finding the 95% interval..

[barydos]

If say a question on exam 2 asks you to find the 95% confidence interval for which ...blah blah..
Are you supposed to use mean+- 2sd or find the 95% interval by using invNorm function?

Like on MAV 2012, they used mean+-2sd for it, but I'm not sure?
What do you think!?

[Special At Specialist]

Although technically 2 standard deviations is closer to 95.45%, they usually follow the 68-95-99.7% rule. If you see any of those 3 numbers, then just use those estimated confidence limits.

[Limista]

Use invNorm function ~ if you've got the Maths Methods Essential textbook look at the example for exercise 18C..that should answer your question.
If not, then know that 95% of the distribution is within 2 standard deviations of the mean, that means that there is 0.05 divided by 2 = 0.025 on the left AND right of this 95% shaded area. If you want to find the EXACT score for (u-2o) use invNorm function with "left" and "right" typing in the mean and standard deviation which you have ALREADY GOT as part of the requirements of the question.

Basically, you can use the (u-2o) formula thing if you want an ESTIMATE of the interval (i.e (x,y) such that x and y are NOT included)
You use the invNorm function if you want the exact score for the interval (i.e. (x,y) such that x and y ARE included)

Hope this helps 

[Shenz0r]

Derrick Ha went over this yesterday, I think two standard deviations away from the mean represent approximately 95% of the interval. Not the exact interval. So it depends on the question, if they ask for an approximately 95% interval, I would use standard deviations, but if they asked for the exact 95% interval I would invnorm it.

[barydos]

Yeah so what I'm saying is that the question in the MAV 2012 exam 2 didn't say approximate nor did it say exact, so I'm not sure which method to go with. Like, the solutions showed the method using the +- 2 standard deviations away from the mean, and I just want to know if that I'd get marks taken off if I did the exact method. :S

[FlorianK]

If say a question on exam 2 asks you to find the 95% confidence interval for which ...blah blah..

This means +- 2sd.

However, this question is probably neither on exam 1 nor on exam 2.
I know it was on my probabiliy SAC (tech-active), but i haven't seen it after that.

[BubbleWrapMan]

To quote the study design:

"This area of study will include:

...

–   property that, for many random variables, approximately 95 per cent of their probability distribution is within two standard deviations of the mean"

Make of that what you will.

[TrueTears]

If say a question on exam 2 asks you to find the 95% confidence interval for which ...blah blah..
Are you supposed to use mean+- 2sd or find the exam 95% interval by using invNorm function?

Like on MAV 2012, they used mean+-2sd for it, but I'm not sure?
What do you think!?

CI for what random variable? CI differs greatly depending on what the distribution is for the random variable you're trying to create a prediction interval for.

Consider a sequence of random variables X_i's which are independent and identically distributed. We can derive a % CI for , the mean of this random variable.

Consider the estimator

Using the central limit theorem:

So where denotes the percentile of the standard normal.

Thus the CI for is

But the CI will differ (although not vastly), if say we didn't know , we would need to estimate it with

Then our distribution would become the student's t-distribution.

So the CI will be where denotes the percentile of a t distribution.

We could also create CI's for variances, ratio of variances, differences in means, the parameter p in a binomial distribution etc

[TrueTears]

Basically, you can use the (u-2o) formula thing if you want an ESTIMATE of the interval (i.e (x,y) such that x and y are NOT included)
You use the invNorm function if you want the exact score for the interval (i.e. (x,y) such that x and y ARE included)

Incorrect, endpoints has nothing to do with the expressions for CI's.

[barydos]

Sorry for a normally distributed variable.