1. (Three parts; 9 marks in total) An introductory Statistics instructor wished
ID: 2928767 • Letter: 1
Question
1. (Three parts; 9 marks in total) An introductory Statistics instructor wished to determine whether there was a difference in lab marks between lab sections. The variable being explored is the mark on lab 1. There were 12 lab sections and 5 graders. Suppose each lab section has 25 students. The graders and the number of lab sections they marked are summarized below: Grader Number of Lab Sections Parameters (subscript is the lab section number) For each question below, you do "NOT" need to show all steps of a hypothesis test (in any case, there is no data presented). Just do the following: clearly define the most appropriate test procedure, state the null and alternative hypothesis in terms of the parameters above, and calculate the degrees of freedom. There is a single appropriate test for each part. Assume all the required assumptions are satisfied. Statistics 252-Homework # 2 Questions-Fall 2017 a) (3 marks) Are there any differences in mean lab marks for the 12 lab sections? b) (3 marks) Do any of the graders have different mean lab marks for their corresponding lab sections? c) (3 marks) Suppose there are three lecture sections, where each lecture section has four lab sections. Suppose lab sections 1-4 are for lecture Q, lab sections 5-8 are for lecture R, and lab sections 9 12 are for lecture S. Do any of the lecture sections have different mean lab marks between their corresponding lab sections?Explanation / Answer
Answer to the part a)
in order to compare all the means , the best test is Two Factor ANOVA
Reason: it compares the mean values of graders and lab sections together
The df (row) = 5 -1 = 4
Df (coulmn) = 12-1 = 11
Null hypothesis: All mean values are equal
Alternate hypothesis: Atlest one mean value is not equal
.
.
Answer to part b)
In this case the best test would be single Factor ANOVA
df = 5-1 =4
Null hypothesis: All mean values are equal
Alternate hypothesis: Atleast one mean value is not equal
.
Answer to part c)
F test works for this case
the df numerator = 3 - 1 = 2
df denominator = 12-3 = 9
df total = 9+2 = 11
Null hypothesis : all lectures have same mean
Alternate hypothesis: all lectures have different mean
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