Question

4. A researcher is interested in whether participating in sports positively influences self-esteem in young girls. She identifies a group of girls who have not played sports before but are now planning to begin participating in organized sports. The researcher gives them a 50-item self-esteem inventory before they begin playing sports and administers the same test again after 6 months of playing sports. The self-esteem inventory is measured on an interval scale, with higher numbers indicating higher self-esteem. In addition, scores on the inventory are normally distributed. The scores follow.

Before=X	After=Y	D = |X-Y|	D - D ̅	(D-D ̅ )^2
44	46	2	0.5	0.25
40	41	1	-0.5	0.25
39	41	2	0.5	0.25
46	47	1	-0.5	0.25
42	43	1	-0.5	0.25
43	45	2	0.5	0.25
		Sum=9		Sum=1.5

= Mean of the difference scores = 9/6 = 1.5

S_D = = 0.548

$7g1i43ZcruuPTQAAAABJRU5ErkJggg==$ =

A) Should H0 be rejected?

B) If significant, compute and interpret the effect size.

C) If significant, draw a graph representing the data.

D) Determine the 95% confidence interval.

I already did the initial analysis, but I am stuck on the parts listed above.

Answer 1

a)

µ_D = Mean of X – Mean of Y

alpha = 0.05

Null and Alternate Hypothesis

H₀: µ_D = 0 (No impact)

H_a: µ_D < 0 (Positive impact of playing sports)

Test Statistic

t = Mean/ Std Dev = -1.5/0.548 = -2.74

P-value = TDIST(2.74,5,1) = 0.020

Result

Since the p-value is less than 0.05, we reject the null hypothesis in favour of alternate hypothesis ie participating in sports positively influences self-esteem in young girls.

b)

SD_pooled = {(s₁²+s₂²)/n₁+n₂-2}^1/2 = 1.15

Effect Size = (x_Q - x_M )/ SD_pooled = 1.5/1.15 = 1.30

The high value of effect size tells us that there is meaningful difference between X and Y.

c)

d)

95% CI of difference of means Mean +/- 1.96 * Std Dev = -1.5+/- 1.96*0.548 = {-2.57, -0.43}