MGMT2263 Tutorial Sheet 2 Solutions

 

1)     Since this is a two-tail test, it doesn’t matter which firm is group 1.

Ho: m1 = m2

Ha: m1 Ή m2

Since we have two independent samples that are both at least 30 in size, we can use the Z test for two means.

Reject Ho if Z < -1.96 or > 1.96.

Z = (480 – 470)/sqrt(142/30 + 102/40) = 3.33.

Reject Ho.

Conclude there is a significant difference in the weekly wages of the two firms.

2)     P-value = 2P(Z > 3.33) = 2(0.0004) = 0.0008. Since this is less than 1%, we have strong support for Ha. So, there is no level of significance between 1% and 10% we could have chosen in which the opposite conclusion would have been reached.

3)     Lower limit = (480 – 470) – 1.96 sqrt(142/30 + 102/40) = 10 – 5.89 = 4.11

Upper limit = 10 + 5.89 = 15.89

4.11 < m(A) - m(B) < 15.89

Since the hypothesized difference of zero does not fall in the 95% confidence interval, we would reject Ho at a 5% level of significance.

4)     We need to solve –1.96 < (xbar1 – xbar2)/ sqrt(142/30 + 102/40) < 1.96 or xbar1 – xbar2 to fall between –5.89 and 5.89  (surprise, surprise, the margin of error from the confidence interval!). So, the largest difference between the two sample means in which we would not have rejected Ho is 5.89.

5)     For this problem, we make the professionals group 1.

Ho: m1 - m2 £ 50

Ha: m1 - m2 > 50

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

Z = (240 – 180 - 50)/sqrt(652/400 + 452/400) = 2.53.

p-value = P(Z > 2.53) = 0.5 – 0.4943 = 0.0057.

Since the p-value is less than 1%, reject Ho.

Conclude the average difference between professionals and the general public is more than 50 hours per month.