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.
Here is the Excel output for this test. Note that the level of significance is set at 0.025 since it is a 2-tail test.
|
F-Test
Two-Sample for Variances |
|
|
|
|
|
|
|
|
Location B |
Location A |
|
Mean |
151.7 |
154.2 |
|
Variance |
909.5667 |
467.2889 |
|
Observations |
10 |
10 |
|
df |
9 |
9 |
|
F |
1.9465 |
|
|
P(F<=f)
one-tail |
0.1677 |
|
|
F
Critical one-tail |
4.0260 |
|
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.
Here is the Excel output for the t test.
|
t-Test:
Two-Sample Assuming Equal Variances |
||
|
|
|
|
|
|
Location A |
Location B |
|
Mean |
154.2 |
151.7 |
|
Variance |
467.2889 |
909.5667 |
|
Observations |
10 |
10 |
|
Pooled
Variance |
688.4278 |
|
|
Hypothesized
Mean Difference |
0 |
|
|
df |
18 |
|
|
t
Stat |
0.2131 |
|
|
P(T<=t)
one-tail |
0.4168 |
|
|
t
Critical one-tail |
1.7341 |
|
|
P(T<=t)
two-tail |
0.8337 |
|
|
t
Critical two-tail |
2.1009 |
|
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.
F = (148.2023)2/(40.3589)2 = 13.4844.
Reject Ho.
Conclude the standard deviations are not equal.
Here is the Excel output for this test. Note that the level of significance is set at 0.025 since it is a 2-tail test.
|
F-Test
Two-Sample for Variances |
||
|
|
|
|
|
|
2 bed |
1 bed |
|
Mean |
823.25 |
651.625 |
|
Variance |
21963.9286 |
1628.8393 |
|
Observations |
8 |
8 |
|
df |
7 |
7 |
|
F |
13.4844 |
|
|
P(F<=f)
one-tail |
0.0014 |
|
|
F
Critical one-tail |
4.9949 |
|
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.
Here is the Excel output for the t test.
|
t-Test:
Two-Sample Assuming Unequal Variances |
||
|
|
|
|
|
|
2 bed |
1 bed |
|
Mean |
823.25 |
651.625 |
|
Variance |
21963.9286 |
1628.8393 |
|
Observations |
8 |
8 |
|
Hypothesized
Mean Difference |
0 |
|
|
df |
8 |
|
|
t
Stat |
3.1604 |
|
|
P(T<=t)
one-tail |
0.0067 |
|
|
t
Critical one-tail |
1.8595 |
|
|
P(T<=t)
two-tail |
0.0134 |
|
|
t
Critical two-tail |
2.3060 |
|
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%. The Excel output gives an exact p-value of 0.0067 = 0.67%. 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
46.40 < m(2 bedroom) - m(1 bedroom) < 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.
If we use Megastat, here is the output.
|
Hypothesis
Test: Paired Observations |
|
||||
|
|
|
|
|
|
|
|
|
0.000 |
hypothesized
value |
|
|
|
|
|
91.692 |
mean
Before |
|
|
|
|
|
97.538 |
mean
After |
|
|
|
|
|
-5.846 |
mean
difference (Before - After) |
|||
|
|
9.118 |
std.
dev. |
|
|
|
|
|
2.529 |
std.
error |
|
|
|
|
|
13 |
n |
|
|
|
|
|
12 |
df |
|
|
|
|
|
|
|
|
|
|
|
|
-2.31 |
t |
|
|
|
|
|
.9803 |
p-value (one-tailed, upper) |
|
||
|
|
|
|
|
|
|
|
|
-11.356 |
confidence
interval 95.% lower |
|
||
|
|
-0.336 |
confidence
interval 95.% upper |
|||
|
|
5.510 |
margin of error |
|
|
|
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. Note that the Megastat output gives the exact p-value of 0.9803 = 98.03%.
Question 7
For the purposes of the confidence interval, I make the difference = after – before since this average difference proves to be positive.
Lower limit = 5.8462 – (2.179) (9.1182/sqrt(13)) = 5.8462 – 5.5105 = 0.3
Upper limit = 5.8462 + 5.5105 = 11.4
0.3 < md < 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. Notice that in the Megastat output, since the difference is before – after, the limits are negative.
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 Megastat, here is the output:
|
Wilcoxon Signed
Rank Test |
|
|
|
||
|
|
|
|
|
|
|
|
|
|
variables: |
After
- Before |
|
|
|
|
|
132 |
sum
of positive ranks |
|
|
|
|
|
21 |
sum
of negative ranks |
||
|
|
|
|
|
|
|
|
|
|
17 |
n |
|
|
|
|
|
76.50 |
expected value |
|
|
|
|
|
21.12 |
standard deviation |
|
|
|
|
|
2.6273 |
z |
|
|
|
|
|
.0043 |
p-value (one-tailed, upper) |
||
Question 10
We choose richer to be group 1.
Ho: median 1 £ median 2
Ha: median 1 > median 2
Reject Ho if T1 is greater than or equal to 50.
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
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 T1.
|
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 |
20 |
21 |
22 |
23 |
24 |
|
Value |
2.35 |
2.45 |
2.64 |
2.75 |
2.75 |
3.14 |
3.23 |
3.25 |
|
Group |
Normal |
Impaired |
Impaired |
Impaired |
Impaired |
Impaired |
Impaired |
Impaired |
|
Rank |
17 |
18 |
19 |
20 |
21 |
22 |
23 |
24 |
T1 = 9.5 + 11 + 14 + 15 + 16 + 18 + 19 + 20 + 21 + 22 + 23 + 24 = 212.5
The mean = (12)(12 + 12 + 1)/2 = 150 and the standard deviation = sqrt(12(12)(25)/12) = 17.3205.
Then Z = (212.5 – 150)/17.3205 = 3.6084.
Reject Ho. Conclude that those who are hearing impaired are more visually acute.
If you use Megastat, here is the output:
|
Wilcoxon -
Mann/Whitney Test |
|
|
|
||
|
|
|
|
|
|
|
|
|
n |
sum of ranks |
|
|
|
|
|
12 |
212.5 |
Impaired |
|
|
|
|
12 |
87.5 |
Normal |
|
|
|
|
24 |
300 |
total |
|
|
|
|
|
|
|
|
|
|
|
|
150.00 |
expected value |
|
|
|
|
|
17.32 |
standard deviation |
|
|
|
|
|
3.5796 |
z |
|
|
|
|
|
.0002 |
p-value (one-tailed, upper) |
||
Unfortunately, there is a bug in the program which gives an incorrect Z value. However, you can use the rest of the output (which is correct) to come up with the correct Z value.