Calculate a 99% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it, and interpret the interval in context. Round to 4 decimal places.

The 99% confidence interval would be given by (0.4238;0.5622)

Step-by-step explanation:

For this case we assume this previous info: "Among a simple random sample of 347 American adults who do not have a four-year college degree and are not currently enrolled in school, 171 said they decided not to go to college because they could not afford school."

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

For this case the estimated proportion is given by:

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 99% 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:  

If we replace the values obtained we got:

The 99% confidence interval would be given by (0.4238;0.5622)

x=6 and x=3

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