Question

A random sample of 1000 people was taken. Four hundred fifty of the people in the sample favored Candidate A. The 95% confidence interval for the true proportion of people who favor Candidate A is:_____

Answer 1

Answer:

The 95% confidence interval would be given (0.419;0.481).

Step-by-step explanation:

Data given and notation  

n=1000 represent the random sample taken    

X=450 represent the people in the sample favored Candidate A

$\hat p=\frac{450}{1000}=0.45$ estimated proportion of people in the sample favored Candidate A  

$\alpha=0.05$ represent the significance level

Confidence =0.95 or 95%

p= population proportion of people in the sample favored Candidate A

Solution to the problem

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 95% confidence interval the value of $\alpha=1-0.5=0.05$ and $\alpha/2=0.025$, with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

And replacing into the confidence interval formula we got:

$0.45 - 1.96 \sqrt{\frac{0.45(1-0.45)}{1000}}=0.419$

$0.45 + 1.96 \sqrt{\frac{0.45(1-0.45)}{1000}}=0.481$

And the 95% confidence interval would be given (0.419;0.481).

We are confident (95%) that about 41.9% to 48.1% of the people are favored Candidate A.