3. A researcher is interested in whether listening to music helps or hinders test-performance. To control for differences in cognitive level, this researcher decides to use a within-participants design. He selects a random sample of participants and has them study different material of equal difficulty in both the music and no music conditions. Participants take a 20-item quiz on the material. The table below shows the scores on the quiz. The study is completely counterbalanced to control for order effects. The scores obtained are measured on an interval-ratio scale and are normally distributed.

Music No Music
17 17
16 18
15 17
16 17
18 19
18 18

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

b. Is this a one- or two tailed test?

c. Identify H0 and Ha for this study.

d. Conduct the appropriate analysis. Should H0 be rejected? What should the researcher conclude?

e. Calculate the 95 confidence interval.

Use processes indicated in your previous post.

a. The appropriate statistical test to analyze the data in this study is a paired t-test.

b. This is a two-tailed test.

c. H0 (null hypothesis): There is no difference in test performance between the music and no music conditions.
Ha (alternative hypothesis): There is a difference in test performance between the music and no music conditions.

d. To conduct the appropriate analysis, we need to calculate the mean difference between the scores obtained in the music and no music conditions, and then perform a paired t-test.

To calculate the mean difference, we subtract the no music score from the music score for each participant, and then take the average of these differences.

Music Scores: 17, 16, 15, 16, 18, 18
No Music Scores: 17, 18, 17, 17, 19, 18

Mean difference = (17-17) + (16-18) + (15-17) + (16-17) + (18-19) + (18-18) / 6 = 0

Next, we calculate the standard deviation of the differences.

Sum of squared differences = (0-0)^2 + (-2)^2 + (-2)^2 + (-1)^2 + (-1)^2 + (0-0)^2 = 2+2+1+1 = 6

Variance of differences = sum of squared differences / (n-1) = 6 / (6-1) = 1.2

Standard deviation of differences = square root of variance = √1.2 ≈ 1.10

Now, we can perform the paired t-test.

t = mean difference / (standard deviation / square root of sample size) = 0 / (1.10 / √6) ≈ 0

The degrees of freedom for this test is n-1, which is 6-1=5.

Using a t-table or statistical software, we find that the critical t-value for a two-tailed test at a significance level of 0.05 with 5 degrees of freedom is ±2.571.

Since the calculated t-value (0) falls within the range of -2.571 to 2.571, we fail to reject the null hypothesis.

Therefore, the researcher should conclude that there is no significant difference in test performance between the music and no music conditions.

e. To calculate the 95% confidence interval, we need to determine the margin of error. The margin of error is equal to the critical value (obtained from the t-table or software) multiplied by the standard deviation of the differences divided by the square root of the sample size.

Margin of error = critical value * (standard deviation / √sample size) = 2.571 * (1.10 / √6) ≈ 1.68

The 95% confidence interval is calculated by subtracting and adding the margin of error to the mean difference.

95% CI = mean difference - margin of error, mean difference + margin of error
= 0 - 1.68, 0 + 1.68
= -1.68, 1.68

Therefore, the 95% confidence interval for the difference in test performance between the music and no music conditions is approximately -1.68 to 1.68.