Question

A random sample of 150 students is chosen from a population of 5,000 students. If the mean IQ in the sample is 110 with a standard deviation of 10, what is the 95% confidence interval for the students' mean IQ score?

Answer 1

how much of this do you already know?

Answer 2

Well i know most of it kind of, im just not sure how to find the 95% interval

Answer 3

would it b 104

Answer 4

No I don't believe so

Answer 5

\[\Large \frac{x}{n}\pm Z_{.95}\sqrt{\frac{x(n-x)}{n^3}}\]

Answer 6

I ended up getting a margin of error of 6.7 which doesnt add up to any of the given answers >_<

Answer 7

or\[\Large \hat p\pm Z_{.95}\sqrt{\frac{\hat p(1-\hat p)}{n}}\]

Answer 8

im getting a little mixed up between proportion and means :)

Answer 9

Ugh this is so confusing

Answer 10

A random sample of 150 students
n = 150

is chosen from a population of 5,000 students.
irrelevant

If the mean IQ in the sample is 110
x bar = 110

with a standard deviation of 10
s = 10

, what is the 95% confidence interval for the students' mean IQ score?
z = 1.96; but we might need to use a t score instead?

Answer 11

\[110\pm1.96\sqrt{\frac{10^2}{150}}\]or change the 1.96 to its t value

Answer 12

I haven't learning anything about t scores yet o.o

Answer 13

then its most likely fine as the z score :)

Answer 14

do you hav a ti83 by chance?

Answer 15

ti83?

Answer 16

its a nice calculator is all

Answer 17

Oh no i wish lol, I only have the calculator on my phone >_<

Answer 18

our E should be:\[E=1.96\sqrt{\frac{10^2}{150}}\]
\[E=1.96\sqrt{\frac{100}{150}}\]
\[E=1.96\sqrt{\frac{2}{3}}\]
\[E=1.96\sqrt{.66666}\]

Answer 19

I got 1.6?

Answer 20

thats good

Answer 21

so, the interval is 110 +- E

Answer 22

soo it should be 108.4-111.6?

Answer 23

well, apart from interval notation ... yes

108.4 to 111.6 is the interval

Answer 24

okay , thank you for your help!

Answer 25

youre welcome :)