Problem 1 Steven collected data from 20 college students on their emotional resp

Question

Problem 1 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 scores are as follows: 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 alpha level was the test significant? b) What is your null and alternative hypothesis? Given the results did you reject or fail to reject the null and why? (Use instructions on page 349 of your textbook, under Hypothesis Tests with the t Distribution to conduct Excel analysis).

Explanation / Answer

a) For these data, the computed statistics would include the following:
n = 20
X = 175/20 = 8.75
Degrees of freedom = n - 1 = 20 – 1 = 19

The standard deviation would be computed thus:
Find the deviation from the mean for all 20 of the scores, square each of those results, and sum them = 425.75. Then divide this result by n – 1, or 19.
This results in 22.4078947.
The standard deviation is the square root of this result, or 4.73369779 Therefore, 4.7 will be used as the standard deviation.

For calculating the t-test score, the following formula is used:
t = (sample mean – population mean) / (standard deviation / square root of “n”)
t = (8.75 – 6) / (4.7 / 4.5) t = 2.75 / 1.04 t = 2.64
The mean for this test would be 8.75 with a t-test score of 2.64.

According to the table provided for Critical Values of Bennett et al, for the Degrees of freedom of 19 at the 0.05 level, a score of 2.093 would be considered significant. Also since the t-test score was 2.64 and obviously greater than the critical value of 2.093, the test would be considered significant.

Finally, by using the P-Value Calculator, the two-tailed P-value would be 0.0161. This difference would be considered to be statistically significant

b) For the above information the null and alternative hypothesis would be
H0 : = 6 H a : 6

This is because the population mean is known to be 6.

Given the results , I reject the null because 2.598 > 2.093.of the t-test, the difference in the sample was statistically significant, so the null hypothesis was rejected

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