Steven collected data from 20 college students on their emotional responses to classical music. Students listened to two 30-second segments from “The Collection from the Best of Classical Music.” After listening to a segment, the students rated it on a scale from 1 to 10, with 1 indicating that it “made them very sad” to 10 indicating that it “made them very happy.” Steve computes the total scores from each student and created a variable called “hapsad.” Steve then conducts a one-sample t-test on the data, knowing that there is an established mean for the publication of others that have taken this test of 6. The following is the scores:
5.0 5.0
10.0 3.0
13.0 13.0
7.0 5.0
5.0 15.0
14.0 18.0
8.0 12.0
10.0 7.0
3.0 15.0
4.0 3.0
a) Conduct a one-sample t-test. What is the t-test score? What is the mean? Was the test significant? If it was significant,at what P-value level was it significant?
b) What is your null and alternative hypothesis? Given the results did you reject or fail to reject the null and why?
the null and alternative hypothesis would be
Ho: ƒÊ = 6 Ha: ƒÊ �‚ 6
To conduct a one-sample t-test, you need to calculate the t-test score, mean, and determine if the test is significant.
a) To find the t-test score, you will follow these steps:
1. Calculate the mean of the data: calculate the average of the total scores (hapsad) from the 20 college students. Add up all the scores and divide by 20.
(5.0 + 5.0 + 10.0 + 3.0 + 13.0 + 13.0 + 7.0 + 5.0 + 15.0 + 14.0 + 18.0 + 8.0 + 12.0 + 10.0 + 7.0 + 3.0 + 15.0 + 4.0 + 3.0) / 20 = 163 / 20 = 8.15
2. Calculate the standard deviation (SD) of the data: subtract the mean from each score, square the result, sum up the squared values, divide by the number of scores minus one (n-1), and take the square root of the result.
First, subtract the mean (8.15) from each score to get the deviations:
(5.0-8.15), (5.0-8.15), (10.0-8.15), (3.0-8.15), (13.0-8.15), (13.0-8.15), (7.0-8.15), (5.0-8.15), (15.0-8.15), (14.0-8.15), (18.0-8.15), (8.0-8.15), (12.0-8.15), (10.0-8.15), (7.0-8.15), (3.0-8.15), (15.0-8.15), (4.0-8.15), (3.0-8.15)
Square each deviation, sum them up, and divide by 19 (n-1 = 20-1):
(5.0-8.15)^2 + (5.0-8.15)^2 + (10.0-8.15)^2 + (3.0-8.15)^2 + (13.0-8.15)^2 + (13.0-8.15)^2 + (7.0-8.15)^2 + (5.0-8.15)^2 + (15.0-8.15)^2 + (14.0-8.15)^2 + (18.0-8.15)^2 + (8.0-8.15)^2 + (12.0-8.15)^2 + (10.0-8.15)^2 + (7.0-8.15)^2 + (3.0-8.15)^2 + (15.0-8.15)^2 + (4.0-8.15)^2 + (3.0-8.15)^2
Sum = 39.27
SD = sqrt(sum/19) = sqrt(39.27/19) ≈ 1.546
3. Calculate the t-score: subtract the established mean (6) from the sample mean (8.15), divide by the standard deviation (1.546), and multiply by the square root of the number of data points (20).
t = (8.15 - 6) / (1.546 / sqrt(20)) ≈ 1.404
The t-test score is approximately 1.404.
4. Determine the significance: To determine if the test is significant, you need to compare the t-test score to a critical value or calculate the p-value.
If you have an alpha level (level of significance, e.g., 0.05), you can look up the critical t-value from a t-distribution table or use statistical software.
If you have a p-value, you can compare it to the alpha level. If the p-value is less than the alpha level, the test is significant.
Unfortunately, you haven't provided an alpha level or p-value, so we cannot determine if the test is significant or at what p-value level it is significant.
b) The null hypothesis (H0) assumes no significant difference between the sample mean and the established mean. The alternative hypothesis (Ha) assumes a significant difference between the sample mean and the established mean.
In this case, the null hypothesis would be: The emotional responses to classical music by college students have a mean equal to 6.
The alternative hypothesis would be: The emotional responses to classical music by college students have a mean different from 6.
To determine whether to reject or fail to reject the null hypothesis, you would typically compare the p-value to the chosen alpha level (level of significance). However, since you haven't provided the p-value, we cannot conclude whether to reject or fail to reject the null hypothesis.