A study was conducted to compare the costs of supporting a family of four Canadi

Question

A study was conducted to compare the costs of supporting a family of four Canadians for a year in different foreign cities. The lifestyle of living in Canada on an annual income of $75,000 was the standard against which living in foreign cities was compared. A comparable living standard in Perth, Australia, and Mexico City was attained for about $64,000. Suppose an executive wants to determine whether there is any difference in the average annual cost of supporting her family of four in the manner to which they are accustomed in Perth and Mexico City. She uses the following data, randomly gathered from 11 families in each city, and an alpha of 0.01 to test this difference. She assumes the annual cost is normally distributed and the population variances are equal. What does the executive find? Round the intermediate values to 2 decimal places. Round your answer to 2 decimal places. Observed t = The decision is to .

Explanation / Answer

1 = Avg. annual cost in Perth, Australia

2 = Avg. annual cost in Mexico city

H0 : 12 = 0

HA : 1 2

Data for Perth: 69500, 63800, 67500, 63800, 66700, 68000, 65000, 69500, 71000, 68500, 67500

Data for Mexico city: 64000, 64000, 65000, 64900, 62000, 60500, 61800, 63000, 64500, 63500, 62900

n1 = 11

n2 = 11

2-sample t-test using pooled variances

Sp = sqrt{[(11 – 1)*(2355.998)2 + (11 – 1)*(1411.962)2]/[11 + 11 – 2]}

= sqrt{3772181.633} = 1942.21

t = (x1 - x2)/(sp*sqrt(1/n1 + 1/n2)

t = 4060.63/828.16 = 4.91

df = n1 + n2 – 2 = 20

p value = .000086 (for 2 tailed)

The result is significant at p < .01

The decision is to reject the null hypothesis

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