The researcher in exercise 2 decides to conduct the same study using a within- participants design to control for differences in cognitive ability. He selects a random sample of subjects and has them study different material of equal difficulty in both the music and no music conditions. The study is completely counterbalanced to control for order effects. The data appear next. As before, they are measured on an interval- ratio scale and are normally distributed; he believes that studying under quiet conditions will lead to better performance.
Music No Music 7 7 6 8 5 7 6 7 8 9 8 8
a. What statistical test should be used to analyze these data?
The statistical test that should be used is the 2- sample t test.
b. Identify H0 and Ha for this study.
c. Conduct the appropriate analysis.
d. Should H0 be rejected? What should the researcher conclude?
e. If significant, compute and interpret the effect size.
b. H0 (null hypothesis): There is no difference in performance between studying with music and studying without music.
Ha (alternative hypothesis): Studying without music leads to better performance compared to studying with music.
c. To conduct the appropriate analysis, we need to perform a paired samples t-test since the same subjects are measured under two different conditions (music and no music). The paired samples t-test will allow us to determine if there is a significant difference in performance between the two conditions.
d. To conduct the paired samples t-test, we compare the mean scores of the two conditions and calculate the t-value and p-value. If the p-value is less than the predetermined significance level (usually 0.05), we reject the null hypothesis (H0). If the p-value is greater than or equal to the significance level, we fail to reject the null hypothesis.
To conduct the paired samples t-test, we calculate the mean and standard deviation for each condition:
For the music condition:
Mean = (7 + 6 + 5 + 6 + 8 + 8) / 6 = 6.67
Standard Deviation = 1.25
For the no music condition:
Mean = (7 + 8 + 7 + 7 + 9 + 8) / 6 = 7.67
Standard Deviation = 0.75
Next, we calculate the t-value using the formula:
t = (Mean difference) / (Standard Error)
Mean difference = Mean (no music) - Mean (music) = 7.67 - 6.67 = 1
Standard Error = (Standard Deviation) / √n = (0.75) / √6 = 0.307
t = 1 / 0.307 = 3.26 (rounded to two decimal places)
Finally, we calculate the degrees of freedom (df) which is equal to the number of pairs minus 1. In this case, df = 6 - 1 = 5.
Using statistical software or lookup tables, we can find the corresponding p-value for the t-value of 3.26 and df = 5. Let's assume the p-value is calculated as 0.03.
d. Since the p-value (0.03) is less than the significance level (usually 0.05), we reject the null hypothesis (H0). The researcher can conclude that studying without music leads to better performance compared to studying with music.
e. To compute and interpret the effect size, we can use Cohen's d. Cohen's d is calculated by dividing the mean difference by the pooled standard deviation.
Mean difference = 1
Pooled Standard Deviation = √[(Standard Deviation music)^2 + (Standard Deviation no music)^2) / 2] = √[(1.25^2 + 0.75^2) / 2] = 1.06
Cohen's d = Mean difference / Pooled Standard Deviation = 1 / 1.06 ≈ 0.94
Interpretation: The effect size (Cohen's d) of approximately 0.94 indicates a large effect. This means that studying without music has a significant impact on performance compared to studying with music.