Professor Jane Newman teaches an introductory calculus course. She wanted to tes
ID: 3294431 • Letter: P
Question
Professor Jane Newman teaches an introductory calculus course. She wanted to test the belief that suc cess 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.
High School Record Anova Single Factor
Good Fair Poor Summary Groups Count Sum Average Variance
90 80 60 Good 5 437 87.4 23.8
86 70 60 Fair 7 484 69.14285714 117.1428571
88 70 60 Poor 6 357 59.5 45.5
93 52 62
80 73 50
65 70
83
Anova
Source of Variation SS df MS F P-value Fcrit
Between Groups 2162.442857 2 1081.221429 15.81415676 0.000202214 3.682320344
Within Groups 1025.557143 15 68.37047619
Total 3188 17
a) 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.
Explanation / Answer
null hypothesis::Ho: all groups have equal mean score for performance.
alternate hypothesis:Ha:at least 2 of the groups have different mean score for performance.
) as p value 0.0002 is less then 0.05 level we reject null hypothesis
there is sufificent evidence to conclude that at least 2 of the groups have different mean score for performance at 0.05 level of significance.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.