Question

Nielsen Media Research collects information on daily TV viewing time, in hours, and publishes it findings in Time Spent Viewing. A random sample of daily viewing times for men, women, teens and children is obtained. The data are contained in the table below.

a) Provide an appropriate graphical summary for the data.

b) Do the data provide sufficient evidence to conclude that a difference exists among the means of the four view groups? Provide a formal hypothesis test as part of the supporting evidence for your conclusion.  

Men	Women	Teens	Children
5.31	6.65	1.80	1.10
7.25	6.19	3.22	1.07
1.11	4.64	1.28	3.93
5.79	5.54	1.43	2.93
4.98	3.92	3.09	2.91
2.54	6.43	5.40	5.33
3.85	3.38	1.95	3.64
5.01	3.50	3.66	3.50
9.87	3.89	2.76	5.58
8.65	0.68	4.78	1.99
2.48	7.26	1.47	4.24
9.05	5.59	3.15	4.45
5.59	3.97	3.93	2.83
4.41	7.16	4.75	3.52
5.57	2.53	5.42	1.45
4.19	4.95	3.23	1.48
4.31	4.74	0.86	3.36
6.73	5.58	1.99	4.28
6.61	5.57	1.51	7.08
6.63	3.80	6.63	2.20
5.51	5.05	2.18	3.48
2.69	4.85	4.22	3.08
5.58	6.04	2.81	0.30
6.95	6.74	4.43	2.54
5.23	6.76	1.95	0.51

Answer 1

(a)

(b)

Null hypothesis, H₀: No difference exists among the means of the four view groups

Alternative hypothesis, H₁: difference exists among at least two means of the four view groups

One-way ANOVA:

Source DF SS MS F P
C5 3 115.90 38.63 13.16 0.000
Error 96 281.89 2.94
Total 99 397.79

From one way ANOVA we observe that F statistic=13.16 and the corresponding p-value=0.000<0.05,

hence we reject H₀ at level 0.05 and conclude that the difference exists among at least two the means of the four view groups.

Now we perform post hoc test i.e. Tukey's HSD:

Tukey 95% Simultaneous Confidence Intervals
All Pairwise Comparisons among Levels of C5

Individual confidence level = 98.97%

Children subtracted from:

Lower Center Upper --------+---------+---------+---------+-
Men 1.096 2.364 3.632 (------*-----)
Teens -1.223 0.045 1.313 (-----*------)
Women 0.677 1.945 3.213 (------*-----)
--------+---------+---------+---------+-
-2.0 0.0 2.0 4.0

Here we see that C.I. of (Men-Children) does not contain zero an contains only positive value, hence mean of Men>Mean of Children. C.I. of (Teens-Children) contains zero hence mean of Teens=Mean of Children. Again, C.I. of (Women-Children) does not contain zero an contains only positive value, hence mean of Women>Mean of Children.
C5 = Men subtracted from:

C5 Lower Center Upper --------+---------+---------+---------+-
Teens -3.588 -2.320 -1.052 (-----*------)
Women -1.687 -0.419 0.849 (-----*-----)
--------+---------+---------+---------+-
-2.0 0.0 2.0 4.0

C.I. of (Women-Men) contains zero hence mean of Women=Mean of Men. Again, C.I. of (Teens-Men) does not contain zero an contains only negative value, hence mean of men>Mean of Teens.
C5 = Teens subtracted from:

C5 Lower Center Upper --------+---------+---------+---------+-
Women 0.632 1.900 3.168 (------*-----)
--------+---------+---------+---------+-
-2.0 0.0 2.0 4.0

C.I. of (Women-Teens) does not contain zero an contains only positive value, hence mean of Women>Mean of Teens.

Therefore mean of Men=mean of Women>Mean of Teens=Mean of Children.