A research institute poll asked respondents if they felt vulnerable to identity theft. In the poll, n=916 and x=564 who said "yes." Use a 99 %99% confidence level.
(a) Find the best point estimate of the population proportion p.
(b) Identify the value of the margin of error E.
(c) Construct the confidence interval.

a)

b) For the 99% confidence interval the value of and , with that value we can find the quantile required for the interval in the normal standard distribution.

The margin of error is given by:

c)

And the 90% confidence interval would be given (0.575;0.657).

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

Part a

For this case the best point of estimate of the population proportion is:

Part b

The confidence interval would be given by this formula

For the 99% confidence interval the value of and , with that value we can find the quantile required for the interval in the normal standard distribution.

The margin of error is given by:

Part c

And replacing into the confidence interval formula we got:

And the 90% confidence interval would be given (0.575;0.657).

cheater

step-by-step explanation: