Question

A random sample of 384 people in a mid-sized city (city one) revealed 112 individuals who worked at more than one job. A second random sample of 432 workers from another mid-sized city (city two) found 91 people who work at more than one job. Find a 99% confidence interval for the difference between the proportions of workers in the two cities who work at more than one job.Select one:a. (0.003, 0.159)b. (0.021, 0.141)c. (-0.159, 0.004)d. (0.031, 0.131)e. Sample sizes aren't large enough to justify using z-procedures

Answer 1

Answer:

99% confidence interval is:

(0.00278 < P1 - P2< 0.15921)

Step-by-step explanation:

For calculating a confidence intervale for the difference between the proportions of workers in the two cities, we calculate the following:

$[(p_{1} - p_{2}) \pm z_{\alpha/2} \sqrt{\frac{p_{1}(1-p_{1})}{n_{1}} + \frac{p_{2}(1-p_{2})}{n_{2}} }$

             Where  $p_{1}$ : proportion sample of individuals who worked      

                                                     at more than one job in the city one

                           $n_{1}$: Number of respondents in the city one

                           $p_{1}$ : proportion sample of individuals who worked      

                                                     at more than one job in the city two

                           $n_{1}$: Number of respondents in the city two

Then

α = 0.01 and α/2 = 0.005

and $z_{\alpha/2} = 2.575$

$p_{1} = \frac{112}{384} = 0.2916$

$p_{2} = \frac{91}{432} = 0.2106$

$n_{1}= 384$  and $n_{2}= 432$

The confidence interval is:

$[(0.2916 - 0.2106) \pm 2.575 \sqrt{\frac{0.2916(1-0.2916)}{384} + \frac{0.2106(1-0.2106)}{432} }$

(0.00278 < P1 - P2< 0.15921)