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.
Enter your answer by the appropriate identifying letter.
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.
Enter your answer by the appropriate identifying letter.
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
Enter your answer by the appropriate identifying letter.
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.
6)What is your decision?
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 µ_{D}, and σ_{D} be the mean and standard deviation of D.
Then, D ~ N(µ, σ^{2}) where σ^{2} 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: H_{0}: µ = µ_{0} = 0 Vs Alternative: H_{A}: µ > 0
Test Statistic:
t = (√n) (Dbar - µ_{0})/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) |
2.927700219 |
√n |
2 |
α |
0.05 |
1 - α |
0.95 |
tcal |
-0.87831007 |
tcrit) |
2.353363435 |
p-value |
0.777780957 |
Distribution, Significance level (α), Critical Value and p-value:
Under H_{0}, t ~ t_{n - 1}. Hence, for level of significance α%, Critical Value = upper α% point of t_{n - 1} and
p-value = P(t_{n - 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, H_{0} is accepted.
Conclusion:
There is not sufficient evidence to suggest that the claim is valid.
Complete
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