Question

One of the greatest debates in the history of great debates: Who’s the G.O.A.T (greatest of all time): Michael Jordan or LeBron James? Although this is a highly complex question, let’s try to get at it by comparing the two on just one performance measure: average points per game (PPG). Below is a table with the average PPG per season for each Jordan and James:

1. This is a non-traditional example that will require a t-test. However, the book only discusses the use of t-tests in terms of comparing sample means. Here, we are going to apply the t-test to two sets of data, each collected from one individual (i.e., we’re comparing multiple data points from one individual to multiple data points from another individual). With this said, which type of t-test do you believe to be most appropriate here: a related samples or independent samples? Please explain why.

2. State the appropriate alternative and null hypotheses (using the statistical notation for stating mathematical relationships) to test the hypothesis that Jordan and James will significantly differ with respects to average PPG.

3. Assuming an alpha level of 0.05, provide the critical and obtained values for this hypothesis test.

4. Assuming an alpha level of 0.05, make a decision as to whether the result of the hypothesis test is significant or insignificant, as well as whether you should reject or fail to reject the null hypothesis, being sure to explain why. Also, interpret the results of the hypothesis test in everyday language (i.e., do Jordan and James differ with respects to average PPG?).

SEASON	Jordan PPG	James PPG
1	28.2	20.9
2	22.7	27.2
3	37.1	31.4
4	35	27.3
5	32.5	30
6	33.6	28.4
7	31.5	29.7
8	30.1	26.7
9	32.6	27.1
10	26.9	26.8
11	30.4	27.1
12	29.6	25.3
13	28.7	25.3
14	22.9	26.4
15	20

Answer 1

This is t-test of a related samples Null hypothesis: That is, Jordan and James will be same with respects to average PPG Alternative hypothesis: Thatis, Jardan and Jaml fectsto aweragPpG. เป้ EXCEL (PH Stat) output:

Paired t Test Data Hypothesized Mean Difference Level of significance 0 0.05 Intermediate Calculations Sample Size DBar Degrees of Freedom 15 4.1467 14 5.5993 1.4457 2.8682 Standard Error t Test Statistic Two-Tail Test Lower Critical Value 2.1448 2.1448 0.0124 Upper Critical Value Value Reject the null hypothesis The critical values arett2.1448, 2.1448 The value of obtained test statistic is t 2.8682

Conclusion: Use a significance level, α 0.05 Here the value of test statistic is 2.8682, which is greater than the value of tt 2.1448 That is, tcaic(-2.8682) >tcrit-2.1448. Therefore, by the rejection rule, reiect the null hvpothesis Ho Interpretation There is sufficient evidence at level of significance α = 0.05 That is, Jordan and James will significantly differ with respects to average PPG.