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:

Question

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?
(Use instructions on page 349 of your textbook, under Hypothesis Tests with the t Distribution to conduct SPSS or Excel analysis).

Answer 1

Have followed the instructions on page 349?

Answer 2

To conduct a one-sample t-test and answer the questions, we need to follow these steps:

Step 1: Set up the hypotheses.
The null hypothesis (H0) states that there is no difference between the sample mean and the established mean of 6. The alternative hypothesis (Ha) states that there is a significant difference between the sample mean and the established mean of 6. So, in this case:
H0: The population mean = 6
Ha: The population mean ≠ 6

Step 2: Calculate the mean of the sample.
For the given data, calculate the mean by adding up all the scores and dividing by the total number of scores. In this case, we have 20 scores, so the mean would be:

Mean = (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

Mean = 179 / 20 = 8.95

So, the mean of the sample is 8.95.

Step 3: Calculate the t-test score.
To calculate the t-test score, we need to use the formula:

t = (mean - established mean) / (standard deviation / √sample size)

In this case, the established mean is 6, the mean from the sample is 8.95, and the sample size is 20. However, we still need to calculate the standard deviation. So, let's do that next.

Step 4: Calculate the standard deviation.
To calculate the standard deviation, we need to find the squared difference between each individual score and the mean, sum them up, divide by the sample size minus 1 (since we are calculating the sample standard deviation), and then take the square root. The formula for the sample standard deviation is:

s = √((Σ(xi - x̄)^2) / (n - 1))

Using the given data, the calculations for each step are as follows (rounded to one decimal place):

(5.0 - 8.95)^2 = 14.6
(5.0 - 8.95)^2 = 14.6
(10.0 - 8.95)^2 = 1.1
(3.0 - 8.95)^2 = 31.4
(13.0 - 8.95)^2 = 15.9
(13.0 - 8.95)^2 = 15.9
(7.0 - 8.95)^2 = 6.0
(5.0 - 8.95)^2 = 14.6
(15.0 - 8.95)^2 = 37.8
(14.0 - 8.95)^2 = 24.3
(18.0 - 8.95)^2 = 83.5
(8.0 - 8.95)^2 = 0.9
(12.0 - 8.95)^2 = 10.2
(10.0 - 8.95)^2 = 1.1
(7.0 - 8.95)^2 = 6.0
(3.0 - 8.95)^2 = 31.4
(15.0 - 8.95)^2 = 37.8
(4.0 - 8.95)^2 = 21.2
(3.0 - 8.95)^2 = 31.4

Sum of squared differences = Σ(xi - x̄)^2 = 441.4

Now we can calculate the standard deviation:

s = √(441.4 / (20 - 1)) = √(441.4 / 19) = √23.23 = 4.82

So, the standard deviation (s) is 4.82.

Step 5: Calculate the t-test score (continued).
Using the mean (x̄ = 8.95), established mean (μ = 6), and standard deviation (s = 4.82), we can calculate the t-test score:

t = (mean - established mean) / (standard deviation / √sample size)
t = (8.95 - 6) / (4.82 / √20)
t = 2.95 / (4.82 / 4.47)
t = 2.95 / 1.08 = 2.74

So, the t-test score is 2.74.

Step 6: Determine the significance level.
To determine the significance level or p-value, we need to consult a t-table or use statistical software like SPSS or Excel. The p-value represents the probability of obtaining results as extreme as the observed data, assuming the null hypothesis is true.

Using SPSS or Excel, enter the given data and perform a one-sample t-test with a null hypothesis of 6 and a mean of 8.95, along with the sample standard deviation (4.82) and sample size (20). The t-test result will provide you with the exact p-value.

Step 7: Evaluate the results.
Compare the obtained p-value with the desired significance level (usually denoted as α). If the p-value is less than α (typically 0.05), then the test is significant, indicating strong evidence against the null hypothesis. Otherwise, the test is not significant, and we fail to reject the null hypothesis.

Once you have the exact p-value from the t-test analysis (e.g., from SPSS or Excel), compare it to α (e.g., 0.05) to determine whether the test is significant or not.

Note: Only with the given data, I can walk you through steps 1 to 5 to find the t-test score. However, conducting the actual t-test and obtaining the p-value requires statistical software or a t-table.