Question

A local country club golf tournament organizer is attempting to estimate the average number of strokes for the 13th hole. On a particular day, 64 players completed the play on the 13th hole, with an average of 4.25 strokes and a population standard deviation of 1.6 strokes. Determine the 95 percent confidence interval for the average number of strokes.
Group of answer choices
A. [3.94, 4.56]
B. [3.86, 4.64]
C. [4.05, 4.45]
D. [3.92, 4.58]

Answer 1

Answer:

B. [3.86, 4.64]

Step-by-step explanation:

To find the 95% confidence interval for the average number of strokes, we need to have these following informations:

Sample size

Average of the sample

Standard deviation of the sample

Is this problem, we have that

Sample size = 64

Average = 4.25

Standard deviation = 1.6

Now, we need to find the degree of freedom and the value of $/alpha$, so we can find the a value that is needed for this calculus in the t-distribution table

The degree of freedom is the sample size subtracted by 1. So:

$v = 64 - 1 = 63$

$/alpha$ is the result of the subtraction of 1 by the confidence interval(decimal) divided by two. So

$\alpha = \frac{1 - 0.95}{2} = 0.025$

Looking at the t-distribution table, with $v = 63, \alpha = 0.025$, we find that the value is 1.998.

Now we need to multiply this value by the standard deviation of the sample and divide by the square root of the sample size. So:

$A = \frac{1.998*1.6}{\sqrt{64}} = \frac{3.1968}{8} = 0.39$

Now

For the lower range of the interval, we subtract A from the mean

$L = 4.25 - 0.39 = 3.86$

For the upper end of the interval, we add A to the mean

So

$H = 4.25 + 0.39 = 4.64$

So the correct answer is

B. [3.86, 4.64]