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.
C. Identify H0 and Ha for this study.
It would help to use the same name.
"whether listening to music helps or hinders test-performance"
Ho: mean music = mean no music
Ha: mean music ≠ mean no music
How many tails would that have?
Find the mean first = sum of scores/number of scores
Subtract each of the scores from the mean and square each difference. Find the sum of these squares. Divide that by the number of scores to get variance.
Standard deviation = square root of variance
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√n
If only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to your Z score.
I'll let you do the calculations.
a. The appropriate statistical test to analyze the data would be the paired samples t-test.
b. This is a two-tailed test because the researcher is interested in whether listening to music helps OR hinders test performance.
c. H0 (null hypothesis): There is no difference in test performance between listening to music and not listening to music.
Ha (alternative hypothesis): There is a difference in test performance between listening to music and not listening to music.
d. To conduct the appropriate analysis, we will perform a paired samples t-test. By comparing the scores of the participants in the music condition with their scores in the no music condition, we can determine if there is a significant difference in test performance.
H0: Mean difference (μd) = 0
Ha: Mean difference (μd) ≠ 0
Using statistical software, we can input the scores in both conditions and calculate the mean difference, standard deviation of the differences, and the t-statistic. Based on the calculated t-value and the degrees of freedom, we can then compare it with the critical value from the t-distribution table to determine if H0 should be rejected.
If the calculated t-value falls within the critical value range, we reject the null hypothesis. If it falls outside the critical value range, we fail to reject the null hypothesis.
The researcher should conclude whether there is a statistically significant difference in test performance between the two conditions based on the outcome of the t-test.
e. To calculate the 95% confidence interval, we can use the formula:
CI = x̄d ± (t * SE)
where x̄d is the mean difference, t is the t-value from the t-distribution table for the desired confidence level, and SE is the standard error of the mean difference.
By plugging in the values from the analysis, we can calculate the upper and lower bounds of the 95% confidence interval. The confidence interval provides a range of values within which the true population mean difference is likely to fall.
Note: The exact calculations for the t-test and confidence interval would require the actual data values provided in the table.