Question

5. William’s golfing group is terrific for a group of amateurs. Are they ready to turn pro?

Here’s the data. (Remember that the lower the score in golf the better!)

	Size	Average Score	Standard Deviation
William's Group	9	82	2.6
The PGA Pro's	500	71	3.1

Answer 1

Step-1:

Let William's group dented by '1' and The PGA Pro's as "2".

Hence;

Sample size : n1 = 9 ; n2=500

Average score : x-bar1=82 ; x-bar2=71

Variance : s1-square = 2.6^2 = 6.76 ; s2-square=3.1^2 = 9.61

Step-2:

Hypothesis :

Null Hypothesis

Alternative Hypothesis : ( Want to test that Willaims scroe is less than PGA )

So, it is a Left tailed test .

Let Alpha = 0.05 ( 5% level of significance )

Step-3:

Test statistic will be t-test for two indepednent means ( because population standard deviation is UNKNOWN)

The formula of t-test as below;

Step-4:
Calculation of t-test statistic;

t = =(82-71)/SQRT((6.76/9)+(9.61/500)) = 12.53..............(1)

Now, find t-critical value from below t-table for degrees of freedom = 9+500-2 = 507 and alpha=0.05

t-critical value = -1.648 ( negative because it is left tailed test ) ..........(2)

From (1) and (2), we get

t-statistic is in the Acceptance region; hence accept Ho. ( Refer figure below)

Conclusion : Willaims scroe is greater than or equal to  PGA score. So, they are ready to turn pro.

### End of answer