Question

# A golf instructor is interested in determining if her new technique for improving players’ golf scores...

A golf instructor is interested in determining if her new technique for improving players’ golf scores is effective. She takes four new students and records their 18-hole scores before learning the new technique and then after having taken her class. She then conducts a hypothesis test at a significance level of 5%.

Player 1   Player 2   Player 3   Player 4

Mean score before class                         83          78           93          87

Mean score after class                           80          80           86          86

What is the Null Hypotheses?

a The mean scores before the class and the mean scores after the class are the same.

b The mean scores before the class are greater than the mean scores after the class.

2) What is the Alternate Hypothesis?

a The mean scores before the class and the mean scores after the class are the same.

b The mean scores before the class are greater than the mean scores after the class.

3)What is the significance level of the test?

Enter answer as a 2 place decimal with a 0 to the left of the decimal point.

4)What is the appropriate distribution for performing this test?

a. z distribution

b. t distribution

c. Neither of the above distributions

d. Either of the above distributions

5)What is the critical value?

Enter answer rounded to 3 place accuracy.

What is the numerical value of the test statistic?

Enter answer as decimal rounded to 2 decimal places.

a Reject the null hypothesis

b Do not reject the null hypothesis Enter your answer by the appropriate identifying letter.

7)What is the Pvalue for this hypothesis test.

Enter answer as a percent without the percent sign with 0 decimal places or as a decimal rounded to 2 decimal places.

Solution

Final answers are given below. Back-up theory and Details of Calculations follow at the end.

1) Null Hypothesis: Option a Answer 1

2) Alternate Hypothesis: Option b Answer 2

3) Significance level of the test: 0.05 Answer 3

4) Appropriate distribution for performing this test: Option b Answer 4

5) C ritical value: 2.353 Answer 5

Numerical value of the test statistic: - 0.88 Answer 6

6) Decision: Option b Answer 7

7) P-value for this hypothesis test: 77.78 Answer 8

DONE

Back-up theory and Details of Calculations

Paired t-test

Let X = Mean score before class

Y = Mean score after class

And D = Y – X.

Let µD, and σD be the mean and standard deviation of D.

Then, D ~ N(µ, σ2) where σ2 is unknown.

Claim: New technique for improving players’ golf scores is effective

[Hint: Given, ‘A golf instructor is interested in determining if her new technique for improving players’ golf scores is effective’]

Hypotheses:

Null: H0: µ = µ0 = 0 Vs Alternative: HA: µ > 0

Test Statistic:

t = (√n) (Dbar - µ0)/s where

Dbar and s are respectively, sample average and sample standard deviation based on n observations on X and Y.

Calculations

Summary of Excel calculations is given below:

 n 4 Dbar -1.3 s(D) 2.9277 √n 2 α 0.05 1 - α 0.95 tcal -0.87831 tcrit) 2.35336 p-value 0.777781

Distribution, Significance level (α), Critical Value and p-value:

Under H0, t ~ tn - 1. Hence, for level of significance α%, Critical Value = upper α% point of tn - 1 and

p-value = P(tn - 1 > tcal).

α = 5% [i.e., 0.05 (given)

Using Excel Function: Statistical TINV TDIST, the above are found to be as given in the above table.

Decision:

Since tcal < tcrit, or equivalently, since p-value > α, H0 is accepted.

Conclusion:

There is not sufficient evidence to suggest that the claim is valid.

Complete

#### Earn Coins

Coins can be redeemed for fabulous gifts.