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

`z=\frac{0.74-0.56}{\sqrt{0.64(1-0.64)((1)/(100)+(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<\alpha` 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.

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)=(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)=(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) \\eq 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)((1)/(n_(1))+(1)/(n_(2)))}}` (1)

Where
`\hat p=(X_(1)+X_(2))/(n_(1)+n_(2))=(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)((1)/(100)+(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<\alpha` 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.