00:02
Part a, 82 college students were asked whether they had completed the required english 101 course, and 53 said yes, find the best point estimate for the proportion of students at the college who have completed the required english 101 course.
00:20
So p -hat is going to be 53 divided by 82, which would be 0 .6363.
00:30
For part b, giving a test to a group of students, their grades and gender, are summarized.
00:38
Let p represent the population proportion of all female students who would receive a grade of a on this test.
00:44
Use a 95 % confidence interval to estimate p to four decimal places, if possible.
00:52
So our point estimate is going to be 10 out of 54, which is 0 .1852.
01:01
So we'll take 0 .1852 plus or minus 1 .96 times the square root of 0 .1852 times 0 .8148 divided by 54.
01:15
That's going to give us 0 .1852 plus or minus a margin of error of 0 .1036 for an interval of 0 .0816 to 0 .28888.
01:31
So that was part b, part c.
01:40
Out of 300 people sampled, 144 head kids, which means we have a p hat of 0 .48, constructed 90 % confidence interval for the true population proportion of people with kids...