MGMT2263 Tutorial Sheet 3 Solutions
Question 1
Since these are independent samples that are assumed to be normal, we first need to establish if the standard deviations are equal or not:
Ho: s1 = s2
Ha: s1 ¹ s2
Since Location B has the higher standard deviation, we will make this group 1. The numerator d.f. = denominator d.f. = 9. Reject Ho if F > 4.03.
F = (30.159)2/(21.6169)2 = 1.9465.
Do not reject Ho.
Conclude the standard deviations are equal.
We can now proceed with the t test assuming equal variances.
Ho: m1 = m2
Ha: m1 ¹ m2
d.f. = 10 + 10 – 2 = 18. Reject Ho if t > 2.101 or < -2.101
Sp2 = [(9)(30.159)2 + (9)(21.6169)2]/18 = [(30.159)2 + (21.6169)2]/2 = 688.4278.
Since Location A has the higher mean, we will make this group 1.
t = (154.2 – 151.7)/sqrt(688.4278/10 + 688.4278/10) = 0.2131.
Do not reject Ho.
Conclude there is no significant difference in the average weight between the two locations.
Question 2
Since these are independent samples that are assumed to be normal, we first need to establish if the standard deviations are equal or not:
Ho: s1 = s2
Ha: s1 ¹ s2
Since the 2-bedroom apartments has the higher standard deviation, we will make this group 1. The numerator d.f. = denominator d.f. = 7. Reject Ho if F > 4.99 or if F < 1 / 4.99 = 0.2.
F = (148.2023)2/(40.3589)2 = 13.4844.
Reject Ho.
Conclude the standard deviations are not equal.
We can now proceed with the t test assuming unequal variances. Since we want to see if 2-bedroom apartments cost more, we make this group 1.
Ho: m1 £ m2
Ha: m1 > m2
d.f. = ((148.2023)2/8 + (40.3589)2/8)2/[((148.2023)2/8)2/7 + ((40.3589)2/8)2/7] = 8.03 which we round to 8.
Reject Ho if t > 1.86
t = (823.25 – 651.625)/sqrt(148.20232/8 + 40.35892/8) = 3.1604.
Reject Ho.
Conclude that 2-bedroom apartments cost significantly more on average than 1-bedroom apartments.
Question 3
From the t table, we see that the 1% critical value is 2.896. Thus P(t > 2.896) = 1%. The p-value is P(t > 3.1604). Since 3.1604 is further to the right than 2.896, this means P(t > 3.1604) < 1%. Thus, under the general rule of thumb, this provides strong support for Ha and we would reject Ho.
Question 4
Lower limit = (823.25 – 651.625) – (2.306)sqrt(148.20232/8 + 40.35892/8) = 171.63 – 125.23 = 46.40
Upper limit = 171.63 + 125.23 = 296.86
With 95% confidence, two-bedroom apartments cost from $46.40 to $296.86 more per month on average than one-bedroom apartments.
Question 5
Since we have dependent samples and the data is assumed to be normal, we conduct the paired t test.
In theory, the times before the changes should be higher than the times after. When we compute the differences, we subtract after from before.
Ho: md £ 0
Ha: md > 0
The d.f. = 12. Reject Ho if t > 1.782.
t = (-5.8462 – 0)/(9.1182/sqrt(13)) = -2.3117
Do not reject Ho.
Conclude the system is not effective.
Question 6
Since this was set up as a right-tail test, the p-value would be P(t > -2.3117). We know that P(t > 0) = 50%. Since -2.3117 is to the left of zero, we can infer than P(t > -2.3117) > 50%. If you had set this up as a left tail test, your test statistic would be 2.3117 and the p-value would be P(t < 2.3117) with equal results.
Question 7
Lower limit = 5.8462 – (2.179) (9.1182/sqrt(13)) = 5.8462 – 5.5105 = 0.3
Upper limit = 5.8462 + 5.5105 = 11.4
With 95% confidence, the agents spend as much as 11 minutes more per week on average on paperwork after the program was introduced than before.
Question 8
We’ll let Movie A be the first group.
Ho: d = 0
Ha: d ¹ 0
Since there are 2 ties, the sample size reduces to 8. Reject Ho if T £ 4.
First we compute the differences:
|
Person |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
|
Movie A |
5 |
6 |
9 |
4 |
8 |
7 |
7 |
2 |
6 |
8 |
|
Movie B |
4 |
4 |
8 |
3 |
7 |
7 |
6 |
3 |
5 |
8 |
|
Difference |
1 |
2 |
1 |
1 |
1 |
0 |
1 |
-1 |
1 |
0 |
Then we rank the absolute values of the differences from lowest to highest, keeping note of the original sign:
|
Number |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
|
|d| |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
2 |
|
Sign |
- |
+ |
+ |
+ |
+ |
+ |
+ |
+ |
|
Rank |
4 |
4 |
4 |
4 |
4 |
4 |
4 |
8 |
We have T- = 4 and T+ = 32. Since, this is a 2-tail test, T is the smaller of T+ and T-, so T = 4.
