Question

The owner of a golf course wants to determine if his golf course is more difficult than the one his friend owns. He has 20 golfers play a round of 18 holes on his golf course and records their scores. Later that week, he has the same 20 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 8.9 and the standard deviation of the differences are 11.986. Calculate a 99% confidence interval to estimate the average difference in scores between the two courses.

Answer 1

(1.985, 15.815)

Explanation:

Given that:

Sample size (n) = 20

Standard deviation (s) = 11.986

Mean (m) = 8.9

α = 99%

Using the relation :

Confidence Interval = Mean ± Error

Where ;

Error = Zcritical * standard deviation / sqrt(n)

Zcritical at 99% = 2.58

Hence,

Error = 2.58 * (11.986/sqrt(20))

Error = 2.58 * 2.6801510

Error = 6.91478958

Hence,

Lower boundary = 8.9 - 6.91478958 = 1.98521042

Upper boundary = 8.9 + 6.91478958 = 15.81478958

(1.985, 15.815)