"Calculate a 90% 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 90% confidence interval would be given by (0.435;0.525)

Step-by-step explanation:

Assuming this previous  info: "Exercise 6.16 presents the results of a poll where 48% of 331 Americans who decide to not go to college do so because they cannot afford it".

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

The estimated proportion for this case is 0.48 and n = 331

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 90% 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 90% confidence interval would be given by (0.435;0.525)