00:01
Okay, so i see that you need help with this question.
00:02
It says, as part of a statistic assignment in october to estimate the percentage of voters who would like to vote for mayoral candidate, each of 500 students collects his or her own random sample of likely voters.
00:16
There are 400 voters in each of the students sample.
00:21
Each student then constructs a 95 % confidence interval for the population proportion who will vote for the candidate using his or her own random sample.
00:30
Considering the 500 intervals constructed by the students, the expected number of intervals that will contain the population proportion who will vote for the candidate will be approximately, so 95 % of the 500, so you do 500 times 0 .95, that's 475.
00:56
So that would be c.
00:59
Suppose that the sample proportion of one of the students in the previous questions, beth's sample, is 46%.
01:05
Beth's 95 % interval estimate proportion of the population of voters is, so the first thing that you're going to do, so the sample size is 500 and it's 0 .46 and it's a 95 % confidence interval.
01:28
So you're going to do 0 .46 plus or minus 1 .96 times the square root of 0 .46 times 0 .54 divided by 500.
01:48
So that's 0 .46 plus or minus 1 .96 times 0 .2229.
02:00
So 0 .46 plus or minus 0 .04369.
02:07
And so the confidence interval, if you just round to two decimal places, it's going to be 0 .42, 0 .52, 0 .50.
02:25
And if i did go out further, i would do 0 .41, 0 .50.
02:44
So 4150.
02:46
So that would be, i would say about a.
02:53
It is estimated that 80 % of americans go out to eat at least once per week with a margin of error of 0 .04 in a 95 % confidence interval.
03:06
1 .09, not 95, 96.
03:15
So a 95 % confidence interval for the population proportion of americans who go out to eat once per week.
03:23
So then that is margin of error is that, oh, so times 0 .04...