Question

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

Answer 1

Answer:

$z=\frac{0.74-0.56}{\sqrt{0.64(1-0.64)(\frac{1}{100}+\frac{1}{125})}}=2.795$  

$p_v =2*P(Z>2.795)= 0.005$  

So if we compare the p value and using any significance level for example $\alpha=0.05$ we see that $p_v$ 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  

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

$X_{2}=70$ represent the number of people suburban residents are in favor

$n_{1}=100$ sample 1 selected

$n_{2}=125$ sample 2 selected

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

$p_{2}=\frac{70}{125}=0.56$ represent the proportion of suburban residents are in favor

z would represent the statistic (variable of interest)  

$p_v$ 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:$p_{1} = p_{2}$  

Alternative hypothesis:$p_{1} \neq p_{2}$  

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

$z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}$   (1)

Where $\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{74+70}{100+125}=0.64$

Calculate the statistic

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

$z=\frac{0.74-0.56}{\sqrt{0.64(1-0.64)(\frac{1}{100}+\frac{1}{125})}}=2.795$  

Statistical decision

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

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

$p_v =2*P(Z>2.795)= 0.005$  

So if we compare the p value and using any significance level for example $\alpha=0.05$ we see that $p_v$ 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.