In a study to estimate the proportion of residents in a certain city and its suburbs who favor the construction of a nuclear power plant, it is found that 68 of 100 urban residents favor the construction while only 58 of 125 suburban residents are in favor. Is there a significant difference between the proportions of urban and suburban residents who favor construction of the nuclear plant? Make use of a P-value.

Assuming 0.05 significance level, there is significant difference between the proportions of urban and suburban residents who favor construction of the nuclear plant.

Step-by-step explanation:

Let p(u) be the urban proportion who support nuclear power plant construction

and p(s) be the suburban proportion who support nuclear power plant construction. Then

[tex]H_{0}[/tex] : p(u) = p(s)

[tex]H_{a}[/tex] : p(u) ≠ p(s)

The formula for the test statistic is given as:

z=[tex]\frac{p1-p2}{\sqrt{{p*(1-p)*(\frac{1}{n1} +\frac{1}{n2}) }}}[/tex] where

p1 is the sample proportion of urban population who support nuclear power plant construction ([tex]\frac{68}{100}[/tex] =0.68)

p2 is the sample proportion of suburban population who support nuclear power plant construction ([tex]\frac{58}{125}[/tex] =0.464)

p is the pool proportion of p1 and p2 ([tex]\frac{68+58}{100+125}=0.56[/tex])

n1 is the sample size of urban population (100)

n2 is the sample size of suburban population (125)

Then z=[tex]\frac{0.68-0.464}{\sqrt{{0.56*0.44*(\frac{1}{100} +\frac{1}{125}) }}}[/tex] ≈ 3.24

P-value of test statistic is ≈ 0.0012

Since p-value (0.0012 ) < significance level (0.05) we can reject the null hypothesis.