MGMT2263
Worksheet #3
Solutions
1) Since the other region has the larger standard deviation, we make this group 1.
Ho: s1 £ s2
Ha: s1 > s2
The degrees of freedom are 12 and 15. Reject Ho if F > 2.48.
F = 7.22/5.32 = 1.8455.
Do not reject Ho.
Conclude the age standard deviation of the region with the high percentage of seniors is not significantly less than that of the other region.
2) Since the standard deviation for store 2 is larger, it might be easier to make this group 1.
Ho: s1 = s2
Ha: s1 ¹ s2
The degrees of freedom are 9 and 8.
F(9,8,0.025) = 4.36.
F(8,9,0.025) = 4.10. So, F(9,8,0.975) = 1 / 4.10 = 0.2439. Reject Ho if F < 0.2439 or > 4.36.
F = 9.22/4.12 = 5.0351.
Reject Ho and conclude the standard deviations for the two stores are significantly different.
3) The first thing we need to determine is whether the standard deviations for the two coils are significantly different or not. Since the standard deviation for Coil B is larger, it might be easier to make this group 1.
Ho: s1 = s2
Ha: s1 ¹ s2
The degrees of freedom are 15 and 8.
F(15,8,0.025) = 4.10.
F(8,15,0.025) = 3.20. So, F(15,8,0.975) = 1 / 3.20 = 0.3125. Reject Ho if F < 0.3125 or > 4.10.
F = 242/172 = 1.9931.
Do not reject Ho and conclude the standard deviations for the two coils are not significantly different.
Now that we have established the equality of the standard deviations, we can proceed to the t test for two means assuming equal variances. There is no harm in keeping Coil B as group 1.
Ho: m1 = m2
Ha: m1 ¹ m2
The degrees of freedom = 9 + 16 – 2 = 23.
Reject Ho if t < -2.069 or > 2.069
Sp2 = (8*172 + 15*242)/23 = 476.1739
t = (143 – 118)/sqrt(476.1739/9 + 476.1739/16) = 2.7496.
Reject Ho.
Conclude the tensile strengths of the two coils are not equal.
4) We see that the test statistic falls between the 1% and 0.5% critical values on the t table. Since the p-value = 2P(t > 2.7496), we conclude the p-value lies between 1% and 2%. Thus, if the level of significance had been set at 1%, we would not have rejected Ho.
5) Lower limit = (148 – 118) – 2.069 sqrt(476.1739/9 + 476.1739/16) = 25 - 18.8119 = 6.1881.
Upper limit = 25 + 18.8119 = 43.8119.
6.1881 < m(coil B) - m(coil A) < 43.8119.
We would reject Ho at a 5% level of significance since the hypothesized difference of zero does not fall in the 95% confidence interval.
6) In question 2, we concluded that there is a significant difference in the standard deviations of the two stores. So, we proceed with the t test for two means assuming unequal variances. There is no harm is keeping Store 2 as group 1.
Ho: m1 = m2
Ha: m1 ¹ m2
The degrees of freedom = (4.12/9 + 9.22/10)2/((4.12/9)2/8 + (9.22/10)2/9) = 12.7 which we round to 12.
Reject Ho if p-value < 1%. Do not reject Ho if p-value > 10%.
t = (48 - 45)/sqrt(4.12/9 + 9.22/10) = 0.9333.
P-value = 2P(t > 0.9333). From the t table, we see the 10% critical value is 1.356. Since P(t > 1.356) = 10%, it follows that P(t > 0.9333) > 10% and that the p-value exceeds 20%.
Do not reject Ho.
Conclude there is no significant difference in the average sales of the two stores.
7) Since we have dependent samples which are assumed to follow a normal distribution, we proceed with the paired t test. The easiest way to proceed is to subtract the 2007 bonuses from the 2006 bonuses and use the t test on KPK on the differences:
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t Test for Population Mean |
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Number of Observations |
10 |
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Sample Standard Deviation |
14.128380 |
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Sample Mean |
-0.500000 |
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Ho:m = 0 |
Ha:m ¹ 0 |
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T* |
-0.111912 |
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2 * P[T ³ ½T*½] two tail |
0.913349 |
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| T Critical |, a = 0.05 |
2.262159 |
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95% CI for Pop. Mean |
-10.606842 |
to |
9.606842 |
Ho: md = 0
Ho: md ¹ 0
Reject Ho if t < -2.262 or > 2.262.
t = -0.1119.
Do not reject Ho.
Conclude there is no significant difference in the average bonuses between the two years.
8) From the KPK output, we see the 95% confidence interval ranges from –10.6068 to 9.6068. If we multiply by 1000 (since this is in thousands of dollars) and round to the nearest hundred, we get
–10,600 < m(2005) - m(2006) < 9,600. The interpretation is: with 95% confidence, the average bonus for 2006 ranges from $10,600 less to $9,600 more than the average bonus for 2007. For example, if the average bonus in 2007 was $200,000, then the average bonus for 2006 would range from $189,400 to $209,600.
9) From the KPK output, the p-value is 0.913349. This provides overwhelming support for Ho. Therefore Ho should not be rejected.
10) Ho: median(A) = median(B)
Ha: median(A) ¹ median(B)
Reject Ho if p-value < 5%.
