A large state university wishes to know how many of its...

A large state university wishes to know how many of its students have more than 1 major, and to do so the true proportion of such students must first be estimated. Of 300 students randomly surveyed, it was found that 93 had more than 1 major. a) What is the 90% confidence interval for true proportion of students with more than 1 major? Round to 3 decimal places.

Transcript

00:01 For this problem, we want to figure out what proportion of students receive some sort of financial aid.

00:08 To do this, we have a random sample of 199 students.

00:14 So n is 199, and i'll use a for aid.

00:19 Of these 199, we find that 67 of them are receiving financial aid.

00:27 And given this information, we want to construct a 95 % confidence interval for the actual proportion of students who receive financial aid.

00:38 So let's recall how we construct these confidence intervals.

00:42 So you take your point estimate for the actual proportion, which we will calculate in a moment, and then to get the endpoints of the confidence interval, you add or subtract the margin of error term, which is given by a z score associated with the 95 % level of confidence times the square root of p hat times 1 minus p hat over n.

01:02 So we just need to work out this z score and this point estimate p hat.

01:09 Point estimate is simple enough.

01:11 We just take the ratio of students in our sample who are receiving aid, that's 67, over the total number of students sampled, which is 199.

01:23 Point estimate works out to be approximately 0 .337, two three decimal places, though i will keep the exact answer in my calculator till the end of the calculation.

01:38 Ok, and now we need this z score.

01:41 So let me just explain how we get that.

01:44 Take yourself a standard normal distribution.

01:47 So mean zero, standard deviation of one.

01:51 And then this z score is such that the area between minus z sub c and plus z sub c is equal to 95 % of the area under the curve, since we want a 95 % confidence interval.

02:05 So this area i'm shading in red here, this is 0 .95.

02:12 And due to the symmetry of the normal curve, that means the remaining 0 .05 area is equally distributed in the two tails here, which is 0 .025 in each tail.

02:28 And now we can say that the area to the left of the z score is 0 .975, which i'll now write down as a cumulative probability, which can easily be turned into the z score by using z score tables or calculators...

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