MGMT2263 Tutorial Sheet 4 Solutions

 

Question 1

Let the members be group 1.

Ho: p1 £ p2

Ha: p1 > p2

Reject Ho if Z > 1.645

p-hat = (144 + 115)/(400) = 0.6475. As well, p1 = 144/200 = 0.72, p2 = 0.575

Z = (0.72 – 0.575)/sqrt(0.6475(0.3525)/200 + 0.6475(0.3525)/200) = 3.035

Reject Ho. Conclude the members are more likely to visit new exhibits than non-members.

 

Question 2

P-value = P(Z > 3.04) = 0.5 – 0.4988 = 0.0012. Since this is less than 1%, we have strong support for Ha. The answer is no.

 

Question 3

Lower limit = (0.72 – 0.575) – 1.96sqrt(0.72(0.28)/200 + 0.575(0.425)/200) = 0.145 – 0.0926 = 0.0524 = 5.24%

Upper limit = 0.145 + 0.0926 = 0.2376 = 23.76%

The percentage of members who will visit new exhibits ranges from 5.24% to 23.76% above the corresponding percentage of non-members.

 

Question 4

Ho: m1 = m2 = m3

Ha: not all the means are equal

The degrees of freedom are 2 and 15. Reject Ho if F > 3.68.

Here is the ANOVA table:

Source of Variation

SS

df

MS

F

Between Groups

1.8678

2

0.9339

0.2360

Within Groups

59.3640

15

3.9576

 

Total

61.2318

17

 

 

F = 0.236. Do not reject Ho. Conclude there is no significant difference among the 3 counters in average sales.

 

Question 5

Ho: s1 = s2 = s3

Ha: not all the standard deviations are equal

Reject Ho if H > 10.8

H = 8.083937/0.794057 = 10.18

Do not reject Ho.

Conclude there is no significant difference among the standard deviations.

 

Question 6

Q = 3.67; D = 3.67sqrt(3.9576/6) = 2.9806. The largest mean is 16.68 from counter 3, the smallest, 15.95 from counter 1. Since 16.68 – 15.95 = 0.73 is less than D, we can conclude there is no significant difference between any of the pairs of means which is consistent with the results of ANOVA in which we concluded that none of the means are significantly different.

 

Question 7

The t value is 2.131 since we have 15 degrees of freedom.

Lower limit = (16.68 – 15.95) – 2.131sqrt(3.9576/6 + 3.9576/6) = 0.73 – 2.45 = -1.72

Upper limit = 0.73 + 2.45 = 3.18

-1.72 < m3 - m1 < 3.18

Since m3 - m1 = 0 falls inside the 95% confidence interval, we conclude there is no significant difference between m1 and m3. This is consistent with the results of ANOVA in which we concluded that none of the means are significantly different.

 

Question 8

a)      Ho: m1 = m2 = m3

Ha: not all the factor means are equal

The degrees of freedom are 2 and 8. Reject Ho if F > 4.46

Here is the ANOVA table:

Source of Variation

SS

df

MS

F

P-value

Rows

33.7333

4

8.4333

1.5521

0.2762

Columns

62.5333

2

31.2667

5.7546

0.0283

Error

43.4667

8

5.4333

 

 

Total

139.7333

14

 

 

 

F = 5.7546. Reject Ho. Conclude that at least 2 of the shifts are significantly different.

b)     Ho: m1 = m2 = m3 = m4 = m5

Ha: not all the block means are equal

The degrees of freedom are 4 and 8. Reject Ho if F > 3.84

F = 1.5521. Do not reject Ho. Conclude there is no significant difference among the lines’ means.

c)      Q = 4.04; D = 4.04sqrt(5.4333/5) = 4.2114. The means of the day, afternoon and night shifts are 30, 26 and 30.6 respectively. The difference between the night and afternoon shifts is 4.6 which is significant. However, the other differences are less than 4.2114 and so are not significant.

d)     The t value is 2.306 since we have 8 degrees of freedom.

Lower limit = (30.6 – 26) – 2.306sqrt(5.4333/5 + 5.4333/5) = 4.6 – 3.4 = 1.2

Upper limit = 4.6 + 3.4 = 8.0

1.2 < mnight - mafternoon < 8; the afternoon shifts produces between 1.2 and 8 more widgets per hour on average than the afternoon shift

 

Question 9

First, here is the ANOVA table:

Source of Variation

SS

df

MS

F

Education

587.6763

2

293.8381

26.3252

Age

437.6763

2

218.8381

19.6059

Interaction

292.2170

4

73.0543

6.5450

Error

200.9133

18

11.1619

 

 

 

 

 

 

Total

1518.483

26

 

 

a)      Ho: age is not significant

Ha: age is significant

The degrees of freedom are 2 and 18. Reject Ho if F > 3.55

F = 19.6059

Reject Ho and conclude that income depends on age.

b)     Ho: education is not significant

Ha: education is significant

The degrees of freedom are 2 and 18. Reject Ho if F > 3.55

F = 26.3252

Reject Ho and conclude that income depends on education levels.

c)      Ho: no significant interaction

Ha: significant interaction

The degrees of freedom are 4 and 18. Reject Ho if F > 2.93

F = 6.545

Reject Ho and conclude there is significant interaction between age and education levels.