Question

There is a margin of error on a food survey of +/– 5%. What is the range of the number of students who do not want hot dogs if 83 of 127 students said they did not want hot dogs on Wednesdays and there are 900 students at the school?

Answer 1

A. 529 to 617 students
B. 559 to 647 students
C. 559 to 617 students
D. 559 to 600 students
these are the options

Answer 2

The confidence interval for a proportion takes on the form
\[\left(\hat{p}-Z_{\alpha/2}\sqrt\frac{\hat{p}(1-\hat{p})}{n},\,\hat{p}+Z_{\alpha/2}\sqrt\frac{\hat{p}(1-\hat{p})}{n}\right)\]
where \(Z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\) is the margin of error, and \(\hat{p}\) is the estimated proportion.

In this case, you're given the MOE to be \(\pm5\%=\pm0.05\), and \(\hat{p}=\dfrac{83}{127}\approx0.65\).

What would the interval be?

Answer 3

i think C maybe?!

Answer 4

In particular, the interval will be a range of proportions, say \((a,b)\). Given this confidence interval, your answer will simply be the range of proportions of the total number of students, i.e. if \((a,b)\) is the proportion confidence interval, then for the entire school of \(n\) students the desired range would be \((an,bn)\).

Answer 5

ok can u explain as if u were explaining to a 2 year old plz im not that smart @SithsAndGiggles

Answer 6

I doubt a 2 year old could grasp the notion of a confidence interval, but sure I'll give it a shot.

Confidence interval (for proportions) = \((\hat{p}-MOE,\,\hat{p}+MOE)\).

Here, \(\hat{p}\) is the estimated proportion, which in this case is \(\dfrac{83}{127}\approx0.65\).

MOE is given to be 5%, or \(0.05\).

So, the interval should be about \((0.60,\,0.70)\). Got it so far?

Answer 7

yeah

Answer 8

so out of these witch would it be A. 529 to 617 students
B. 559 to 647 students
C. 559 to 617 students
D. 559 to 600 students

Answer 9

Alright, so we have a set range of proportions. We want to apply this to a student body of 900 students. How many students make up 60% of 900? How many make up 70%?

For whatever reason, the right answer isn't listed above... Are you sure you provided the right \(\hat{p}\) and MOE?

Answer 10

so 630

Answer 11

Right, the upper limit should be 630.

Answer 12

ok so what would the lower limit be

Answer 13

You found the upper limit by multiplying the upper limit of the interval by 900, i.e. 70% of 900. The lower limit is found similarly by finding 60% of 900.

Answer 14

umm 540

Answer 15

but those arnt any of the answers