STAT217 Tutorial sheet #3 Solutions

Question 1

We make manufacturing group 1.

Ho: m1 = m2

Ha: m1 ¹ m2

Reject Ho if Z < -1.96 or > 1.96

Z = (580 – 540)/sqrt(54²/30 + 48²/40) = 3.215

Reject Ho.

Conclude there is a significant difference in average weekly wages between employees in the manufacturing and service industries.

Question 2

P-value = 2P(Z > 3.215)

P(Z > 3.215) = 1 – P(Z < 3.215) = 1 – 0.9994 = 0.0006

P-value = 2(0.0006) = 0.0012 = 0.12%

We would reject Ho since the p-value is less than 1%.

Question 3

Lower limit = (580 – 540) – 1.96 sqrt(54²/30 + 48²/40) = 40 – 24.39 = 15.61

Upper limit = 40 + 24.39 = 64.39

With 95% confidence, the average weekly wage in manufacturing is $15.61 to $64.39 higher than that in the service industry.

If this interval were used to test the hypothesis in question 1, we would reject Ho since the hypothesis difference of zero is not in the interval.

Question 4

We first conduct the F test for the equality of variances. Since the old neighbourhood has the higher standard deviation, we choose this as group 1.

Ho: s1 = s2

Ha: s1 ¹ s2

Numerator df = denominator df = 11. Reject Ho if F > 3.4737. Note that the lower critical value is not needed.

To keep it simple, I divide everything by 1000.

F = 15.4²/12.6² = 1.4938

Do not reject Ho.

Conclude no significant difference in the variances. We assume s1 = s2 and proceed with the t test assuming equal variances.

We choose old as group 1.

Ho: m1 £ m2

Ha: m1 > m2

Reject Ho if t > 1.7171

Sp² = [11(15.4)² + 11(12.6)²]/22 = 197.96

t = (104 – 95)/sqrt(197.96/12 + 197.96/12) = 1.5669

Do not reject Ho.

Conclude the average annual household incomes in the old neighbourhood is not significantly higher than that of the new subdivision.

Question 5

P-value = P(t > 1.5669) = 1 – P(t < 1.5669) = 1 – 0.9343 = 0.0657 = 6.57% based on 22 degrees of freedom.

Since we did not reject the null hypothesis, we would reject Ho for levels of significance between 6.57% and 10%.

Question 6

Lower limit = (104 – 95) – 2.0739sqrt(197.96/12 + 197.96/12) = 9 – 11.9 = -2.9

Upper limit = 9 + 11.9 = 20.9

We multiply by 1000 to get the confidence interval limits:

-$2,900 < mean(old) – mean(new) < $20,900

If the average annual household income in the new neighbourhood is $95,000, then with 95% confidence, the average annual household income in the old neighbourhood ranges between $92,100 and $115,900.

Question 7

Since we expect before to be greater, we subtract after from before.

Ho: md £ 0

Ha: md > 0

Reject Ho if t > 1.7823

t = -5.8462/(9.1182/sqrt(13)) = -2.3117

Do not reject Ho.

Conclude the experimental drug does not significantly reduce the average time.

Question 8

P-value = P(t > -2.3117) = 1 – P(t < -2.3117) = 1 – 0.0197 = 0.9803 = 98.03% based on 12 degrees of freedom. If a level of significance had not been chosen, we would reach the same conclusion since the p-value is greater than 10%.

Question 9

Lower limit = -5.8462 – 2.1788(9.1182)/sqrt(13) = -5.8462 – 5.51 = -11.4

Upper limit = -5.8462 + 5.51 = -0.3

-11.3 < md < -0.3

With 95% confidence, the average time before the drug was administered was 0.3 to 11.4 seconds less than after the drug was administered.

Question 10

We choose members as group 1.

Ho: p1 £ p2

Ha: p1 > p2

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

phat = (170 + 144)/400 = 0.785

Z = (0.85 – 0.72)/sqrt(0.785(0.215)/200 + 0.785(0.215)/200) = 3.1644

P-value = P(Z > 3.1644) = 1 – P(Z < 3.1644) = 1 – 0.9992 = 0.0008 = 0.08%

Reject Ho.

Conclude the members are significantly more likely to visit new exhibits than those from the general public.

Question 11

Lower limit = (0.85 – 0.72) – 2.5758sqrt(0.85(0.15)/200 + 0.72(0.28)/200) = 0.13 – 0.1045 = 0.0255

Upper limit = 0.13 + 0.1045 = 0.2345

2.55% < p(members) – p(public) < 23.45%

With 95% confidence, the percentage of members who will visit new exhibits is 2.55% to 23.45% higher than that of the general public.