The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

S3 Confidence intervals

Could someone possibly help with the second and third parts?
I've got 0.3108 for the first part
image.jpg560.5KB
For the first part of the question I did:

0.4−01=0.4[br]ϕ(0.4)=0.6554[br]1−0.6554=0.3446[br]so,1−(2×0.3446)=0.3108[br] \frac {0.4-0}{1} =0.4[br]\phi (0.4) = 0.6554[br]1-0.6554=0.3446[br]so, 1-(2 \times 0.3446) = 0.3108[br]
Original post by Mutleybm1996
For the first part of the question I did:

0.4−01=0.4[br]ϕ(0.4)=0.6554[br]1−0.6554=0.3446[br]so,1−(2×0.3446)=0.3108[br] \frac {0.4-0}{1} =0.4[br]\phi (0.4) = 0.6554[br]1-0.6554=0.3446[br]so, 1-(2 \times 0.3446) = 0.3108[br]


Looks good so far.

So, we have E∼N(0,0.12)E\sim N(0,0.1^2)

What then is the distribution of the sample mean for 9 observations?

The mean will be 0, but what about the variance?
(edited 9 years ago)
Original post by ghostwalker
Looks good so far.

So, we have E∼N(0,0.12)E\sim N(0,0.1^2)

What then is the distribution of the sample mean for 9 observations?

The mean will be 0, but what about the variance?


X9? I think....


Posted from TSR Mobile
(edited 9 years ago)
Original post by Mutleybm1996
X9? I think....


Posted from TSR Mobile


Needs to be divided by 9, so, 0.12/90.1^2/9

You should be familiar with this - it's fairly basic for sampling.
Original post by ghostwalker
Needs to be divided by 9, so, 0.12/90.1^2/9

You should be familiar with this - it's fairly basic for sampling.


Wait, sorry, I read it as 9 different butterflies! In this case would it be x9?


Posted from TSR Mobile
Original post by Mutleybm1996
Wait, sorry, I read it as 9 different butterflies! In this case would it be x9?


Posted from TSR Mobile


If it was 9 different butterflies, then they'd have different weights. The error on the mean weight would have the same distribution I think, but I'm not 100% sure.
Original post by ghostwalker
Looks good so far.

So, we have E∼N(0,0.12)E\sim N(0,0.1^2)

What then is the distribution of the sample mean for 9 observations?

The mean will be 0, but what about the variance?



So, we have E∼N(0,0.129)E\sim N(0, \frac{0.1^2}{9})

Then do I do the same again as for the first question?

What about the third part?:smile:




Posted from TSR Mobile
Original post by Mutleybm1996
So, we have E∼N(0,0.129)E\sim N(0, \frac{0.1^2}{9})

Then do I do the same again as for the first question?

Yes.

Eˉ∼N(0,0.129)\bar{E}\sim N(0, \frac{0.1^2}{9})



What about the third part?:smile:


Standard, constructing a confidence interval.
Original post by ghostwalker
Yes.

Eˉ∼N(0,0.129)\bar{E}\sim N(0, \frac{0.1^2}{9})



Standard, constructing a confidence interval.


Doing the same again gives z=45? I think. For part b


Posted from TSR Mobile
Sorted it! Thanks :smile:


Posted from TSR Mobile

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you