Question

Problem regarding confidence intervals and probability (screenshot attached)

Answer 1

I know for the expected value, I need n and p. I think I can define n as the observed population, 1000. Is 95% p in this case?

Answer 2

Yes, p = 0.95

Answer 3

Okay, so then the expected value should be 950, right?

Answer 4

Yes, this is a sampling distribution
confidence level of 95% means that 95% of the samples will capture the population mean

Answer 5

Okay, so how might I proceed to solve the second question?

Answer 6

The statement, " 95% of the samples will capture the population mean"

is same as saying "each sample has 95% probability to containt he population mean"

yes ?

Answer 7

the samples were selected independently and we have binomial distrabution :

\(n=1000\) and \(p=0.95\)

Answer 8

use the normal approximation formulas to find the probability for 950 to 970 samples to contain the population mean

Answer 9

Normal approximation as in standardizing to Z?

Answer 10

since the number of samples are large enough (1000), the sampling distribution would be approxmately normal. so we can use the approximation formulas

Answer 11

basically we're converting binomial distribution to the normal distribution so that we can use zscores to compute the probability easily

Answer 12

you could also work the probability in binomial distribution, but computing 20 binomial coefficients and adding them is a pain

Answer 13

That makes sense
So \(P(950<Y<970)\)
will become\[P(\frac{ 950-\mu }{ \sigma / \sqrt{n} }<Z<\frac{970-\mu}{\sigma / \sqrt{n}})\]

Answer 14

right?

Answer 15

Or is the \(\sqrt{n}\) not needed here.. I'm not certain when it is needed and when it is not

Answer 16

nope, you need to use the formulas for approximating "binomial distribution" as normal distribution

Answer 17

Okay but isn't it the same formula though (but without the \(\sqrt{n}\)) ?

Answer 18

zscore formula remains same, you need to find standard deviation using that approximation formula

Answer 19

Oh, okay! I wasn't going to do that but I wanted to make sure that the formula I was using was correct. I thought you were stating that the formula in general was completely wrong, with or without the sqrt(n).

I got that \(\sigma=6.89\)
So then\[P (\frac{ 950-950 }{ 6.89 }<\text{Z}<\frac{970-950}{6.89})\]\[=P(0<\text{Z}<2.9)\]\[=\Phi(2.9)-\Phi(0)\]

Answer 20

That looks good, but there is a slight mistake

Answer 21

binomial distribution is discrete
normal distribution is continuous

when converting form binomial to normal, we need to account for that

Answer 22

simply widen the interval :

949.5 < X < 970.5

Answer 23

Darnit X)
So
\[P (\frac{ (950-0.5)-950 }{ 6.89 }<\text{Z}<\frac{(970+0.5)-950}{6.89})\]

Answer 24

Looks good !

Answer 25

Okay perfect, thank you! X)

Answer 26

np :)