Question

6. You are curious if two types of music, classical or rock, have different effects on the ability of college students to perform a series of mental tasks requiring concentration. You gather two groups of participants and have them perform a series of tasks while listening to music and record the number of errors they make in completing the tasks. Group 1 listens to classical music and Group 2 listens to rock. You believe classical music will have a more calming effect and produce less errors. The data for each group are below. Assuming an alpha level of .05 and standard deviations of 2.07 (Group 1) and 1.97 (Group 2), perform the appropriate t-test for this data set. (8 points)

a. What are your hypotheses?

b. What is tcrit?

c. What is tobt?

d. What do you conclude?

Classical	Rock
3	2
4	4
5	5
1	6
0	2
	1

Answer 1

Answer 2

a. Hypotheses: Null Hypothesis (H0): There is no significant difference in the mean number of errors made by college students while listening to classical music and rock music during the mental tasks. Alternative Hypothesis (Ha): There is a significant difference in the mean number of errors made by college students while listening to classical music and rock music during the mental tasks.

b. Tcritical (tcrit): Since the sample size for both groups is small (less than 30), we will use a two-sample t-test with a significance level (alpha) of 0.05. For a two-tailed test at alpha = 0.05 with degrees of freedom (df) = n1 + n2 - 2, tcrit can be found using a t-distribution table or calculator. The degrees of freedom in this case are df = 5 + 5 - 2 = 8.

Using a t-distribution table, tcrit for alpha = 0.05 and df = 8 is approximately ±2.306.

c. Tobt: To perform the t-test, we need to calculate the pooled standard error and then the t-value.

The pooled standard error (SEp) can be calculated using the formula: SEp = sqrt[(s1^2/n1) + (s2^2/n2)]

where s1 and s2 are the standard deviations of the two groups, and n1 and n2 are the sample sizes of Group 1 (classical) and Group 2 (rock), respectively.

SEp = sqrt[(2.07^2/5) + (1.97^2/5)] SEp = sqrt[4.2985 + 3.8805] SEp = sqrt(8.179) SEp = 2.86 (rounded to two decimal places)

The t-value (tobt) can be calculated using the formula: tobt = (x̄1 - x̄2) / SEp

where x̄1 and x̄2 are the sample means of the two groups.

Group 1 (classical) mean (x̄1) = (3 + 4 + 5 + 1 + 0 + 1) / 6 = 14 / 6 ≈ 2.33 (rounded to two decimal places) Group 2 (rock) mean (x̄2) = (2 + 4 + 5 + 6 + 2 + 9) / 6 = 28 / 6 ≈ 4.67 (rounded to two decimal places)

tobt = (2.33 - 4.67) / 2.86 tobt = -2.34 (rounded to two decimal places)

d. Conclusion: The calculated t-value (tobt) is -2.34. Since the absolute value of tobt (2.34) is greater than tcrit (2.306) at the 0.05 significance level (two-tailed test), we reject the null hypothesis (H0). There is sufficient evidence to suggest that there is a significant difference in the mean number of errors made by college students while listening to classical music and rock music during the mental tasks.

Answer 3

To perform the appropriate t-test for the data set, we'll use a two-sample independent t-test, assuming unequal variances since the standard deviations for both groups are different. The t-test will help us determine if there is a significant difference in the mean number of errors between the two groups (Classical and Rock).

a. Hypotheses: Null Hypothesis (H0): There is no significant difference in the mean number of errors between the two groups (Classical and Rock). Alternative Hypothesis (Ha): There is a significant difference in the mean number of errors between the two groups (Classical and Rock).

b. Calculate tcrit: To find tcrit at a 5% significance level (alpha level of 0.05) and degrees of freedom (df) equal to the smaller sample size minus 1 (df = n1 - 1), we can use a t-distribution table or a statistical tool. In this case, df = 10 - 1 = 9.

Using a t-distribution table or a t-distribution calculator, tcrit for a two-tailed test with 9 degrees of freedom and alpha = 0.05 is approximately ±2.262.

c. Calculate tobt: To calculate tobt (the test statistic), we'll use the formula for the independent samples t-test:

tobt = (x̄1 - x̄2) / √[(s1^2 / n1) + (s2^2 / n2)]

where: x̄1 and x̄2 are the sample means for Group 1 (Classical) and Group 2 (Rock), respectively. s1 and s2 are the sample standard deviations for Group 1 (Classical) and Group 2 (Rock), respectively. n1 and n2 are the sample sizes for Group 1 (Classical) and Group 2 (Rock), respectively.

Given the data: Classical: 3, 4, 5, 1, 0 (n1 = 5, x̄1 = 2.6, s1 = 2.07) Rock: 2, 4, 5, 6, 2 (n2 = 5, x̄2 = 3.8, s2 = 1.97)

tobt = (2.6 - 3.8) / √[(2.07^2 / 5) + (1.97^2 / 5)] tobt = -1.2 / √[(4.2849 / 5) + (3.8809 / 5)] tobt = -1.2 / √[0.85698 + 0.77618] tobt = -1.2 / √1.63316 tobt = -1.2 / 1.27941 tobt ≈ -0.9378

d. Conclusion: Comparing the calculated tobt value (-0.9378) with the critical t-value (±2.262), we see that the absolute value of tobt is less than tcrit. Therefore, we fail to reject the null hypothesis (H0).

In conclusion, there is no significant difference in the mean number of errors between the two groups (Classical and Rock) based on the data provided at the 5% significance level.