Question

A magazine provided results from a poll of 500 adults who were asked to identify their favorite pie. Among the 500 ​respondents, 12 ​% chose chocolate​ pie, and the margin of error was given as plus or minus 5 percentage points. What values do ModifyingAbove p with caret ​, ModifyingAbove q with caret ​, ​n, E, and p​ represent? If the confidence level is 90 ​%, what is the value of alpha ​?

Answer 1

Answer:

n=500 represent the random sample taken    

$\hat p=0.12$ estimated proportion of people that chose chocolate​ pie

$\hat q =1-\hat p=1-0.12=0.88$ represent the people that NOT chose chocolate​ pie

E=0.05 represent the error or margin of error given by the following formula:

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

p= true population proportion of people that chose chocolate​ pie

If the confidence level is 90 ​%, what is the value of alpha ​?

$\alpha=1-0.9 =0.1$ and the value of $\alpha/2 =0.05$,

$z_{\alpha/2}=-1.64$ and $z_{1-\alpha/2}=1.64$

$ME=1.64 \sqrt{\frac{0.12(1-0.12)}{500}}=0.0238$

Step-by-step explanation:

Data given and notation

What values do ModifyingAbove p with caret ​, ModifyingAbove q with caret ​, ​n, E, and p​ represent?

n=500 represent the random sample taken    

X represent the people that chose chocolate​ pie

$\hat p=0.12$ estimated proportion of people that chose chocolate​ pie

$\hat q =1-\hat p=1-0.12=0.88$ represent the people that NOT chose chocolate​ pie

E=0.05 represent the error or margin of error given by the following formula:

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

z would represent the quantile of the normal standard distribution

p= true population proportion of people that chose chocolate​ pie

The confidence interval for the population proportion is given by this formula :

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

If the confidence level is 90 ​%, what is the value of alpha ​?

On this case the value for the significance would be $\alpha=1-0.9 =0.1$ and the value of $\alpha/2 =0.05$, we can find the quantiles of the normal standard distribution given by:

$z_{\alpha/2}=-1.64$ and $z_{1-\alpha/2}=1.64$

And with the following excel codes:

"=NORM.INV(0.05,0,1)" "=NORM.INV(1-0.05,0,1)"

And we can find the margin of error like this:

$ME=1.64 \sqrt{\frac{0.12(1-0.12)}{500}}=0.0238$