Question

When the 2010 General Social Survey asked subjects in the US whether they would be willing to accept cuts in their standard of living to protect the environment, 486 of 1374 subjects said yes. a. Estimate the population proportion who would say yes. b. Construct and interpret a 99% confidence interval for this proportion. c. Conduct a significance test to determine whether a majority or minority of the population would say yes. Report and interpret the P-value.

Answer 1

Answer:

a) $\hat p=\frac{486}{1374}=0.354$ estimated proportion of adults that they would be willing to accept cuts in their standard of living to protect the environment

b) $0.354 - 2.58 \sqrt{\frac{0.354(1-0.354)}{1374}}=0.321$

$0.354 + 2.58 \sqrt{\frac{0.354(1-0.354)}{1374}}=0.387$

And the 90% confidence interval would be given (0.321;0.387).

c) $z=\frac{0.354 -0.5}{\sqrt{\frac{0.5(1-0.5)}{1374}}}=-10.82$  

$p_v =P(z<-10.82)=1.38x10^{-27}$  

So the p value obtained was a very low value and using the significance level given $\alpha=0.01$ we have $p_v<<\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the proportion of the population that would say yes is significantly lower than 0.5.

Step-by-step explanation:

Data given and notation

n=1374 represent the random sample taken

X=486 represent the adults that said that they would be willing to accept cuts in their standard of living to protect the environment

Part a

$\hat p=\frac{486}{1374}=0.354$ estimated proportion of adults that they would be willing to accept cuts in their standard of living to protect the environment

Part b

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$, with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.58$

And replacing into the confidence interval formula we got:

$0.354 - 2.58 \sqrt{\frac{0.354(1-0.354)}{1374}}=0.321$

$0.354 + 2.58 \sqrt{\frac{0.354(1-0.354)}{1374}}=0.387$

And the 90% confidence interval would be given (0.321;0.387).

Part c

$p_o=0.5$ is the value that we want to test

$\alpha=0.01$ represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)  

We need to conduct a hypothesis in order to test the claim that whether a majority or minority of the population would say yes:  

Null hypothesis:$p\geq 0.5$  

Alternative hypothesis:$p < 0.5$  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)  

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$.

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

$z=\frac{0.354 -0.5}{\sqrt{\frac{0.5(1-0.5)}{1374}}}=-10.82$  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level assumed is $\alpha=0.01$. The next step would be calculate the p value for this test.  

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

$p_v =P(z<-10.82)=1.38x10^{-27}$  

So the p value obtained was a very low value and using the significance level given $\alpha=0.01$ we have $p_v<<\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the proportion of the population that would say yes is significantly lower than 0.5.