The owner of a golf course wants to determine if his golf course is more difficult than the one his friend owns. He has 8 golfers play a round of 18 holes on his golf course and records their scores. Later that week, he has the same 8 golfers play a round of golf on his friend's course and records their scores again. The average difference in the scores (treated as the scores on his course - the scores on his friend's course) is 9.582 and the standard deviation of the differences is 15.9274. Calculate a 90% confidence interval to estimate the average difference in scores between the two courses.1) (13.73,-0.65).2) (-10.359,-4.021).3) (-9.259,-5.121).4) (-13.745,-0.635).5) (13.745, -0.635).

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01-07-2021
Mathematics

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The owner of a golf course wants to determine if his golf course is more difficult than the one his friend owns. He has 8 golfers play a round of 18 holes on his golf course and records their scores. Later that week, he has the same 8 golfers play a round of golf on his friend's course and records their scores again. The average difference in the scores (treated as the scores on his course - the scores on his friend's course) is 9.582 and the standard deviation of the differences is 15.9274. Calculate a 90% confidence interval to estimate the average difference in scores between the two courses.1) (13.73,-0.65).2) (-10.359,-4.021).3) (-9.259,-5.121).4) (-13.745,-0.635).5) (13.745, -0.635).

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joaobezerra joaobezerra · Answer 1 · 2021-07-01T21:10:13+02:00

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 [tex]1 - \frac{1 - 0.9}{2} = 0.95[/tex]. So we have T = 1.8946

The margin of error is:

[tex]M = T\frac{s}{\sqrt{n}} = 1.8946\frac{15.9274}{\sqrt{8}} = 10.67[/tex]

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).