Question

In Table 7.4 on page 173, all ten top hitters in the major league baseball in 2011 had lower batting averages in 2012, supporting regression toward the mean. Treating averages as whole numbers (without decimal points) and subtracting their batting averages for 2012 from those for 2011 (so that positive difference scores support regression toward the mean), we have the following ten difference scores: 14, 39, 61, 60, 13, 21, 50, 93, 16, 61.

Question:				Calculations or Logic:	Answer:
Test the null hypothesis (that the hypothetical population mean difference equals zero for all sets of top ten hitters over the years) at the .05 level of significance.	Step 1	What is the research problem?
	Step 2	What is the null hypothesis?
		What is the alternative hypothesis?
	Step 3	What is the decision rule?
	Step 4	What is the critical t?
		What is the value of t? (you will need to calculate this)
	Step 5	What is the decision? (retain or reject the null hypothesis at the specified level of significance; note the relationship between the observed and critical t scores)
	Step 6	What is your interpretation of the decision in relation to the original research problem?
Specify the p -value for this test result.
Construct a 95% confidence interval.
Calculate Cohen’s d .
How might these findings be reported?

Answer 1

Given,

difference scores -

14, 39, 61, 60, 13, 21, 50, 93, 16, 61

Mean of the difference = 42.8

Standard deviation of the difference = 26.8

Sample size = 10

degrees of freedom = 10 - 1 = 9

1.

Do All ten top hitters in the major league baseball in 2011 have significantly lower batting averages in 2012?

2.

Null hypothesis -

All ten top hitters in the major league baseball in 2011 do not have significantly lower batting averages in 2012

Alternate hypothesis -

All ten top hitters in the major league baseball in 2011 have significantly lower batting averages in 2012

3.

Reject null hypothesis if all ten top hitters in the major league baseball in 2011 have significantly lower batting averages in 2012, or in other words reject null hypothesis if the t value is greater than critical t value

4.

Critical t value = 1.83 (For a significance level of 0.05, degrees of freedom = 9 and one tail hypothesis test)

t value = mean difference / standard deviation of the difference / sqrt (sample size)

= 42.8 / 26.8 / sqrt (10)

= 5.04

5.

Since, t value > critical t value we reject the null hypothesis

6.

All ten top hitters in the major league baseball in 2011 have significantly lower batting averages in 2012

7.

p value for this test result is 0.00035 (From t table, t = 5.04, df = 9)

8.

Confidence interval = mean difference +/- margin of error

margin of error = critical t value * standard error

standard error = 26.8 / sqrt (10)

= 8.5

=> margin of error = 15.5

Confidence interval = 42.8 +/- 15.5

= 27.3 to 58.3

9.

Cohen's d = mean difference / pooled standard deviation * (N - 3) / (N - 2.25) * sqrt [(N - 2) / N]

= 1.6 * 0.9 * 0.9

= 1.3

10.

As a rule of thumb, a cohen's d score larger than 0.8 is meant to have a large effect on the difference. Hence, there is significantly large effect on the mean batting averages of the players in two seasons.