Reject Ho. Conclude the two movies are not rated equally.
Question 9
Since we want to see if there is an improvement, we let After be group 1.
Ho: d £ 0
Ha: d > 0
Since there are 3 ties, the sample size reduces to 17. Since the number of observations is greater than 15, we can convert this to a Z test. Reject Ho if Z > 1.645. As with question 1, we rank the absolute values of the differences from lowest to highest:
|
Number |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
|
|d| |
1 |
1 |
1 |
1 |
1 |
2 |
2 |
2 |
2 |
|
sign |
- |
- |
+ |
+ |
+ |
+ |
+ |
+ |
+ |
|
rank |
3 |
3 |
3 |
3 |
3 |
9 |
9 |
9 |
9 |
|
Number |
10 |
11 |
12 |
13 |
14 |
15 |
16 |
17 |
|
|
|d| |
2 |
2 |
2 |
3 |
3 |
3 |
3 |
3 |
|
|
sign |
+ |
+ |
+ |
- |
+ |
+ |
+ |
+ |
|
|
rank |
9 |
9 |
9 |
15 |
15 |
15 |
15 |
15 |
|
We need T+. However, T- = 3 + 3 + 15 = 21. Since T+ plus T- = 17(18)/2 = 153, we can compute that T+ = 153 – 21 = 132. The mean is 17(18)/4 = 76.5 and the standard deviation is sqrt(17(18)(35)/24) = 21.1246. Then Z = (132 – 76.5)/21.1246 = 2.6273.
Reject Ho. Conclude there is a significant improvement in the ratings.
If you do this with KPK, you need to have before in one column and after in the next column. You would choose this to be a left tail test (since before < after). Here is the output:
|
Wilcoxon Test |
|
|
T(+) |
21 |
|
T(-) |
132 |
|
Mean (Z) |
76.5000 |
|
Std. Dev. (Z) |
21.1246 |
|
Z Value |
-2.6273 |
|
p-value (Using Z) |
.0043 |
|
p-value (Using Table) |
> .05 |
Notice that the output is the opposite of the calculations above.
Question 10
Since the sample sizes are the same, we can choose richer to be group 1.
Ho: median 1 £ median 2
Ha: median 1 > median 2
Reject Ho if p-value < 1%; do not reject Ho if p-value > 10%.
We combine all the data and sort from lowest to highest:
|
Number |
1 |
2 |
3 |
4 |
5 |
6 |
|
Value |
4 |
5 |
5 |
8 |
8 |
9 |
|
Group |
Poorer |
Poorer |
Poorer |
Richer |
Poorer |
Poorer |
|
Rank |
1 |
2.5 |
2.5 |
4.5 |
4.5 |
6.5 |
|
Number |
7 |
8 |
9 |
10 |
11 |
12 |
|
Value |
9 |
10 |
12 |
13 |
15 |
16 |
|
Group |
Poorer |
Richer |
Richer |
Richer |
Richer |
Richer |
|
Rank |
6.5 |
8 |
9 |
10 |
11 |
12 |
T1 = 4.5 + 8 + 9 + 10 + 11 + 12 = 54.5
U1 = (6)(6) + (6)(7)/2 – 54.5 = 2.5 which we round up to 3.
Looking in Table A.10, where n1=6 and n2=6, we find the p-value to be 0.0076.
Reject Ho. Conclude those in the richer neighbourhood eat out more.
Question 11
Since the sample sizes are the same, we choose the hearing impaired to be group 1.
Ho: median 1 £ median 2
Ha: median 1 > median 2
Reject Ho if Z > 1.645
As with the previous question, we combine the data and sort from lowest to highest. Since this is a Z test, we need U2.
|
Number |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
|
Value |
1.15 |
1.23 |
1.37 |
1.43 |
1.56 |
1.64 |
1.65 |
1.75 |
|
Group |
Normal |
Normal |
Normal |
Normal |
Normal |
Normal |
Normal |
Normal |
|
Rank |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
|
Number |
9 |
10 |
11 |
12 |
13 |
14 |
15 |
16 |
|
Value |
1.83 |
1.83 |
1.95 |
1.96 |
2.03 |
2.17 |
2.23 |
2.3 |
|
Group |
Impaired |
Normal |
Impaired |
Normal |
Normal |
Impaired |
Impaired |
Impaired |
|
Rank |
9.5 |
9.5 |
11 |
12 |
13 |
14 |
15 |
16 |
|
Number |
17 |
18 |
19 |