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Yeti
04-25-2005, 04:25 PM
Meh, I hate Stats.

'A 95% confidence interval fo the mean amount of gas consumed per household during the month of January is (152.4,160.2) therms.

This interval is based on results from a random sample of 25 households. Find a 99% confidence interval for the mean amount of gas consumed per household during the month of January, assuming that gas consumption per household is Normally distributed with unknown variance.

Give the sample mean and population variance of the sample of 25 households.'

I'm trying but don't really have a clue what I'm doing. My, likely wrong, answers to follow soonish.

Any help appreciated.

Pokerscott
04-25-2005, 05:21 PM
ok I'll take a quick shot...

95% confidence range is mean +/- 1.96 stds.
99% confidence range is mean +/- 2.575 stds.

You need to find mean and std from data given.

since normal is symmetric your mean is the middle of the range given which is 156.3

The range given should be 1.96*2 stds, so a standard deviation estimate in this case is (160.2-152.4)/(2*1.96) = 1.99

Finally your 99% confidence region would be mean +/- 2.575*std = mean +/- 2.575*1.99

So I get a 99% confidence region of about (151.18,161.42)

It is concerning that I don't use the fact that you only have 25 data points, but I can't see how that would matter in translating from a 95% confidence interval estimate to a 99% estimate.

Pokerscott

Siegmund
04-25-2005, 05:44 PM
[ QUOTE ]

It is concerning that I don't use the fact that you only have 25 data points, but I can't see how that would matter in translating from a 95% confidence interval estimate to a 99% estimate.

[/ QUOTE ]

The original poster said a population with unknown variance. That is, we're using Student's T with 24 degrees of freedom, not the Z table.

You will still be looking up two numbers and making the interval wider accordingly, but both numbers will be just a tad bigger than 1.96 and 2.58.

Yeti
04-25-2005, 05:48 PM
Ah, well at least I'm now heading in the right direction.

Cheers /images/graemlins/smile.gif

Pokerscott
04-25-2005, 06:14 PM
[ QUOTE ]
[ QUOTE ]

It is concerning that I don't use the fact that you only have 25 data points, but I can't see how that would matter in translating from a 95% confidence interval estimate to a 99% estimate.

[/ QUOTE ]

The original poster said a population with unknown variance. That is, we're using Student's T with 24 degrees of freedom, not the Z table.

[/ QUOTE ]

Aha thanks. I should read more closely before I get to the writing phase lol.

Pokerscott

Yeti
04-25-2005, 07:45 PM
Confused.

I'm unsure as to what method to use for different questions. If the question doesn't specify that it's a Normal distribution can I not use the 'Normal' methods?

eg. the question after that is about caffeine consumption in a random sample of 125 women.

I'm given a total (from which i can work out the mean obviously), and sum of the squares. I'm asked to find a 99% confidence interval for the population (ie. all women) mean.

No mention of the word Normal. I'm guessing this isn't as simple as it seems, or is it?

Cheers.