Question

The researcher in exercise number 5 from the Mod. 11 assignment decides to conduct the same study using a within-subjects design in order to control for differences in cognitive ability. He selects a random sample of participants and has them study different material of equal difficulty in both the music and no music conditions. They are measured on an interval-ratio scale and are normally distributed. The scores are:

Music           No Music

6                   10

7                  7

6                  8

5                  7

6                  7

8                  9

8 8

a. What statistical test should be used to analyze these data?

b. Identify H₀ and H_a for this study.

c. Conduct the appropriate analysis.

d. Should H₀be rejected? What should the researcher conclude?

e. If significant, compute the effect size and interpret this?

h. Calculate the power of the test.

Answer 1

Result:

The researcher in exercise number 5 from the Mod. 11 assignment decides to conduct the same study using a within-subjects design in order to control for differences in cognitive ability. He selects a random sample of participants and has them study different material of equal difficulty in both the music and no music conditions. They are measured on an interval-ratio scale and are normally distributed. The scores are:

a. What statistical test should be used to analyze these data?

for this study using a within-subjects design, we use paired t test.

b. Identify H₀ and H_a for this study.

= mean of difference( x1-x2)

Ho: µ_d=0 H1: µ_d ≠ 0

c. Conduct the appropriate analysis.

MINITAB used

d. Should H₀be rejected? What should the researcher conclude?

Calculated t= -2.71, P=0.035 which is < 0.05 level of significance. Ho is rejected.

We conclude that there is significant difference in music and no music conditions.

e. If significant, compute the effect size and interpret this?

Effect size = mean difference/ sd = 1.429/1.397

= 1.022

The effect is large.

h. Calculate the power of the test.

Power and Sample Size

Paired t Test

Testing mean paired difference = 0 (versus ≠ 0)

Calculating power for mean paired difference = difference

α = 0.05 Assumed standard deviation of paired differences = 1.397

Results

Difference	Sample Size	Power
1.429	7	0.619519

Power Curve for Paired t Test

Power = 0.6195