We first need to compute T1 and T2:
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|
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
|
Value |
0 |
0 |
1 |
2 |
2 |
3 |
3 |
4 |
4 |
5 |
|
Group |
A |
B |
B |
A |
B |
A |
B |
A |
B |
A |
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Rank |
1.5 |
1.5 |
3 |
4.5 |
4.5 |
6.5 |
6.5 |
8.5 |
8.5 |
10 |
T1 = 1.5 + 4.5 + 6.5 + 8.5 + 10 = 31
T2 = 1.5 + 3 + 4.5 + 6.5 + 8.5 = 24
Note that T1 + T2 = 55 = 10(11)/2
n1 = 5; n2 = 5
U1 = 5(5) + 5(6)/2 – 31 = 9
U2 = 5(5) + 5(6)/2 – 24 = 16
Thus, U = 9
From Table A.10 we see that p-value/2 = 0.2738. Thus p-value = 2(0.2738) = 0.5476.
Do not reject Ho.
Conclude there is no significant difference between the two secretaries in the number of typing mistakes.
11) Ho: median(A) = median(B)
Ha: median(A) ¹ median(B)
Reject Ho if Z < -1.96 or Z > 1.96
We first need to compute T2:
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|
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
12 |
13 |
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Value |
1012 |
1095 |
1102 |
1102 |
1106 |
1107 |
1109 |
1110 |
1134 |
1150 |
1185 |
1193 |
1185 |
|
Group |
A |
A |
A |
B |
B |
A |
A |
B |
A |
A |
B |
B |
B |
|
Rank |
1 |
2 |
3.5 |
3.5 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
12 |
13 |
|
|
14 |
15 |
16 |
17 |
18 |
19 |
20 |
21 |
22 |
23 |
24 |
25 |
26 |
|
Value |
1203 |
1208 |
1210 |
1251 |
1255 |
1265 |
1290 |
1295 |
1313 |
1375 |
1384 |
1401 |
1405 |
|
Group |
A |
A |
B |
B |
A |
A |
B |
B |
A |
A |
B |
A |
B |
|
Rank |
14 |
15 |
16 |
17 |
18 |
19 |
20 |
21 |
22 |
23 |
24 |
25 |
26 |
We use Brand B for T2:
T2 = 3.5 + 5 + 8 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 20 + 21 + 24 + 26 = 205.5
n1 = 12; n2 = 14
U2 = 12(14) + 14(15)/2 – 205.5 = 67.5
m(U2) = 12(14)/2 = 84; s(U2) = sqrt(12*14*27/12) = 19.4422
Z = (67.5 – 84)/19.4422 = -0.8487
Do not reject Ho
Conclude there is no significant difference in how long the two brands last.
12) We choose after as group 1.
Ho: d £ 0
Ha: d > 0
Reject Ho if T- £ 8
First we compute the absolute value of the differences (after – before):
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|
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
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Before |
70 |
72 |
75 |
61 |
82 |
55 |
43 |
51 |
84 |
|
After |
74 |
88 |
71 |
62 |
89 |
58 |
41 |
63 |
80 |
|
d = Difference |
4 |
16 |
-4 |
1 |
7 |
3 |
-2 |
12 |
-4 |
|
|d| |
4 |
16 |
4 |
1 |
7 |
3 |
2 |
12 |
4 |
Now we sort |d| from lowest to highest, keeping note if the difference was positive or negative:
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|
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
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|d| |
1 |
2 |
3 |
4 |
4 |
4 |
7 |
12 |
16 |
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Sign |
+ |
- |
+ |
+ |
- |
- |
+ |
+ |
+ |
|
Rank |
1 |
2 |
3 |
5 |
5 |
5 |
7 |
8 |
9 |
T- = 2 + 5 + 5 = 12
Do not reject Ho.
Conclude the median from after is not significantly higher than the median from before. Thus, the seminar’s methods are not effective.
13) We choose before as group 1.
Ho: d £ 0
Ha: d > 0
Reject Ho if Z > 1.645
We need to compute T+.
We first compute the absolute value of the differences (before – after)
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|
#1 |
#2 |
#3 |
#4 |
#5 |
#6 |
#7 |
#8 |
#9 |
#10 |
|
Before |
34 |
35 |
43 |
46 |
16 |
26 |
68 |
38 |
61 |
52 |
|
After |
31 |
31 |
44 |
44 |
15 |
28 |
63 |
39 |
63 |
54 |
|
d = Difference |
3 |
4 |
-1 |
2 |
1 |
-2 |
5 |
-1 |
-2 |
-2 |
|
|d| |
3 |
4 |
1 |
2 |
1 |
2 |
5 |
1 |
2 |
2 |
|
|
#11 |
#12 |
#13 |
#14 |
#15 |
#16 |
#17 |
#18 |
#19 |
#20 |
|
Before |
42 |
19 |
20 |
26 |
40 |
51 |
24 |
29 |
48 |
19 |
|
After |
38 |
21 |
19 |
22 |
38 |
50 |
21 |
33 |
51 |
18 |
|
d = Difference |
4 |
-2 |
1 |
4 |
2 |
1 |
3 |
-4 |
-3 |
1 |
|
|d| |
4 |
2 |