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: m_d £ 0

Ha: m_d > 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 < m_d < 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.