Question

Help please

Answer 1

I attested the picture did it not post?

Answer 2

I don't see the question. Try again.

Answer 3

Photo

Answer 4

Did it work this time :/

Answer 5

I don't understand why it's not posting

Answer 6

I see it.

Answer 7

Oh yay do you have any idea how to do it?

Answer 8

The picture is the question?

Answer 9

There I posted it again

Answer 10

Do you see the parts to it now? :)

Answer 11

Yes, but I want to study it also. Let's tag @Zarkon

Answer 12

Can you help me or not? 😂

Answer 13

@raffle_snapple can you help?

Answer 14

*raffle_snaffle

Answer 15

To me, for the first question, if point estimator is \(\bar Y\) , then \(E(\bar Y)= \mu\), then \(\bar Y\) is an unbiased point estimate of \(\mu\)

Answer 16

I'm super stuck on b I've done it two ways and I dunno which one is right any clue?

Answer 17

Show me your work, please.

Answer 18

That's one way be the other way I just got the answer 4.84 but for all I know both ways could be wrong

Answer 19

To me, for b) if we use \(\hat\sigma^2=\dfrac{1}{9}\sum_{i=1}^9 (Y_i-\bar Y)^2\), then we can get \(\hat\sigma^2\) is point estimate for \(\sigma^2\)

Answer 20

Do you use n or n-1 as the divisor?

Answer 21

Proof: \(\sum_{i=1}^9 (Y_i-\bar Y)^2 =\sigma^2X_8^2\) and we know that \(Var X_8^2=16\) and \(E(X_8^2)=8\)
Hence \(E(\hat\sigma^2) = E((1/9)(\sigma^2 X_8^2)= \sigma^2/9 E(X_8^2)=8 \sigma^2/9\)
and then
\((B_\sigma^2 \hat\sigma^2) = (8\sigma^2/9-\sigma^2)^2=(-1/9) \sigma^2 <0\)

Answer 22

That shows \(\hat\sigma^2\) is unbiased for \(\sigma^2\)

Answer 23

Thanks I think I get it :) any clue on C ? 😂

Answer 24

Ok, on C,
for a) we have \(\bar Y\) is estimator, right?
Now, find \(P(a< \bar Y<b)=0.95\)

Answer 25

Sorry I meant d I actually did c

Answer 26

If you know how to do c, then the same for d. Assume that mu is known. just sigma is unknown

Answer 27

Okay never mind thanks

Answer 28

I need your explanation for this problem @Zarkon please