Question

A PGA (Professional Golf Association) tournament organizer is attempting to estimate the average number of strokes for the 13th hole on a given golf course. 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.

Answer 1

Answer:

The 95 percent confidence interval for the average number of strokes is (3.858, 3.642).

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

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

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$.

So it is z with a pvalue of $1-0.025 = 0.975$, so $z = 1.96$

Now, find M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.96*\frac{1.6}{\sqrt{64}} = 0.392$

The lower end of the interval is the mean subtracted by M. So it is 4.25 - 0.392 = 3.858

The upper end of the interval is the mean added to M. So it is 4.25 + 0.392 = 4.642

The 95 percent confidence interval for the average number of strokes is (3.858, 3.642).