Hello this is my question On the basis of his experience with student-athletes,
ID: 3047492 • Letter: H
Question
Hello this is my question
On the basis of his experience with student-athletes, a college director bleves that academic average is influenced by participation in team athletics. He decides to test this idea. He randomly selects 25 athletes from various college teams. The academic average for the sample athletes is 68 with a standard deviation of 20. The academic average for the students in general at his college is 76 (p) Does academic average differ between student-athletes and students in general? Solve this problem explicitly using the appropriate null hypothesis testing procedure, and set the alpha at 5% Formulate the null and research hypothesis Research Null. 1 = 2 What are the characteristics of the comparison distribution? Mean: Standard deviation Determine the cutoff sample score (critical test statistic value) on the comparison distribution at which the null hypothesis should be rejected. If your test is two-tailed just give the absolute value. Determine your sample score on the comparison distribution (i.e. Calculate the appropriate test statistic value(s).) Decide whether or not to reject the null hypothesis The test statistic is student-athletes and students in general ( ; we therefore | the null hypothesis. The academic average | significantly betweenExplanation / Answer
Solution:
Here, we have to use one sample t test for the population mean.
Null: µ1 = 76
Research: µ1 76
(Two tailed test.)
Mean = Xbar = 68
Standard deviation = S = 20
Sample size = n = 25
df = n – 1 = 25 – 1 = 24
Level of significance = alpha = 5% = 0.05
Critical value = 2.0639 (and -2.0639)
Test statistic = t = (Xbar - µ) / [S/sqrt(n)]
Test statistic = t = (68 – 76) / [20/sqrt(25)]
Test statistic = t = -8/4 = -2.00
(by using t-table)
The test statistic is greater than lower critical value. Absolute test statistic value is less than absolute critical value. So, we therefore do not reject the null hypothesis.
The academic average is not different significantly between student athletes and students in general at 5% level of significance.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.