A total of 23 High School students were admitted to State University. Of those students, 7 were offered athletic scholarships. The school’s guidance counselor looked at their composite ACT scores (given in the table below), wondering if State U might admit people with lower scores if they also were athletes
composite ACT test scores
non atheletes
25
22
19
25
24
25
24
23
21
27
29
26
30
27
26
23
atheletes
22
21
24
27
19
23
17
Test an appropriate hypothesis and state your conclusion. Use alpha = .05 for your test. Show the null and alternative hypothesis, the p-value, your conclusion to reject or not, and finally a summarizing statement regarding your conclusion.
Then create a 90% confidence interval and tell me what it means in terms of the problem
I DON'T WANT THE ANSWER JUST NEED CALCULATIONS AND SET UP
To test an appropriate hypothesis and state the conclusion, we can perform a two-sample t-test to compare the composite ACT test scores of non-athletes and athletes at State University.
Null hypothesis (H0): There is no significant difference in the composite ACT test scores between non-athletes and athletes at State University.
Alternative hypothesis (Ha): There is a significant difference in the composite ACT test scores between non-athletes and athletes at State University.
To conduct this test, we need to calculate the t-statistic and the associated p-value. The t-statistic can be calculated using the following formula:
t = (mean1 - mean2) / sqrt((s1^2 / n1) + (s2^2 / n2))
Where:
mean1 and mean2 are the sample means of the non-athletes and athletes, respectively.
s1 and s2 are the sample standard deviations of the non-athletes and athletes, respectively.
n1 and n2 are the sample sizes of the non-athletes and athletes, respectively.
The p-value can then be determined from the t-statistic and the degrees of freedom.
To calculate the 90% confidence interval, we can use the following formula:
CI = (mean1 - mean2) ± (t * SE)
Where:
mean1 and mean2 are the sample means of the non-athletes and athletes, respectively.
t is the critical t-value for a 90% confidence level and the given degrees of freedom.
SE is the standard error of the difference, calculated as sqrt((s1^2 / n1) + (s2^2 / n2)).
The confidence interval will provide us with a range within which we can be 90% confident that the true difference in means lies.
Once we have the p-value and confidence interval, we can make conclusions based on the results. If the p-value is less than the chosen significance level (alpha = 0.05), we can reject the null hypothesis and conclude that there is a significant difference in the composite ACT test scores between non-athletes and athletes at State University. If the p-value is greater than the significance level, we fail to reject the null hypothesis and conclude that there is no significant difference.
The confidence interval will provide a range of plausible values for the difference in means. If the interval does not include zero, it suggests that there is a significant difference between the two groups. If the interval includes zero, it suggests that the difference could be due to random variation, and there is no significant difference.