Confidence Intervals

Can someone please explain how to find the CI when σ isn't given?

Construct a 95% CI for the mean for the following set of data:
2.2, 1.8, 3.1, 2.0, 2.4, 2.0, 2.1, 1.2.

Since σ isn't given I know I have to use E(S^2), except he never really explained what it was. Google isn't being very helpful...

Reply 1

aznkid66

Could he be referring to: σ^2 = variance = mean of square minus square of mean = E[X^2] - (E[X])^2 ?

Reply 2

Tom722

Original post by aznkid66

Could he be referring to: σ^2 = variance = mean of square minus square of mean = E[X^2] - (E[X])^2 ?

To be honest I'm not 100% sure, I'm working it out from the sample. The lecturer has taught this so badly I'm having to look through an S3 textbook. I've got S^2 as 0.289 but it says E(S^2) = variance. I'm getting all sorts of formulas for E(S^2)?! This should be really easy...

Reply 3

Hedgeman49

Original post by Tom722

You have a mean of 16.8/8 = 2.1.
Variance = 1/8(0.1^2 + 0.3^2 + 1 + 0.1^2 + 0.3^2 + 0.1^2 + 0.9^2) = 0.2525
SD = sqrt(0.2525) = 0.5025
CI = (2.1 - 1.96*0.5025, 2.1 + 1.96*0.5025) = (1.1151, 3.0849)

Reply 4

Tom722

Original post by Hedgeman49

You have a mean of 16.8/8 = 2.1.
Variance = 1/8(0.1^2 + 0.3^2 + 1 + 0.1^2 + 0.3^2 + 0.1^2 + 0.9^2) = 0.2525
SD = sqrt(0.2525) = 0.5025
CI = (2.1 - 1.96*0.5025, 2.1 + 1.96*0.5025) = (1.1151, 3.0849)

Shouldn't that +- 1.96*0.5025 be divided by sqrt(n)? Thanks for the help though, I'm thinking this is how it should be done, think I'll ignore my lecturer from now on...

Reply 5

aznkid66

Original post by Tom722

As Hedgeman49 implied, if S is the difference between the mean and each data point, then σ^2 the variance is equal to E(S^2) the sum of S^2 (for each data point) divided by the number of data points.

Reply 6

Tom722

I used Minitab (Statistical software package) and it's given me this:

One-Sample T: C1

Variable: C1
N: 8
Mean: 2.100
StDev: 0.537
SE Mean: 0.190
95% CI: (1.651, 2.549)

(edited 10 years ago)

Reply 7

Hedgeman49

Original post by Tom722

I used Minitab (Statistical software package) and it's given me this:

One-Sample T: C1

Variable: C1
N: 8
Mean: 2.100
StDev: 0.537
SE Mean: 0.190
95% CI: (1.651, 2.549)

Got it - divide by 7 instead of 8 when calculating variance (because one of the differences is 0, apparently you don't count it in the sample size - this is news to me)

Reply 8

Tom722

Original post by Hedgeman49

Got it - divide by 7 instead of 8 when calculating variance (because one of the differences is 0, apparently you don't count it in the sample size - this is news to me)

Thank you very much, I think I'll just make up data sets and play around until I understand it completely!

Reply 9

aznkid66

Original post by Hedgeman49

Got it - divide by 7 instead of 8 when calculating variance (because one of the differences is 0, apparently you don't count it in the sample size - this is news to me)

For the unbiased estimate of population variance of a sample, you divide by n-1 instead of n.

Reply 10

Hedgeman49

Original post by aznkid66

For the unbiased estimate of population variance of a sample, you divide by n-1 instead of n.

There you go, thought it didn't make sense

Reply 11

Tom722

After much research I finally like to bring this to a close!

Since we don't know the standard deviation of the population we use the T distribution.
So mean of the sample = 2.1
Variance of the sample, S^2 = 0.289 => SD of the sample, S = 0.587 (Same as Minitab which is promising)
Now since we're dealing with T-Distribution it isn't 1.96, we need to find (-t < T < t) = 0.95 where we have n - 1 = 7 degrees of freedom, finding this online the value is t = 2.3646. To standardize (Similar to Normal Distribution) the formula is T = [(Mean of sample) - (Mean of population)]/(S/sqrt(n))

Thus the 95% CI of the Mean of the population is given by (2.1 +- 2.3646(0.587)/sqrt(8)), giving (1.651, 2.549), the same as Minitab.

Thanks for the help guys, most frustrating two hours of my life!