Question

A new program of imagery training is used to improve the performance of basketball players shooting free-throw shots. The first group did an hour imagery practice, and then shot 30 free throw basket shots with the number of shots made recorded. A second group received no special practice, and also shot 30 free throw basket shots. The data are below. Did the imagery training make a difference? Set alpha = .05.

Group 1: 15, 17, 20, 25 26, 27

Group 2: 5, 6, 10, 15, 18, 20

A) You must use all five steps in hypothesis testing:

• Restate the question as a research hypothesis and a null hypothesis about the populations.
• Determine the characteristics of the comparison distribution.
• Determine the cutoff sample score on the comparison distribution at which the null hypothesis should be rejected.
• Determine your sample’s score on the comparison distribution.
• Decide whether to reject the null hypothesis.

B) Solve for and evaluate the effect size of this study using Cohen's D.

C) Create a 90% confidence interval for this problem.

Answer 1

(A)

H0: Null Hypothesis: ( the imagery training did not make a difference )

HA: Research Hypothesis: ( the imagery training made a difference ) (Claim)

From the given data, the following statistics are calculated:

n1 = 6

1 = 21.6667

s1 = 5.0465

n2 = 6

2 = 12.3333

s2 = 6.2823

df = n1 + n2 - 2 = 6 + 6 - 2 = 10

The comparison distribution. is t distribution with degrees of freedom = 10

= 0.05

By Technology,critical value of t = 1.812

The cutoff sample score on the comparison distribution at which the null hypothesis should be rejected. :

Reject H0 if t > 1.812

Pooled Standard Deviation is given by:

Test Statistic is given by:

Since calculated value of t =2.837 is greater than critical value of t= 1.812, the difference is significant. Reject null hypothesis.

Conclusion:
The data support the claim that the imagery training made a difference

(B)

Effect Size: Cohen's d is given by:

So, we conclude: Large effect size.

(C)

df = 6 + 6 - 2 = 10

= 0.10

From Table, critical values of t = 0.129

90% Confidence Interval:

So,

Answer is:

(8.909 9.757)