Please find questions in attached word document related to inferential statistics (Mathematical statistics)

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Quiz 1

1. One football player was tired of teachers and students making comments under the general assumption that student-athletes were less intelligent or inferior students to those not playing sports. To get rid of the “dumb jock” labels, he compared the results on a college readiness assessment of student-athletes and non-athletes. On this particular assessment, a student could receive the following scores: “under-prepared,” “on-track,” or “college-ready.” He recorded the following chart:

	Under-Prepared	On-Track	College-Ready
Athlete	88	180	224
Non-Athlete	127	250	300

0. Which statistical test would you use?

0. Give a brief one-sentence statement that led you to choose that test.

0. Provide your null (H0) and alternative (Ha) hypothesis.

0. Perform the test.

0. What is your statistic?

0. Which value are you comparing it to (critical value)?

0. What decision does this inform you to make?

0. Write a one or two sentence conclusion.

0. What limitations or other considerations would you point out for this study?

Answers:

(a)

According to the given scenario, I will use Chi Square Test of Independence.

(b)

Here, the objective is to test whether college readiness assessment of student - athletes and non-athletes are independent.

(c)

H0: Null Hypothesis is: College readiness assessment (Under-prepared, On-track and College-ready) is independent of Athlete and Non-athlete

Ha: Alternative Hypothesis is: College readiness assessment (Under-prepared, On-track and College-ready) is dependent of Athlete and the Non-athlete

(d) Test and its statistics (E)

Observed Frequencies:

Under-prepared On-track College-ready Total

Athlete 88 180 224 492

Non-athlete 127 250 300 677

Total 215 430 524 1169

Assuming H0, the Expected Frequencies are calculated as follows:

Under-prepared On-track College-ready Total

Athlete 180.98 220.54 492

Non-athlete 124.51 249.02 303.46 677

Total 215 430 524 1169

O E (O- E) 2/E

88 90.49 0.07

180 180.98 0.01

224 220.54 0.05

127 124.51 0.05

250 249.02 0.00

300 303.46 0.04

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$\chi ^{2}$ = 0.221

By Technology, p-value = 0.8954.

Since p-value is greater than $\alpha$ = 0.05, Fail to reject the null hypothesis.

Conclusion: The data support the claim that College readiness assessment (Under-prepared, On-track and College-ready) is independent of Athlete and Non-athlete

(f)

Take $\alpha$ = 0.05

ndf = 2

From Table,

Critical value of $\chi ^{2}$ = 5.9915

(g)

Since the calculated value of $\chi ^{2}$ = 0.221 is less than critical value of $\chi ^{2}$ = 5.9915. Fail to reject null hypothesis.

(h) Conclusion: The data support the claim that College readiness assessment (Under-prepared, On-track and College-ready) is independent of Athlete and Non-athlete

(i) According to the given scenario, the limitation for this study is that intelligence of a student depends not only on sports but many other factors, like his attendance, etc.