Question

Margin error of sample mean

A juice shop owner wants to know the average amount of juice his staff puts into each glass. A sample of 50 glasses was found to have an average (mean) of 245 ml. If the known standard deviation of the amount of juice in each glass is 4 ml for the entire population, what is the margin of error of the sample mean?

A. 1.13 ml
B. 1.24 ml
C. 1.35 ml
D. 1.49 ml

I start with 1.96 and then I get lost @amistre64

Answer 1

margin of error is just the z(SE) formulation

mean +- Error

mean +- z(SE)

SE is standard error, which is just the standard deviation of a distribution of sample means

SE = sigma/sqrt(n)

Answer 2

A sample of 50 glasses was found to have an average (mean) of 245 ml. If the known standard deviation of the amount of juice in each glass is 4 ml for the entire population, what is the margin of error of the sample mean?

sample size is 50, sigma in this case is 4

Answer 3

so 4/sqrt(50)

Answer 4

yeah, one question, why are you starting with z=1.96? i dont see why that would be in post

Answer 5

I thought to find standard error 1.96 is the start of the formula. but thats another formula

Answer 6

1.96 relates to a ... confidence interval ... for lack of a better reference.

Answer 7

if we want to know the 95% confidence interval that the sample would indicate.

then 1.96(SE) gives the interval in which we would be 95% confident that the true population mean actually falls within.

Answer 8

I'm not getting any of the answer options. how did you go through the formula?

Answer 9

i don thtink we have a full queston here. can you provide a screen shot or picture of it? i feel theres something missing

Answer 10

I typed the question in exactly as it was written.

Answer 11

2.576(SE) gives us a margin of error that makes us 99% confident ....

1.96(SE) gives us a margin of error that makes us 95% confident ....

1.645(SE) gives us a margin of error that makes us 90% confident ....

Answer 12

Do you think the formula 4/sqrt(50) is wrong?

Answer 13

nope

Answer 14

what answer did you get from that formula?

Answer 15

well, without knowing what z score to use .... z(4/sqrt(50)) = z(.5657)

Answer 16

i'll try tagging @perl and see what she thinks. she has helped me on these question before.

Answer 17

they should ask you for a margin of error that is within some sort of range

if we wanted a margin of error of 2% .... etc

.02 = z(.5657)

z = 2/(56.57) or about .04

but thats usually not what is asked. they usually want to know what size sample we should get to be say 95% confident that the margin of error is with 2%

.02 = 1.96(4/sqrt(n))

and we can solve for n

your question simply is not making any sense in what you have typed out.

Answer 18

ok, perl might see something that im not :) im fine with that

Answer 19

Does the average mean of 245 ml not mean anything at all?

Answer 20

so its A) 1.13

Answer 21

is that true cause i just dont know