Question

The employees at the East Vancouver office of a multinational company are demanding higher salaries than those offered at the company office located in Oshawa Ontario. Their justification for the pay difference is that the difference between the average price of single-family houses in East Vancouver and that in Oshawa is more than $60,000. Before making a decision, the company management wants to study the difference in the prices of single-family houses for sale at the two locations. The results of their search of recent house sales are as follows (in $1000, rounded to the nearest thousand): East Vancouver 345 290 279 259 410 174 252 455 228 369 Oshawa 219 122 200 134 179 204 129 132 174 142 136 159 168 170 227 Assuming that the population distributions are approximately normal, can we conclude at the 0.05 significance level that the difference between the two population means is greater than $60,000? What is the degree of freedom? Multiple Choice 9 4 15 23 5

Answer 1

Ans:

degree of freedom=10-1=9

*(we use smaller of n₁-1 or n₂-1)

Assume population variances are not same.

	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	mean	std. dev
East Vancouver	345	290	279	259	410	174	252	455	228	369						306.1	87.00
Oshawa	219	122	200	134	179	204	129	132	174	142	136	159	168	170	227	166.3333	34.22
																139.7667

Test statistic:

t=(139.77-60)/SQRT((87^2/10)+(34.21^2/15))

t=2.76

p-value=tdist(2.76,9,1)=0.0110

As,p-value is less than 0.05,we reject the null hypothesis.

There is sufficient evidence to conclude that  that the difference between the two population means is greater than $60,000.