Question

One mark is awarded to a correct answer for a multiple choice question; zero mark, otherwise.
Denote the probability that a student knows the correct answer to the question by p.

A test consists of n such multiple choice questions. Assume that the questions are independent and equally difficult to the student. How large does n need to be so that the length of a 95% confidence interval for p is smaller than 10%, if we believe p(1 − p) < 0.2?

[Hint: consider only the situations with large n.]

Answer 1

do you recall the formula for an error?

Answer 2

the length of 95% is smaller than 10% ? thats hard to parse

Answer 3

in general. the setup for a confidence interval is:
\[\bar x\pm Z_\%\sqrt{\frac{p(1-p)}{n}}\]
or written another way
\[\bar x\pm E\]
so:\[E=Z_\%\sqrt{\frac{p(1-p)}{n}}\]
i spose they want E to be .10 by that other stuff, solve for n

Answer 4

or would it be .10/2 ?

Answer 5

95% confidence interval mean something else
10% meaning that for example i have X and in this X +-10/2 is in interval where 95% of data is placed.

Answer 6

I guess 10 = +-5% on each side of the mean

Answer 7

yeah, that does seem like a better read :)

Answer 8

and how to understand that p(1 − p) < 0.2?
as far as i know it is variance but why it is <0.2? and how it works than?

Answer 9

do we put 0.2 in and that it?

Answer 10

just leave that part as an unknown for now and work out the n, then we can use that to find a range with

Answer 11

ok

Answer 12

I got that\[n=p(1-p)/0.00065077\]

Answer 13

since they ask for 95% CI than z=1.96 is it right?

Answer 14

(1.96/.05)^2 = 1536.64, times p(1-p)
\[n = 1536.64p(1-p)\]

so n = 0 to 1536.64(.02)

Answer 15

yeah, 0 to 31 is what i get

Answer 16

i see you did (E/z)^2 for a denominator which is fine too

Answer 17

ok than answer in from 0 to 307.4?

Answer 18

i used .02 instead of .2 didnt i ... yeah, that fine. id rnd up to 308

Answer 19

ok, thank you i think a get the point.