Community college students conduct a survey at their college. They ask "Do you plan to transfer to a university to pursue a baccalaureate degree?" In a random sample of 100 students, 80 indicate that they plan to transfer to a university. Find the 99% confidence interval for estimating the proportion of all of the community college's students who plan to transfer. Group of answer choices 0.734 to 0.866 0.722 to 0.878 0.697 to 0.903 We should not calculate the 99% confidence interval because normality conditions are not met.

The 99% confidence interval would be given by (0.697;0.903)

And the best option is : 0.697 to 0.903

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

For this case the conditions we have the conditions to assume a normal distribution for the population proportion since:

np=100*0.8=80 >10 , n(1-p) = 10*(1-0.8) =20 >10

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:  

The estimated proportion is given by . If we replace the values obtained we got:

The 99% confidence interval would be given by (0.697;0.903)

And the best option is : 0.697 to 0.903

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b). 4320 have a nice day