Question

Suppose that a MBA level stat course is taught using four different methods of instruction: (1) 100% online; (2) a “half and half” format where one week the class meets for a lecture, the next week, material is posted online, etc.; (3) traditional weekly lecture meetings plus supplementary material posted online; and (4) traditional weekly lecture meeting with no use of the web.     Twenty students are surveyed from each course and are asked to estimate the average number of hours per week that they spent on the course, including time spent attending lectures if the course had any. The results appear in the included data file.

(a)    Create boxplots for these four sets of data (all on the same graph). Based on the plots, what do you think about the ANOVA assumptions of normal populations and equal variances? (You’ll test these in Part (d), I’m just interested in a visual interpretation here.)   Your answer should include justification….. don’t just say ‘yup’ or ‘nope.’

(b)    Create interval plots for this data (four intervals on the same plot) so that we can visually compare the four groups. This is the plot that shows a confidence interval for each unknown population mean. Based on the plot, do you believe that the population mean time spent is the same for all groups? Again, show the reasoning behind your answer.

(c)    Let μ1 represent the population mean time spent for method (1), μ2 the mean time spent for method (2), and so on. Test the null hypothesis that all means are equal at the 0.05 level of significance, versus the usual ANOVA alternative.

(d)    Is there evidence of violations of the usual ANOVA assumptions of equal variances and normal populations? Set up and perform appropriate TESTS at the α = 0.05 level of significance.  

(e)    If your answer in Part (c) was to “reject H0” then perform an appropriate statistical procedure to determine which means are different from which other means (a visual inspection is not sufficient.).   If differences exist, be sure to report the ‘direction’ of the d

Time	Method
6.0	OnLine
7.1	OnLine
5.9	OnLine
8.9	OnLine
7.3	OnLine
6.1	OnLine
7.7	OnLine
7.1	OnLine
6.3	OnLine
8.4	OnLine
7.9	OnLine
6.9	OnLine
6.9	OnLine
6.3	OnLine
6.7	OnLine
6.0	OnLine
6.4	OnLine
8.3	OnLine
7.5	OnLine
8.2	OnLine
5.0	Half&Half
5.9	Half&Half
8.1	Half&Half
7.9	Half&Half
7.1	Half&Half
7.9	Half&Half
7.6	Half&Half
4.7	Half&Half
6.8	Half&Half
6.2	Half&Half
7.4	Half&Half
5.0	Half&Half
5.8	Half&Half
6.9	Half&Half
6.9	Half&Half
5.7	Half&Half
5.8	Half&Half
6.7	Half&Half
7.4	Half&Half
6.9	Half&Half
5.7	LecturePlus
5.5	LecturePlus
7.0	LecturePlus
5.9	LecturePlus
4.1	LecturePlus
7.1	LecturePlus
6.6	LecturePlus
6.4	LecturePlus
5.4	LecturePlus
5.7	LecturePlus
6.1	LecturePlus
4.8	LecturePlus
7.2	LecturePlus
6.2	LecturePlus
4.9	LecturePlus
6.3	LecturePlus
5.4	LecturePlus
6.3	LecturePlus
6.1	LecturePlus
5.5	LecturePlus
5.3	Lecture
5.1	Lecture
0.0	Lecture
6.4	Lecture
4.9	Lecture
7.3	Lecture
5.6	Lecture
6.5	Lecture
6.2	Lecture
4.8	Lecture
7.1	Lecture
7.1	Lecture
5.8	Lecture
5.9	Lecture
5.9	Lecture
5.7	Lecture
7.3	Lecture
5.2	Lecture
5.9	Lecture
6.4	Lecture

Answer 1

(a)

From the above box plots we see that lengths of boxes are not same so assumption of equal variances does not hold.

From the box plots, we see that third quartile-median is almost equal to median-first quartile for "Lecture" only so the assumption of normality does not hold here.

(b)

From the above plots we see that the population mean time spent is not same for all groups since 95% C.I.s of "LecturePlus" and "Online" are disjoint as well as 95% C.I.s of "Lecture" and "Online" are disjoint.

(c)

One-way ANOVA: Time versus Method

Source DF SS MS F P
Method 3 23.97 7.99 6.51 0.001
Error 76 93.26 1.23
Total   79 117.24

Since p-value<0.05 so we reject H0 at 0.05 level and conclude that population means time spent for 4 methods are not same.

(d)

Test for Equal Variances: Time versus Method

95% Bonferroni confidence intervals for standard deviations

Method N Lower StDev Upper
Half&Half 20 0.72618 1.02458 1.67793
Lecture 20 1.09983 1.55177 2.54130
LecturePlus 20 0.56134 0.79200 1.29704
OnLine 20 0.64324 0.90756 1.48628

Bartlett's Test (Normal Distribution)
Test statistic = 10.26, p-value = 0.016

Levene's Test (Any Continuous Distribution)
Test statistic = 0.43, p-value = 0.730
Under normality assumption, the assumptions of equal variances does not hold (since p-value<0.05).

Since p-value of AD test<0.05 so normality assumption does not hold here.

(e)

Tukey 95% Simultaneous Confidence Intervals
All Pairwise Comparisons among Levels of Method

Individual confidence level = 98.97%

Method = Half&Half subtracted from:

Method Lower Center Upper ---------+---------+---------+---------+
Lecture -1.786 -0.865 0.056 (-------*------)
LecturePlus -1.596 -0.675 0.246 (------*-------)
OnLine -0.411 0.510 1.431 (------*-------)
---------+---------+---------+---------+
-1.2 0.0 1.2 2.4

Method = Lecture subtracted from:

Method Lower Center Upper ---------+---------+---------+---------+
LecturePlus -0.731 0.190 1.111 (-------*------)
OnLine 0.454 1.375 2.296 (------*-------)
---------+---------+---------+---------+
-1.2 0.0 1.2 2.4

Method = LecturePlus subtracted from:

Method Lower Center Upper ---------+---------+---------+---------+
OnLine 0.264 1.185 2.106 (-------*-------)
---------+---------+---------+---------+
-1.2 0.0 1.2 2.4
From the obove C.I.s we observed that population means of "Lecture" and "Online" and population means of "LecturePlus" and "Online" are significantly different since the corresponding C.I.s do not contain zero.