Question

The employees at the East Vancouver office of a multinational company are demanding higher salaries than...

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?

Homework Answers

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

df=10-1=9

(we use smaller of n1-1 or n2-1)

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.

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