MGMT2262 Worksheet #5 Solutions

Question 1

Z = (4.3 – 4)/(1.2/sqrt(50)) = 1.77.

P(Xbar > 4.3) = P(Z > 1.77) = 0.5 – 0.4616 = 0.0384.

Question 2

Z = (4.3 – 4)/(1.2/sqrt(100)) = 2.5.

P(Xbar > 4.3) = P(Z > 2.5) = 0.5 – 0.4938 = 0.0062.

Question 3

a) Lower limit = 15.14 – 1.96(2.54)/sqrt(100) = 15.14 – 0.50 = 14.64

Upper limit = 15.14 + 0.5 = 15.64.

With 95% confidence, the average amount that people spend per week on fast food ranges from $14.64 to $15.64.

b) The Z value changes from 1.96 to 2.576. So, the margin of error changes to 2.576(2.54)/sqrt(100) = 0.65.

Lower limit = 15.14 – 0.65 = 14.49

Upper limit = 15.14 + 0.65 = 15.79

Since the margin of error is larger, the confidence interval is wider.

c) Once again, the Z value is 1.96. The margin of error is 1.96(2.54)/sqrt(200) = 0.35.

Lower limit = 15.14 – 0.35 = 14.79

Upper limit = 15.14 + 0.35 = 15.49

Since the margin of error is smaller, the confidence interval is narrower.

d) The margin of error is 1.96(5.08)/sqrt(100) = 1.00

Lower limit = 15.14 – 1= 14.14

Upper limit = 15.14 + 1= 16.14

Since the margin of error is larger, the confidence interval is wider.

e) Sample size = n = (1.96(2.54)/0.5)² = 9.96² = 99.2 which we round up to 100.

Question 4

We use the t distribution since the data is assumed to be normally distributed, the population standard deviation is unknown and the sample size is under 30. The sample size is 8, so the degrees of freedom is 7. The t value is 2.365 from the t.025 column in the t table under 7 degrees of freedom. The sample mean is 16.1125 and the sample standard deviation is 2.9967.

Lower limit = 16.1125 – 2.365(2.9967)/sqrt(8) = 16.11 – 2.51 = 13.60

Upper limit = 16.11 + 2.51 = 18.62

Question 5

The sample size n = (1.96(2.9967)/0.25)² = 23.4944² = 551.99 which we round up to 552.

Question 6

Margin of error = 1.96(2.9967)/sqrt(300) = 0.3391 which rounds to 0.34 or 34 cents.

Question 7

Ho: m £ 1000

Ha: m > 1000

Reject Ho if Z > 1.645

Z = (1120 – 1000)/(250/sqrt(14)) = 1.796

Reject Ho

Conclude average daily sales are significantly greater than $1000 and that the campaign was successful.

Question 8

Ho: m ³ 2.2

Ha: m < 2.2

Reject Ho if Z < -2.326

Z = (1.8 – 2.2)/(1.3/sqrt(50)) = -2.1757

Do not reject Ho

Conclude the average number of mistakes is not significantly below 2.2 per 1000 lines of code. Therefore the program did not significantly reduce the average number of mistakes.

Question 9

Ho: m = 1000 ml

Ha: m ¹ 1000 ml

Reject Ho if Z < -2.17 or Z > 2.17

Z = (1002 – 1000)/(12.5/sqrt(100)) = 1.6

Do not reject Ho

Conclude the average amount of milk is not significantly different than 1 L. Therefore, the specifications are being met.

Question 10

Question 7: p-value = P(Z > 1.8) = 0.5 – 0.4641 = 0.0359. Since we rejected Ho, if we had chosen a level of significance between 1% and 3.59%, we would not have rejected Ho.

Question 8: p-value = P(Z < -2.18) = 0.5 – 0.4854 = 0.0146. Since we did not reject Ho, if we had chosen a level of significance between 1.46% and 10%, we would have rejected Ho.

Question 9: p-value = 2P(Z > 1.6); P(Z > 1.6) = 0.5 – 0.4452 = 0.0548; p-value = 2(0.0548) = 0.1096. Since the p-value is greater than 10%, there is no level of significance between 1% and 10% we could have chosen in which the opposite conclusion would have been reached.

Question 11

Lower limit = 1002 – 2.17(12.5)/sqrt(100) = 1002 – 2.7125 = 999.2875

Upper limit = 1002 + 2.7125 = 1004.7125

Since the hypothesis mean of 1000 ml falls inside the 97% confidence interval, we do not reject Ho at a 3% level of significance.

Question 12

Ho: m = 25

Ha: m ¹ 25

Reject Ho if t < -2.365 or t > 2.365 (7 degrees of freedom)

t = (20.25 – 25)/(4.7132/sqrt(8)) = -2.8505

Reject Ho

Conclude that the average number of hours per week that semi-retired people work is significantly different than 25 hours.

Question 13

P-value = 2P(t > 2.8505). We see that from the t table that the test statistic lies between 2.365 and 2.998. Therefore, P(t > 2.8505) lies between 0.01 and 0.025. By extension, the p-value is between 0.02 and 0.05.

Question 14

Lower limit = 20.25 – 2.365(4.7132)/sqrt(8) = 20.25 – 3.94 = 16.31

Upper limit = 20.25 + 3.94 = 24.19.

With 95% confidence, semi-retired people work between 16.3 and 24.2 hours per week. We would reject Ho at a 5% level of significance since the hypothesis mean of 25 hours falls outside the 95% confidence interval.