suppose that, based on a sample, the 68% confidence interval for the mean of a population is (26, 32). what is the value of s/sqrtn?
A. 5
B. 3
C. 2
D. 4

Question

suppose that, based on a sample, the 68% confidence interval for the mean of a population is (26, 32). what is the value of s/sqrtn?
A. 5
B. 3
C. 2
D. 4

anonymous · Answer

$$\frac{ s }{ \sqrt{n} }$$

ganeshie8 · Answer

68% is 1 standard deviation away from mean

ganeshie8 · Answer

so, $s$ =  (32-26)/2 = 3

anonymous · Answer

so its B?

ganeshie8 · Answer

the question is asking u to find $\large \frac{s}{\sqrt{n}}$,

ganeshie8 · Answer

you got $\large   s =  3$

ganeshie8 · Answer

im not sure how to find $n$

anonymous · Answer

well 3 over something is closer than i was before so thank you

ganeshie8 · Answer

on second thoughts, 

$\large \frac{s}{\sqrt{n}} = 3$ is right

ganeshie8 · Answer

correct option is B

ganeshie8 · Answer

$3$ is the margin of error for given confidence interval, its not standard deviation

ganeshie8 · Answer

and, margin of error for 68% data is simply $\large \frac{s}{\sqrt{n}}$

anonymous · Answer

ahaha thank you, it was 3, i passed the test

ganeshie8 · Answer

yayyy! wonderful xD congrats !!