Question

Office 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. The data are linked below. Construct a 99% confidence interval to estimate the difference in vacancy rates between these two cities. What conclusions can be made?
city 1 city 2
x1=22 x2=14
n1=165 n2=140

Answer 1

Answer:

$(0.13-0.1) - 2.58 \sqrt{\frac{0.13(1-0.13)}{165} +\frac{0.10(1-0.10)}{140}}=-0.0664$  

$(0.13-0.1) + 2.58 \sqrt{\frac{0.13(1-0.13)}{165} +\frac{0.10(1-0.10)}{140}}=0.124$  

We are confident at 99% that the difference between the two proportions is between $-0.0664 \leq p_1 -p_2 \leq 0.124$ . And since the confidence interval cotains the value 0 we don't have enough evidence to conclude that we have significant differences between the to proportions in these two cities

Step-by-step explanation:

$p_1$ represent the real population proportion for 1

$\hat p_1 =\frac{22}{165}=0.13$ represent the estimated proportion for 1

$n_1=165$ is the sample size required for 1

$p_2$ represent the real population proportion for 2

$\hat p_2 =\frac{14}{140}=0.10$ represent the estimated proportion for 2

$n_2=140$ is the sample size required for 2

$z$ represent the critical value for the margin of error  

The confidence interval for the difference of two proportions would be given by this formula  

$(\hat p_1 -\hat p_2) \pm z_{\alpha/2} \sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1} +\frac{\hat p_2 (1-\hat p_2)}{n_2}}$  

For the 99% confidence interval the significance is $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$, and the critical value using the normal standard distribution.  

$z_{\alpha/2}=2.58$  

Replacing we got:

$(0.13-0.1) - 2.58 \sqrt{\frac{0.13(1-0.13)}{165} +\frac{0.10(1-0.10)}{140}}=-0.0664$  

$(0.13-0.1) + 2.58 \sqrt{\frac{0.13(1-0.13)}{165} +\frac{0.10(1-0.10)}{140}}=0.124$  

We are confident at 99% that the difference between the two proportions is between $-0.0664 \leq p_1 -p_2 \leq 0.124$ . And since the confidence interval cotains the value 0 we don't have enough evidence to conclude that we have significant differences between the to proportions in these two cities