MGMT2262 Practice Final Solutions

Question 1

a) Ho: m = 56.50

Ha: m ¹ 56.50

Reject Ho if t < -2.718 or t > 2.718

t = (62.33 – 56.50)/(4.4855/sqrt(12)) = 4.5024

Reject Ho

Conclude there is a significant change in the average closing price on the TSX from $56.50.

b) Lower limit = 62.33 – 2.718(4.4855)/sqrt(12) = 62.33 – 3.52 = 58.81

Upper limit = 62.33 + 3.52 = 65.85

With 98% confidence, the average closing price on the TSX is between $58.81 and $65.85.

We would reject the null hypothesis at a 2% level of significance since the hypothesis mean of $56.50 falls outside the 98% confidence interval.

c) We see that the 0.5% critical value is 3.106 (based on 11 degrees of freedom).

P-value = 2P(t > 4.5024). Since P(t > 3.106) = 0.005, P(t > 4.5024) < 0.005. Consequently the p-value < 2(0.005) or less than 0.01. We would reject the null hypothesis under the general rule of thumb since the p-value is less than 1%.

Question 2

a) Ho: p £ 10%

Ha: p > 10%

Reject Ho if Z > 1.645.

Z = (0.12 – 0.1)/sqrt(0.1*0.9/500) = 1.49

Do not reject Ho.

Conclude the percentage of people who recognize the store’s name is not more than 10%.

b) P-value = P(Z > 1.49) = 0.5 – 0.4319 = 0.0601. Since the p-value > 5%, we do not reject Ho.

c) Lower limit = 0.12 – 1.96sqrt(0.12*0.88/500) = 0.12 – 0.0285 = 0.0915

Upper limit = 0.12 + 0.0285 = 0.1485

With 95% confidence, the percentage of people who recognize the store’s name is between 9.15% and 14.85%.

d) Since the p-value is between 1% and 10%, the results would be inconclusive.

e) Sample size = (1.96/0.01)²(0.12)(0.88) = 4056.7 which we round up to 4057.

f) Margin of error = 1.96sqrt(0.12*0.88/1500) = 0.0164 or 1.64%.

Question 3

a) Ho: s ³ 1300

Ha: s < 1300

Reject Ho if C² < 3.0535

C² = 11(1161.894)²/1300² = 8.787

Do not reject Ho.

Conclude the daily sales are not more consistent.

b) P-value = P(C² < 8.787). We see that the 90% critical value is 5.5778. This means P(C² > 5.5778) = 0.9 or that P(C² < 5.5778) = 0.1. Consequently, P(C² < 8.787) > 0.1. Since the p-value is greater than 10%, we would not reject Ho under the general rule of thumb.

c) Lower limit for s² = 11(1161.894)²/26.7568 = 554,998.144

Upper limit for s² = 11(1161.894)²/2.6032 = 5,704,507.66

Lower limit for s = sqrt(554,998.144) = 744.98

Upper limit for s = sqrt(5,704,507.66) = 2,388.41

With 99% confidence, the standard deviation for daily sales ranges between $744.98 and $2,388.41.

d) Ho: m £ 9200

Ha: m > 9200

Reject Ho if Z > 2.11

Z = (9846.25 – 9200)/(1300/sqrt(12)) = 1.722

Do not reject Ho.

Conclude average daily sales are not significantly increased.

e) P-value = P(Z > 1.72) = 0.5 – 0.4573 = 0.0427. Since we did not reject Ho, choosing a level of significance between 4.27% and 10% would have resulted in the opposite conclusion.

Question 4

a) Ho: m ³ 43.62

Ha: m < 43.62

Reject Ho if Z < -1.9.

Z = (40.38 – 43.62)/(8.25/sqrt(60)) = -3.04

Reject Ho.

Conclude the new method significantly reduces the cost of reworking bolts from $43.62.

b) P-value = P(Z < -3.04) = 0.5 – 0.4988 = 0.0012. Since the p-value is less than 1%, we reject Ho under the general rule of thumb.

c) Lower limit = 40.38 – 1.645(8.25)/sqrt(60) = 40.38 – 1.75 = 38.63

Upper limit = 40.38 + 1.75 = 42.13

With 90% confidence, the average cost of reworking bolts with the new method ranges from $38.63 to $42.13 per 1000 bolts.

d) Sample size = (1.96*8.25/0.5)² = 1045.9 which we round up to 1046.

e) Margin of error = 1.96(8.25)/sqrt(500) = 0.72. The margin of error is 72 cents.