Respuesta :
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:
[tex][(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}} }[/tex]
Where [tex]p_{1}[/tex] : proportion sample of individuals who worked
at more than one job in the city one
[tex]n_{1}[/tex]: Number of respondents in the city one
[tex]p_{1}[/tex] : proportion sample of individuals who worked
at more than one job in the city two
[tex]n_{1}[/tex]: Number of respondents in the city two
Then
α = 0.01 and α/2 = 0.005
and [tex]z_{\alpha/2} = 2.575[/tex]
[tex]p_{1} = \frac{112}{384} = 0.2916[/tex]
[tex]p_{2} = \frac{91}{432} = 0.2106[/tex]
[tex]n_{1}= 384[/tex] and [tex]n_{2}= 432[/tex]
The confidence interval is:
[tex][(0.2916 - 0.2106) \pm 2.575 \sqrt{\frac{0.2916(1-0.2916)}{384} + \frac{0.2106(1-0.2106)}{432} }[/tex]
(0.00278 < P1 - P2< 0.15921)
