Suppose that a fast food restaurant decides to survey its customers to gauge interest in a breakfast menu. after surveying multiple people, the restaurant created a 95% confidence interval for the proportion of customers interested in a breakfast menu. the confidence interval is ( 0.521 , 0.597 ) . use the confidence interval to find the point estimate

So for our case then

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".  

The population proportion have the following distribution

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by and . And the critical value would be given by:

The confidence interval for the mean is given by the following formula:  

For our case we have the interval already given (0.521;0.597)

So we can find the margin of error from this, first we need to find the width of the interval:

And then the margin of error is given by:

Now we can find the point of estimate subtracting the margin of error tot he upper limit or adding the margin of error to the lower limti like this:

So for our case then

-4, -2, 1, 4

step-by-step explanation:

domain = x values

range = y values