MGMT2263
Worksheet #4
Solutions
Question 1
Ho: p1 = p2
Ha: p1 ¹ p2
Reject Ho if Z < -1.96 or > 1.96
phat = (6 + 8)/(1000) = 14/1000 = 0.014
Z = (0.012 – 0.016)/sqrt(0.014(0.986)/500 + 0.014(0.986)/500) = -0.5383
Do not reject Ho.
Conclude there is no significant difference in the percentage of scrap widgets between the 2 plants.
Question 2
Lower limit = (0.016 – 0.012) – 1.96sqrt(0.012(0.988)/500 + 0.016(0.984)/500) = 0.004 – 0.014562 = -0.011
Upper limit = 0.004 + 0.014562 = 0.019
-1.1% < p2 – p1 < 1.9%
With 95% confidence, the percentage of scrap widgets at plant #2 ranges from 1.1% below to 1.9% above the percentage of scrap widgets at plant #1.
Question 3
We begin by making plant #2 group 1.
Ho: p1 – p2 £ 1%
Ha: p1 – p2 > 1%
Reject Ho if p-value < 1%. Do not reject Ho if p-value > 10%
Z = [(0.016 – 0.012) – 0.01]/sqrt(0.012(0.988)/500 + 0.016(0.984)/500) = -0.81
p-value = P(Z > -0.81) = 0.5 + 0.291 = 0.791
Do not reject Ho.
Conclude the difference between the percentage of scrap widgets at plant #2 and plant #1 is not more than 1%.
Question 4
Ho: m1 = m2 = m3
Ha: not all are equal
The degrees of freedom are 2 and 12. Reject Ho if F > 3.89
Here is the ANOVA table:
|
Source of Variation |
SS |
df |
MS |
F |
|
Between
Groups |
9.433333 |
2 |
4.716667 |
3.234286 |
|
Within
Groups |
17.5 |
12 |
1.458333 |
|
|
Total |
26.93333 |
14 |
|
|
F = 3.2343. Do not reject Ho. Conclude there is no significant difference in the average sales among the 3 offices.
Question 5
Ho: m1 = m2 = m3 = m4
Ha: not all are equal
The degrees of freedom are 3 and 28. Reject Ho if F > 2.95
a) Here is the ANOVA table:
|
Source of Variation |
SS |
df |
MS |
F |
|
Between
Groups |
2331.84375 |
3 |
777.28125 |
4.7792 |
|
Within
Groups |
4553.875 |
28 |
162.6384 |
|
|
Total |
6885.71875 |
31 |
|
|
F = 4.7792. Reject Ho. Conclude that the average grades for at least 2 schools are significantly different.
b) The Q value must be between 3.85 and 3.90. Since 28 is closer to 30 than 24, we take Q = 3.85 (as KPK does). Then, D = 3.85sqrt(162.6384/8) = 17.3591. If we compute the difference between the means for each pair of schools, we see that the only difference that is larger than the critical value is the difference between school D and school B. It is consistent with the results of ANOVA since we concluded in ANOVA that at least 2 means are significantly different; based on Tukey’s test, the means for schools B and D are significantly different.
c) The common standard deviation is the square root of 162.6384 = 12.753.
d) The t value is 2.048.
Method by hand:
Lower limit = (87 – 65.25) – 2.048sqrt(162.6384/8 + 162.6384/8) = 21.75 – 13.059 = 8.691
Upper limit = 21.75 + 13.059 = 34.809
8.691 < m(D) - m(B) < 34.809
Method using KPK:
The KPK output gives the interval 8.6883 < m(D) - m(B) < 34.8117. This is more accurate than the interval done by hand since there are no rounding errors (though both are equal if the limits are rounded to one decimal).
If we test the hypothesis:
Ho: m(B) = m(D)
Ha: m(B) ¹ m(D)
we see that m(D) - m(B) = 0 falls outside the 95% confidence interval, indicating a significant difference between the average grades of schools B and D, which are consistent with the results of ANOVA in which we concluded that at least 2 means are significantly different.
Question 6
a) Ho: m1 = m2 = m3
Ha: not all are equal
The degrees of freedom are 2 and 10. Reject Ho if F > 4.10.
Here is the ANOVA table:
|
Source
of Variation |
SS |
df |
MS |
F |
|
Rows |
108.4444 |
5 |
21.6889 |
0.8076 |
|
Columns |
2260.1111 |
2 |
1130.0556 |
42.0790 |
|
Error |
268.5556 |
10 |
26.8556 |
|
|
|
|
|
|
|
|
Total |
2637.1111 |
17 |
|
|
F = 42.1. Reject Ho and conclude that the average time is significantly different for at least 2 of the tasks.
b) The Q value is 3.88. Then D = 3.88sqrt(26.8556/6) = 8.2087. Here are the averages for each task:
|
Paperwork |
22.8333 |
|
Handling
returns |
34.3333 |
|
Meeting
with clients |
50.1667 |
No matter which difference we compute, the difference is always greater than 8.2087. Thus, all 3 tasks are significantly different.
c) Ho: m1 = m2 = m3 = m4 = m5 = m6
Ha: not all are equal
The degrees of freedom are 5 and 10. Reject Ho if F > 3.33.
F = 0.81. Do not reject Ho. Conclude there is no significant difference among the salespeople in the average amount of time spent on the tasks.
Question 7
We begin with the ANOVA table:
|
Source of Variation |
SS |
df |
MS |
F |
|
Sample |
681.4249 |
2 |
340.7125 |
7.5822 |
|
Columns |
34.2726 |
1 |
34.2726 |
0.7627 |
|
Interaction |
115.1366 |
2 |
57.56829 |
1.2811 |
|
Within |
808.8438 |
18 |
44.93577 |
|
|
Total |
1639.678 |
23 |
|
|
a) Ho: m1 = m2
Ha: m1 ¹ m2
The degrees of freedom are 1 and 23. Reject Ho if F > 4.28
F = 0.7627
Do not reject Ho.
Conclude the average amount spent per week eating out does not depend on gender.
b) Ho: m1 = m2 = m3
Ha: not all are equal
The degrees of freedom are 2 and 23. Reject Ho if F > 3.42.
F = 7.5822
Reject Ho.
Conclude the average amount spent per week eating out depends on income.
c) Ho: there is no significant interaction between gender and income
Ha: there is significant interaction between gender and income
The degrees of freedom are 2 and 23. Reject Ho if F > 3.42.
F = 1.2811
Do not reject Ho.
Conclude there is no significant interaction between gender and income.