MGMT2263

Worksheet #2 Solutions

Question 1
a: independent since we have 2 separate groups with different sample sizes
b: independent since we have 2 separate 30-day periods
c: dependent since each person rates both Coke and Pepsi

Question 2
a: Z test since we have independent samples both over 30
b: Wilcoxon rank-sum since neither sample is normally distributed
c: Wilcoxon signed-ranks since the samples are dependent and the scale is ordinal
d: t test assuming equal variances since the samples are independent, the sample sizes are less than 30, the data is assumed to be normal and the standard deviations are equal
e: paired t-test since the samples are dependent and the data is assumed to be normal

Question 3
The null and alternative hypotheses are:
Ho: m(B) < m(A)
Ha: m(B) > m(A)
Reject Ho if Z > 1.645
Z = (5.6 - 4.4)/sqrt(2.3²/100 + 1.6²/100) = 4.28
Reject Ho
Conclude those in Location B watch more videos on average than those in Location A.

Question 4
p-value = P(Z > 4.28) = 0 for all intent and purposes. So, there is no level of significance between 1% and 10% that could have been chosen in which the opposite conclusion would have been reached.

Question 5
Lower limit = (5.6 - 4.4) - 1.96sqrt(2.3²/100 + 1.6²/100) = 1.2 - 0.55 = 0.65
Upper Limit = 1.2 + 0.55 = 1.75
0.65 < m(B) - m(A) < 1.75

Question 6
Ho: m(B) - m(A) < 1
Ha: m(B) - m(A) > 1
Reject Ho if p-value < 1%; do not reject Ho if p-value > 10%
Z = [(5.6 - 4.4) - 1]/sqrt(2.3²/100 + 1.6²/100) = 0.71
p-value = P(Z > 0.71) = 0.5 - 0.2612 = 0.2388
Do not reject Ho
Conclude the difference in the average number of videos watched per month between Location B and Location A does not exceed 1 per month.

Question 7

We choose executives to be group 1.

Ho: m1 = m2

Ha: m1 ¹ m2

Reject Ho if Z < -1.96 or Z > 1.96

Z = (13.5 – 11.5)/sqrt(2.4²/150 + 4.5²/200) = 5.35

Reject Ho

Conclude there is a significant difference between executives and regular workers in the average time spent on email, text messaging, etc.

Question 8

In hours:

Lower limit = (13.5 – 11.5) – 1.96sqrt(2.4²/150 + 4.5²/200) = 2 – 0.7324 = 1.2676 hours

Upper limit = 2 + 0.7324 = 2.7324 hours

When we convert the limits to the nearest minute we get 76 < m1 - m2 < 164

With 95% confidence, executives spend, on average, 76 to 164 more minutes on email, etc. than regular workers.

If this interval were used to test the hypothesis in question 7, we would reject Ho at a 5% level of significance since the hypothesis difference of zero lies outside the 95% confidence interval.

Question 9

P-value = 2*P(Z > 5.35) = 0 for all intent and purposes. Since the p-value is less than 1% under the general rule of thumb, we reject Ho.

Question 10

Since the mean for the executives is greater than that of regular workers, the appropriate one-tail test is:

Ho: m1 < m2
Ha: m1 > m2

with the executives as group 1 as previously.

It is a given that if you reject Ho for a two-tail test, you will also reject Ho for the appropriate one-tail test since the p-value will be half of what it was previously. Since the p-value for the two-tail test was zero, the p-value for the one-tail test is also zero. The conclusion is that executives spend significantly more time on email, etc. than regular workers.

From the textbook:

9.8 (page 274)

Use self-serve as group 1.

(4.2 – 3.7) ± 1.96sqrt(1.6²/100 + 0.9²/100)

= 0.5 ± 0.36

0.14 < µ1 - µ2 < 0.86

Since d0 = 0 falls below the 95% confidence interval, we can be 95% certain that the average time in the express checkout is less than that of the self-serve checkout.

9.83 (page 289)

Part a

Ho indicates no significant difference in average time between the two types of checkouts. Ha indicates that the average time in the express checkout is significantly less than that of the self-serve checkout.

Part b

Reject Ho if Z < -1.645

Z = (3.7 – 4.2)/sqrt(1.6²/100 + 0.9²/100) = -2.72

Reject Ho. Conclude the average time in the express checkout is significantly less than that in the self-serve checkout. Note that this hypothesis was tested using express as group 1. If we used self-serve as group 1, the critical value would be 1.645 and the test statistic would be 2.72. Of course, the conclusion would remain the same.

Part c

P-value = P(Z < -2.72) = 0.5 – 0.4967 = 0.0033 = 0.335. For the 10%, 5% and 1% levels of significance, we would reject Ho. Only for the 0.1% level of significance would we not reject Ho. There is strong evidence that the average time for the express checkout is significantly less.

9.48 (page 290)

Switch the samples so that we have Ho: µ1 - µ2 = 10. Then Z = [(162 – 151) – 10]/sqrt(8²/625 + 6²/625) = 2.5.

P-value = 2*P(Z > 2.5) = 2(0.5 – 0.4938) = 2(0.0062) = 0.0124 = 1.24%. For the 10% and 5% levels of significance we would reject Ho. For the 1% and 0.1% levels of significance, we do not reject Ho.