Five occupancy in a city is an indication of the economic health of the region in which it is located. A random sample of offices in two cities was selected, and the number of vacancies was recorded. Construcr a 95% confidence interval to estimate the difference in vacancy rates between these two cities. What conclusions can be made? Occupancy by City
City 1 City 2
x1 = 22 x2 = 12
n1 = 155 n2 = 135
A. Since the confidence interval does include zero, there is no evidence that the vacancy rates are different between the two cities.
B. Since the confidence interval does not include zero, there is no evidence that the vacancy rates are different between the two cities.
C. Since the confidence interval does not include zero, there is evidence that the vacancy rates are different between the two cities.
D. Since the confidence interval does include zero, there is evidence that the vacancy rates are different between the two cities.

Step-by-step explanation:

Confidence interval for the difference in the two proportions is written as

Difference in sample proportions ± margin of error

Sample proportion, p= x/n

Where x = number of success

n = number of samples

For city 1,

x = 22

n1 = 155

p1 = 22/155 = 0.14

For city 2,

x = 12

n2 = 135

p2 = 12/135 = 0.09

Margin of error = z√[p1(1 - p1)/n1 + p2(1 - p2)/n2]

To determine the z score, we subtract the confidence level from 100% to get α

α = 1 - 0.95 = 0.05

α/2 = 0.05/2 = 0.025

This is the area in each tail. Since we want the area in the middle, it becomes

1 - 0.025 = 0.975

The z score corresponding to the area on the z table is 1.96. Thus, confidence level of 95% is 1.96

Margin of error = 1.96 × √[0.14(1 - 0.14)/155 + 0.09(1 - 0.09)/135]

= 1.96 × √0.00138344086

= 0.073

Confidence interval = 0.12 - 0.09 ± 0.073

= 0.03 ± 0.073

C. Since the confidence interval does not include zero, there is evidence that the vacancy rates are different between the two cities.

i think it would be 42cm