Question

Claim: “Golfers who had one lesson with this instructor had an average increase of more than 15 yards in the distance of their drives.”

A random sample of 5 golfers who had attended a free trial lesson with this golf instructor was selected. Before the lesson, each of these golfers hit several drives from a tee, and the average distance for each golfer was recorded (in yards). This process was repeated after the lesson. Assume that the differences between the “before” and “after” distances are approximately normally distributed. Test the claim at the 0.1 significance level.

Golfer    Aaron Buddy Chloe David Eric
Dist Before Lesson 201.0 195.1 186.4 236.5 250.4
Dist After Lesson 224.7 208.6 200.6 264.3 261.9

A.) Identify the correct HYPOTHESES
B.) Identify the value of the TEST STATISTIC
C.) Identify the value of the CRITICAL VALUE(S)
D.) Identify the P-VALUE
E.) Identify the CONCLUSION

***THE CHART DIDN'T COME OUT RIGHT. EACH COLUMN SHOULD BE ONE GOLFER WITH A DISTANCE BEFORE AND AFTER THE LESSON UNDER EACH***

Answer 1

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u_d> - 15

Alternative hypothesis: u_d < - 15

Note that these hypotheses constitute a one-tailed test.  

Formulate an analysis plan. For this analysis, the significance level is 0.10. Using sample data, we will conduct a matched-pairs t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard deviation of the differences (s), the standard error (SE) of the mean difference, the degrees of freedom (DF), and the t statistic test statistic (t).

s = sqrt [ (\sum (d_i - d)² / (n - 1) ]

s = 7.1654

SE = s / sqrt(n)

S.E = 3.20447

DF = n - 1 = 5 -1

D.F = 4

t = [ (x₁ - x₂) - D ] / SE

t = - 2.667

t_critical = - 1.533

where d_i is the observed difference for pair i, d is mean difference between sample pairs, D is the hypothesized mean difference between population pairs, and n is the number of pairs.

Since we have a one-tailed test, the P-value is the probability that a t statistic having 4 degrees of freedom is less than - 0.98.

Thus, the P-value = 0.191

Interpret results. Since the P-value (0.191) is greater than the significance level (0.10), we have to accept the null hypothesis.

Do not Reject H0. From the above test we do not have sufficient evidence in the favor of the claim that Golfers who had one lesson with this instructor had an average increase of more than 15 yards in the distance of their drives.