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?
To conduct a one-sample t-test, we need to calculate the t-test score, mean, and determine if the test is significant, as well as the p-value level at which it is significant.
To calculate the t-test score, follow these steps:
Step 1: Calculate the mean of the scores.
Sum of scores = 165.0
Mean = Sum of scores / Number of scores = 165.0 / 20 = 8.25
Step 2: Calculate the standard deviation of the scores.
Subtract the mean from each score, square the result, sum all the squared differences, divide by (n-1) for sample standard deviation.
Sum of squared differences = (5.0-8.25)^2 + (5.0-8.25)^2 + (10.0-8.25)^2 + (3.0-8.25)^2 + (13.0-8.25)^2
+ (13.0-8.25)^2 + (7.0-8.25)^2 + (5.0-8.25)^2 + (15.0-8.25)^2 + (14.0-8.25)^2
+ (18.0-8.25)^2 + (8.0-8.25)^2 + (12.0-8.25)^2 + (10.0-8.25)^2 + (7.0-8.25)^2
+ (3.0-8.25)^2 + (4.0-8.25)^2 + (3.0-8.25)^2 = 305.25
Sample standard deviation = sqrt(Sum of squared differences / (Number of scores - 1)) = sqrt(305.25 / 19) ≈ 3.27
Step 3: Calculate the standard error of the mean.
Standard error of the mean = Sample standard deviation / sqrt(Number of scores) = 3.27 / sqrt(20) ≈ 0.73
Step 4: Calculate the t-test score.
t-test score = (Mean - Test mean) / Standard error of the mean = (8.25 - 6) / 0.73 ≈ 3.77
The t-test score is approximately 3.77.
Next, we need to determine if the test is significant and at what p-value level. To do this, we consult the t-distribution table or use statistical software. Assuming a two-tailed test, the critical t-value at a 5% significance level with 19 degrees of freedom is approximately 2.10. Since the calculated t-test score (3.77) is greater than the critical t-value (2.10), the test is significant.
The p-value corresponds to the level of significance at which the test is significant and can be obtained from the t-distribution table or statistical software. In this case, the p-value is less than 0.05, indicating significance at the 5% level. However, the exact p-value is not provided in the given information.
Therefore, the t-test score is approximately 3.77, the mean is 8.25, the test is significant, and it is significant at the p < 0.05 level.
To conduct a one-sample t-test, we can follow these steps:
Step 1: Define the null and alternative hypotheses.
The null hypothesis (H0) states that there is no significant 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.
Step 2: Calculate the sample mean and the standard deviation.
The sample mean can be calculated by summing all the scores and dividing by the number of scores (20 in this case).
Sample 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 = 9.05
The sample standard deviation can be calculated using the following formula:
Standard Deviation = sqrt(sum((score - mean)^2) / (n-1))
Step 3: Calculate the t-test score.
The t-test score can be calculated using the following formula:
t = (sample mean - population mean) / (sample standard deviation / sqrt(n))
In this case, the population mean is 6, the sample mean is 9.05, the sample standard deviation needs to be calculated, and n is the total number of students (20).
Step 4: Determine the p-value.
Using the t-test score and the degrees of freedom (n-1), we can determine the p-value by referring to a t-distribution table or using statistical software.
Step 5: Interpret the results.
If the p-value is less than the chosen significance level (e.g., 0.05), we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
Now, let's calculate the t-test score, mean, and check for significance.
Step 1: Given the null and alternative hypotheses are:
H0: μ = 6 (Population mean is 6)
Ha: μ ≠ 6 (Population mean is not equal to 6)
Step 2: Calculating the sample mean:
The sample mean is already calculated in Step 2.
Sample Mean = 9.05
Step 3: Calculating the sample standard deviation:
Using the formula mentioned earlier:
Standard Deviation = sqrt(sum((score - mean)^2) / (n-1))
Standard Deviation = sqrt((5.0-9.05)^2 + (5.0-9.05)^2 + ... + (4.0-9.05)^2 + (3.0-9.05)^2) / 19)
Step 4: Calculating the t-test score:
Using the formula mentioned earlier:
t = (sample mean - population mean) / (sample standard deviation / sqrt(n))
Step 5: Determining the p-value:
Using the t-test score and degrees of freedom (n-1), we can determine the p-value by referring to a t-distribution table or statistical software.
Once we have the p-value, we compare it to the chosen significance level (e.g., 0.05) to determine if the test is significant.
Unfortunately, the scores provided in the question are incomplete, and we need all 20 scores to calculate the t-test score, mean, and determine significance. Please provide the rest of the scores so that we can help you further.