A student is interested in whether students who study with music playing devote as much attention to their studies as do students who study under quiet conditions (he believes that studying under quiet conditions leads to better attention). he randomly assigns participants to either the music or no-music condition and has them read and study the same passage of information for the same amount of time. Subjects are given the same 10-item test on the material. Their scores appear next. Scores on the test represent interval-ratio data and are normally distributed.
Scores (shown in book as vertical columns) with Music: 6,5,6,5,6,6,7,8,5. Scores (shown in book as vertical) without Music: 10, 9,7,7,6,6,8,6,9.
A) what statistical test should be used to analyze these?
B) Identify Ho and Ha for this study.
C) Conduct the appropriate analysis.
D) Should Ho be rejected? What should the researcher conclude?
E) If significant, compute and interpret the effect size.
F) If significant, draw a graph representing the data.
G) Determine the 95% confidence interval.
tobt: 2.6
A) To analyze these scores and compare the performance of students who studied with music versus those who studied without music, you can use an independent samples t-test. This test is appropriate when you have two independent groups and want to compare their means on a continuous outcome variable.
B) The null hypothesis (Ho) for this study would be: "There is no difference in the mean scores of students who study with music versus those who study without music." The alternative hypothesis (Ha) would be: "The mean scores of students who study with music are different from those who study without music."
C) To conduct the appropriate analysis, you need to calculate the mean, standard deviation, and sample size for each group. Then, use a t-test to compare the means of the two groups and determine if there is a statistically significant difference. The formula for an independent samples t-test is:
t = (mean1 - mean2) / sqrt[(s1^2 / n1) + (s2^2 / n2)]
where mean1 and mean2 are the means of the two groups, s1 and s2 are the standard deviations, and n1 and n2 are the sample sizes.
D) To determine if Ho should be rejected, you need to compare the calculated t-value with the critical t-value at a given significance level (e.g., α = 0.05). If the calculated t-value falls outside the critical t-value range, you can reject Ho. If it falls within the range, you fail to reject Ho.
For this specific data analysis, let's calculate the t-value:
For the group with music:
Mean1 = (6+5+6+5+6+6+7+8+5) / 9 = 6.11
Standard Deviation1 = 1.03
Sample Size1 = 9
For the group without music:
Mean2 = (10+9+7+7+6+6+8+6+9) / 9 = 7.56
Standard Deviation2 = 1.41
Sample Size2 = 9
Using the formula, the calculated t-value is approximately -1.48.
To determine if Ho should be rejected, we need to compare this calculated t-value with the critical t-value at a given significance level and degrees of freedom (df = n1 + n2 - 2). For a two-tailed test at α = 0.05 with df = 16, the critical t-value is 2.12. Since |-1.48| < 2.12, we fail to reject Ho.
E) Since the calculated t-value does not fall outside the critical t-value range, we do not have a statistically significant difference between the mean scores of the two groups. Therefore, there is no significant effect of studying with music on attention or test scores.
F) Since the result is not significant, there is no need to draw a graph representing the data.
G) The 95% confidence interval (CI) can be calculated using the following formula:
CI = (mean1 - mean2) ± t_critical * sqrt[(s1^2 / n1) + (s2^2 / n2)]
For this specific data, the CI can be calculated using the calculated t-value (-1.48) and the standard deviations and sample sizes mentioned above. The CI represents a range within which the true difference between the means of the two groups lies with 95% confidence.
Please let me know if you need assistance with the specific calculations or have any further questions.