Professor Jane Newman teaches an introductory calculus course. She wanted to tes
ID: 3313554 • Letter: P
Question
Professor Jane Newman teaches an introductory calculus course. She wanted to test the belief that success in her course is affected by high school performance. She collected the randomly selected data listed below and ran an ANOVA test as shown below. The data in the "High School Record" table represents performance of the student in Jane's calculus class. Anova: Single Factor High School Record dFair o SUMMARY 90 86 70 88 61 93 52 80 73 80 60 Groups Count Sum Average Variance 87.4 5 437 7 484 69.14285714 117.1428571 6 357 60 Good 23.8 Fair Poor 59.5 45.5 62 50 6570 ANOVA 83 Source of Variation df MS P-value F crit Between Groups 2162.442857 2 1081.221429 15.81415676 0.000202214 3.682320344 Within Groups1025.557143 15 68.37047619 Total 3188 17 If we set at 0.05, what would we conclude about the ANOVA test? State the null and alternative hypotheses and result clearly. Give a reason for your result. a)Explanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: 1 = 2 = 3
Alternative hypothesis: At-least one of the is not equal.
Formulate an analysis plan. For this analysis, the significance level is 0.05.
Analyze sample data.
F statistics is given by:-
F = 15.814
Fcritical = 3.682
The P-value = 0.000202
Interpret results. Since the P-value (0.0002) is less than the significance level (0.05), we have to reject the null hypothesis.
Conclusion:-
Reject H0, There is sufficient evidence for significant differences between the high school record for different success.
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