1. ## Confidence interval

Electricity usage in homes was studied over a period.
For Year 1:
Mean = 1130
Std dev. = 196
n = 16

For Year 2:
Mean = 1350
Std dev. = 302
n = 21

Calculate a 95% confidence interval for the average usage year 2?

x-bar = 1350 +/- ME
where ME = t(n-1), (alpha/2) x s / sq root of n

Am I on the right track here?

2. Originally Posted by Pinto9
Electricity usage in homes was studied over a period.
For Year 1:
Mean = 1130
Std dev. = 196
n = 16

For Year 2:
Mean = 1350
Std dev. = 302
n = 21

Calculate a 95% confidence interval for the average usage year 2?

x-bar = 1350 +/- ME
where ME = t(n-1), (alpha/2) x s / sq root of n

Am I on the right track here?
Since this is exactly the formula your textbook and classnotes would say to use, I don't see why you are usure.

3. Well, thanks. I have another question though.
The question:
Test the hypothesis that the std dev has increased in year 2 compared to year 1?

So H(0): var(x) < or = var(y)
and H(a): var(x) > var(y) ?

But as s(x) < s(y)
(if x= year 1 and y= year 2)
how do you state the rejection of H(0)?
Is it reject H(0) if
F(s(x)^2 / s(y)^2) < or > F n(x)-1, n(y)-1, alpha
?

4. I would state your null hypothesis that there is "no difference between the variances" vs. "the variance in year 2 is greater than in year 1".

Your decision rule should be to reject the null hypothesis using a one-sided test. You have a two-sided test listed there. Since your teacher asked specifically to test "increased in year 2", then this to me would indicate a one-sided test.