MGMT2262 Tutorial Sheet 5 Solutions

1) Using the central limit theorem, Z = (25 – 30)/(12.5/sqrt(50)) = -2.83. P(Xbar < 25) = P(Z < -2.83) = 0.5 – 0.4977 = 0.0023.

2) Lower limit = 29.50 – 1.96(12.5)/sqrt(50) = 29.50 – 3.46 = 26.04

Upper limit = 29.50 + 3.46 = 32.96.

With 95% confidence, the average sale at this stores ranges from $26.04 to $32.96.

3) If the level of confidence is increased to 99%, the Z value changes from 1.96 to 2.576. The margin of error then becomes 2.576(12.5)/sqrt(50) = 4.55. The difference between the new margin of error and the old one is 4.55 – 3.46 = 1.09.

4) Sample size = n = (1.96(12.50)/2.5)² = 9.8² = 96.04 which we round up to 97.

5) The thing to keep in mind is that the Central Limit theorem works regardless of the parent distribution if the sample size is large enough.

Z1 = (75.3 – 75)/(1.2/sqrt(60)) = 1.94

Z2 = (74.7 – 75)/(1.2/sqrt(60)) = -1.94

So P(74.7 < Xbar < 75.3) = P(-1.94 < Z < 1.94). From the Z table, P(0 < Z < 1.94) = 0.4738. So P(-1.94 < Z < 1.94) = 2(0.4738) = 0.9476.

6) Since the sample size is 8, the degrees of freedom is 7. The t value is 2.365. The mean is 72.4625 and the standard deviation is 10.7661.

Lower limit = 72.4625 – 2.365(10.7661)/sqrt(8) = 72.4625 – 9 = 63.5 after rounding to 1 decimal.

Upper limit = 72.4625 + 9 = 81.5

7) If the level of confidence is changed to 99%, the t value changes to 3.499 and the margin of error becomes 3.499(10.7661)/sqrt(8) = 13.3186. The lower limit is 72.4625 – 13.3186 = 59.14 and the upper limit is 72.4625 + 13.3186 = 85.78

8) Ho: m £ 5

Ha: m > 5

Reject Ho if Z > 1.645

Z = (5.3 - 5)/(1.25/sqrt(60)) = 1.86

Reject Ho

Conclude that people spend significantly more than $5 on average.

9) P-value = P(Z > 1.86) = 0.5 – 0.4686 = 0.0314. Since we rejected Ho, if we had chosen a level of significance between 1% and 3.14%, we would not have rejected Ho.

10) Ho: m = 25

Ha: m ¹ 25

Reject Ho if Z < -2.576 or Z > 2.576

Z = (27.9 – 25)/(14.2/sqrt(200)) = 2.89

Reject Ho

Conclude that people no longer spend 25 minutes on average commuting to work.

11) P-value = 2P(Z > 2.89); P(Z > 2.89) = 0.5 – 0.4981 = 0.0019; p-value = 2(0.0019) = 0.0038. Since this is less than 1% under the general rule of thumb, we would still reject Ho and reach the same conclusion.

12) Ho: m ³ 175

Ha: m < 175

Reject Ho if p-value is less than 1%. Do not reject Ho if p-value > 10%.

Z = (173.5 – 175)/(2.05/sqrt(12)) = -2.5347 which we round to –2.53

P-value = P(Z < -2.53) = 0.5 – 0.4943 = 0.0057. Since this is less than 1%, we reject Ho. Conclude the average amount the machine dispenses is significantly less than 175 ml.

13) Ho: m = 0

Ha: m ¹ 0

Reject Ho if p-value < 1%. Do not reject Ho if p-value > 10%.

t = (0.005317 – 0)/(0.022951/sqrt(6)) = 0.5674.

P-value = 2P(t > 0.5674) with 5 degrees of freedom.

From the t table, we see that 0.5675 is less than the 10% critical value of 1.476. Therefore P(t > 0.5675) > 10% and consequently so is the p-value. Therefore, do not reject Ho. Conclude the average time for this clock system is not significantly different than GMT.

14) Lower limit = 0.005317 – 2.571(0.02295)/sqrt(6) = 0.005317 – 0.02409 = -0.02

Upper limit = 0.005317 + 0.02409 = 0.03

With 95% confidence, the average difference between the clock and GMT is between

–0.02 and + 0.03 seconds. We would reach the same conclusion if we had tested at a 5% level of significance since the hypothesis mean of zero falls inside the 95% confidence interval.