Question

Suppose that, based on a sample, the 99.7% confidence interval for the mean of a population is (14, 44). What is the value of s/√n?

A. 4
B. 3
C. 2
D. 5

Answer 1

remember how margin of error \(m\) is your critical value for that confidence level times the SD? I'll call the critical value \(z^*\), but replace it with whatever variable name is relevant

\[m = z^* \sigma = z^* \frac{s}{\sqrt{n}} \\
\frac{s}{\sqrt{n}} = \frac{m}{z^*} = \frac{(14 + 44)/2}{z^*} = \frac{29}{z^*} = \frac{29}{3}\]

Your \(z^*\) value is of course 3 since 99.7% of data lies within 3 SDs of the mean on a normal distribution (assuming this is normal).

Answer 2

@KevinOrr so the answer is B. 3?

Answer 3

hmm, your margin of error seems good but i want to know why you would do mean/z

lets take the uppper half of the confidence interval:

\[mean +z(s/\sqrt n)=max\]
\[z(s/\sqrt n)=max-mean\]
\[s/\sqrt n=\frac{max-mean}{z}\]
since mean = (min+max)/2

max - (min+max)/2

(2max - min-max)/2

(max - min)/2 which isnt quite the mean.

Answer 4

\[s/\sqrt n = \frac{(44-14)/2}{3}\]

Answer 5

oops, complete lapse of thought: 44 - 14; not 44+14 lol

Answer 6

@amistre64 D. 5 ?

Answer 7

using the last formula you put I got 5

Answer 8

@kjsofreshmurray can you reproduce it? Say, 95% of the data lies between 10 and 22.

Answer 9

5 is what im going for as well