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

Answer 1

I tried 4/sqrt(50) but that is none of the answers.

Answer 2

ME= Zc * sigma / sqrt(n)

ME = 1.96 * 4 / sqrt(40)

Answer 3

how did you get sqrt(40)?

Answer 4

typo

Answer 5

ME = 1.96 * 4 / sqrt(50)= 1.108 at 95 % significance
ME = 2.58 * 4 / sqrt(50)=1.459 at 99% significance

Answer 6

I tried all the signifiance percents but none worked.

Answer 7

I tried d first and it was wrong.

Answer 8

I don't know how to choose between a b and c.

Answer 9

im not sure, the significance level is not given.
can you try to upload the question

Answer 10

well I typed in the question exactly as written.

Answer 11

oh they used Zc = 2

Answer 12

ME = 2 * 4 / sqrt(50)

Answer 13

that comes out exactly 1.13

Answer 14

How did you get 2*4?

Answer 15

whoever made this question must have thought this was clear or obvious (it isn't)

Answer 16

$$ \bf \rm margin~ of~ error = ( critical ~z ~score ) * \frac{ standard~ deviation~ of pop.}{ \sqrt{sample size} }

$$

Answer 17

the critical z score is 2 here

Answer 18

okay ill go with 1.13