Question

The office occupancy rates were reported for 4 Texas metropolitan areas. Do the following data suggest that the office vacancies were independent of metropolitan area? Choose one alpha value from 0.05, 0.01, 0.1. What is your conclusion? Please include all the work.

Occupancy Status	Dallas	Houston	Austin	San Antonio
Occupied	160	116	192	174
Vacant	40	34	33	26

Answer 1

Answer Date: 30/07/2020 To test the claim is that the office vacancies were independent of metropolitan area at 5% significance level. The null and alternative hypothesis is, The null hypothesis is that the office vacancies were independent of metropolitan area. The alternative hypothesis is that the office vacancies were not independent of metropolitan area. The chi-square test statistics is, First, compute expected frequencies then find the chi-square test statistics.

The expected frequency for Occupied and Dallas is, Row total x Column total Expected frequency = N 642 x 200 775 = 165.677 = The expected frequency for Occupied and Houston is, Row total x Column total Expected frequency = N 642 x150 775 = 124.258

The expected frequency for Occupied and Austin is, Row total > Column total Expected frequency = N 642 x 225 775 = 186.387 The expected frequency for Occupied and San Antonio is, Row total > Column total Expected frequency = N 642 x 200 775 = 165.677

The expected frequency for Vacant and Dallas is, Row total > Column total Expected frequency = N 133 x 200 775 = 34.323 The expected frequency for Vacant and Houston is, Expected frequency Row total > Column total N 133 x 150 775 = 25.742

The expected frequency for Vacant and Austin is, Row total x Column total Expected frequency = N 133 x 225 775 = 38.613 = The expected frequency for Vacant and San Antonio is, Row total x Column total Expected frequency = N 133 x 200 775 = 34.323

The chi-square test statisticis, r? =($. - f.)? + + + fe (160 – 165.677)2 (116-124.258)2 (192 – 186.387)2 (174-165.677) 165.677 124.258 186.387 165.677 (40–34.323)2 (34 – 25.742)2 (33 – 38.613)2 (26–34.323)2 + 34.323 25.742 38.613 34.323 = 7.753 + + + The chi-square test statistic is 7.753 The degrees of freedom are, The information in given matrix is two rows and two columns for compute degrees of freedom. d.f. =(r-1)×(C-1) =(2-1)(4-1) = 3

The p-value is, From the chi-square table, show chi-square value 7.753 and degrees of freedom is 3, so the p-value is between 0.05 and 0.10. The p-value is 0.05< p <0.10 Conclusion The significant p-value is higher than 0.05, so the null hypothesis is not rejected at 5% level of significance. It cannot conclude that the office vacancies were not independent of metropolitan area. The result is not statistically significant. So, the data does suggest that the office vacancies were independent of metropolitan area.