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Answer:
The 90% confidence interval to estimate the average difference in scores between the two courses is (-1.088, 20.252).
Step-by-step explanation:
We have the standard deviation for the differences, which means that the t-distribution is used to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 8 - 1 = 7
90% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 7 degrees of freedom(y-axis) and a confidence level of . So we have T = 1.8946
The margin of error is:
In which s is the standard deviation of the sample and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 9.582 - 10.67 = -1.088
The upper end of the interval is the sample mean added to M. So it is 9.582 + 10.67 = 20.252
The 90% confidence interval to estimate the average difference in scores between the two courses is (-1.088, 20.252).
=
504