Question

A random sample of 1000 people was taken. Four hundred fifty of the people in the sample favored Candidate AT. The 95% confidence interval for the true proportion of people who favor Candidate A is a. .424 to .476. b. .419 to .481. c. .40 to .50. d. .45 to .55.

Answer 1

Answer:

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

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

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

b. .419 to .481

Step-by-step explanation:

We know the following info:

$n = 1000$ sample size selected

$X= 450$ represent the number of people who favored Candidate AT

The sample proportion would be:

$\hat p=\frac{450}{1000}=0.45$

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 $\alpha=1-0.95=0.05$ 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).   And the best option would be:

b. .419 to .481