 83.2 88 73.5 65

Test grades are believed to be normally distributed.

Use a significance level of 5%.
State the alternative hypothesis:
H
A
:

μ
>
65

μ
≠
65

μ
<
65

State the mean of the sample:
State the standard error of the sample means:
State the test statistic:
t
=

State the p-value:
Decision:
Fail to reject the null hypothesis.
Reject the null hypothesis.

Alternate hypothesis: u>65

Sample mean: 78.22

The estimated population standard deviation is: 13.10

The estimated standard deviation of the sample mean is: 4.14

The t value is: 3.19

The p value is: 0.005

Reject the null hypothesis

Step-by-step explanation:

It is not clear what is the null hypothesis, but the phrasing of the question suggests that the null hypothesis is that the mean is 65.

In this case, the alternate hypothesis is that the mean is greater than 65.

The sample mean is

m=(63.9+68.4+96+85.5+96+62.7+83.2+88+73.5+65)/1=

m=78.22

The standard error of the sample is the estimated population standard deviation. This is the square root of the estimated population variance.

The estimated population variance is the sum of the squared differences of the data points from their mean, all divided by n-1=9, in this case.

Taking the square root, I get

s=13.10

The estimated standard deviation of the sample mean is the estimated population standard deviation divided by the square root of the sample size, in this case the square root of 10

su=13.10/sqrt(10)=4.14

The t statistic is the (m-u)/su

t=(78.22-65)/4.14=3.19

The p value is the probability of seeing a t value this high or higher.

P(t>=3.19)=0.005

This is way less than the desired test size of 0.05, so we reject the null hypothesis.

s =17