Arandom 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:

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

estimated proportion of people in the sample favored Candidate A  

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

For the 95% confidence interval the value of and , with that value we can find the quantile required for the interval in the normal standard distribution.

And replacing into the confidence interval formula we got:

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.  

answer: there are 453.592 grams in a pound.