2.06. 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 74 of 100 urban residents favor the construction while only 70 of 125 suburban residents are in favor. Is there a significant difference between the proportions of urban and suburban residents who favor the construction of the nuclear plant at 5% significance level

So if we compare the p value and using any significance level for example [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the proportions are different at 5% of significance.

Step-by-step explanation:

Data given and notation

[tex]X_{1}=74[/tex] represent the number of residents in a certain city and its suburbs who favor the construction of a nuclear power plant

[tex]X_{2}=70[/tex] represent the number of people suburban residents are in favor

[tex]n_{1}=100[/tex] sample 1 selected

[tex]n_{2}=125[/tex] sample 2 selected

[tex]p_{1}=\frac{74}{100}=0.74[/tex] represent the proportion of residents in a certain city and its suburbs who favor the construction of a nuclear power plant

[tex]p_{2}=\frac{70}{125}=0.56[/tex] represent the proportion of suburban residents are in favor

z would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the value for the test (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the proportions are different, the system of hypothesis would be:

Null hypothesis:[tex]p_{1} = p_{2}[/tex]

Alternative hypothesis:[tex]p_{1} \neq p_{2}[/tex]

We need to apply a z test to compare proportions, and the statistic is given by:

[tex]z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}[/tex] (1)

Where [tex]\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{74+70}{100+125}=0.64[/tex]

Calculate the statistic

Replacing in formula (1) the values obtained we got this:

[tex]z=\frac{0.74-0.56}{\sqrt{0.64(1-0.64)(\frac{1}{100}+\frac{1}{125})}}=2.795[/tex]

Statistical decision

The significance level provided is [tex]\alpha=0.05[/tex] ,and we can calculate the p value for this test.

Since is a two tailed test the p value would be:

[tex]p_v =2*P(Z>2.795)= 0.005[/tex]

So if we compare the p value and using any significance level for example [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the proportions are different at 5% of significance.