STAT217
Tutorial sheet #4 Solutions
Question 1
a) Ho: m1 = m2 = m3
Ha: not all means are equal
The degrees of freedom are 2 and 15. Reject Ho if F > 3.6823.
|
Source
of Variation |
SS |
df |
MS |
F |
|
Between
Groups |
1.8678 |
2 |
0.9339 |
0.2360 |
|
Within
Groups |
59.3640 |
15 |
3.9576 |
|
|
Total |
61.2318 |
17 |
|
|
F = 0.236. Do not reject Ho. Conclude there is no significant difference among the 3 counters in average sales.
b) Ho: s1 = s2 = s3
Ha: not all the standard deviations are equal
Reject Ho if H > 10.8
H = 8.083937/0.794057 = 10.18
Do not reject Ho.
Conclude there is no significant difference among the standard deviations.
c) D = 3.67sqrt(3.9576/6) = 2.9806. The largest mean is 16.68 from counter 3, the smallest, 15.95 from counter 1. Since 16.68 15.95 = 0.73 is less than D, we can conclude there is no significant difference between any of the pairs of means which is consistent with the results of ANOVA in which we concluded that none of the means are significantly different.
d) Lower limit = (16.68 15.95) 2.9806 = 0.73 2.98 = -2.25
Upper limit = 0.73 + 2.98 = 3.71
-2.25 < m3 - m1 < 3.71
Since m3 - m1 = 0 falls inside the 95% confidence interval, we conclude there is no significant difference between m1 and m3. This is consistent with the results of ANOVA in which we concluded that none of the means are significantly different.
Question 2
a) a) Ho: m1 = m2 = m3
Ha: not all factor means are equal
The degrees of freedom are 2 and 8. Reject Ho if F > 4.459.
|
Source
of Variation |
SS |
df |
MS |
F |
P-value |
|
Rows |
33.7333 |
4 |
8.4333 |
1.5521 |
0.2762 |
|
Columns |
62.5333 |
2 |
31.2667 |
5.7546 |
0.0283 |
|
Error |
43.4667 |
8 |
5.4333 |
||
|
Total |
139.7333 |
14 |
|
|
|
F = 5.7546. Reject Ho. Conclude that at least 2 of the shifts are significantly different.
b) Ho: m1 = m2 = m3 = m4 = m5
Ha: not all block means are equal
The degrees of freedom are 4 and 8. Reject Ho if F > 3.8379
F = 1.5521. Do not reject Ho. Conclude there is no significant difference among the lines means.
c) D = 4.04sqrt(5.4333/5) = 4.2114. The means of the day, afternoon and night shifts are 30, 26 and 30.6 respectively. The difference between the night and afternoon shifts is 4.6 which is significant. However, the other differences are less than 4.2114 and so are not significant.
d) d) Lower limit = (30.6 26) 4.2114 = 4.6 4.2114 = 0.4
Upper limit = 4.6 + 4.2114 = 8.8
0.4 < mnight - mafternoon < 8.8; the afternoon shifts
produces between 0.4 and 8.8 more widgets per hour on average than the
afternoon shift
Question 3
a) Ho: m1 = m2 = m3
Ha: not all means are equal
The degrees of freedom are 2 and 21. Reject Ho
if F > 3.4668.
|
Source |
df |
SS |
MS |
F |
|
Between
Groups |
2 |
1601.5833 |
800.7917 |
4.4556 |
|
Within
Groups |
21 |
3774.2500 |
179.7262 |
|
|
Total |
23 |
5375.8333 |
|
|
F = 4.4556. Reject Ho. Conclude that at least 2
of the methods are significantly different.
b)
P-value
= P(F > 4.4556) = 1 P(F < 4.4556) = 1 0.9756
= 0.0244 = 2.44%. Since we rejected Ho, we would not reject Ho for levels of
significance between 1% and 2.44%.
c)
D
= 3.58sqrt(179.7262/8) = 16.9685. The average
difference between A and C is 17.75, indicating a significant difference. The
average difference between A and B is 16.875, indicating that technically there
is not a significant difference. The average difference between B and C is
0.875, indicating that there is not a significant difference.
d) Lower
limit = 17.75 16.9685 = 0.7815
Upper limit = 17.75 + 16.9685 = 34.7185
0.7815 < m(A) - m(C) < 34.7185
The confidence interval is consistent with the
results of ANOVA since m(A) - m(C) = 0 lies outside the confidence
interval.
Question 4
a)
Ho:
΅1 = ΅2 = ΅3
Ha: not all factor means are equal
Reject Ho if F > 3.7389
|
Source |
df |
SS |
MS |
F |
|
Factors |
2 |
19872.6558 |
9936.3279 |
13.8010 |
|
Blocks |
7 |
6555.2117 |
936.4588 |
1.3007 |
|
Error |
14 |
10079.6108 |
719.9722 |
|
|
Total |
23 |
36507.4783 |
|
|
F = 13.801
Reject Ho
Conclude there is a significant difference
between at least 2 of the plants in the average sulphur
dioxide emissions.
b)
D
= 3.7sqrt(719.9722/8) = 35.1
The respective means are:
|
Plant
#1 |
Plant
#2 |
Plant
#3 |
|
301.425 |
353.3875 |
286.1625 |
Plant #2 Plant
#3 = 67.225 -> significant difference
Plant #2 Plant #1 = 51.9625 -> significant
difference
Plant #1 Plant #3 = 15.2625 -> no
significant difference
c)
Lower
limit = 67.225 35.1 = 32.125
Upper limit = 67.225 + 35.1 = 102.325
32.125 < m(Plant #2) - m(Plant #3) < 102.325
The confidence interval is consistent with the
results of ANOVA since m(Plant #2) - m(Plant #3) = 0 lies outside the confidence
interval.