Suppose that Y ? Bin(100, ?) and you have observed y1, . . . , y20 such that (?(from i=1 to 20)_yi )/20=45.2
Compute numerically, showing all your working, the 95% confidence interval for ?, where you may assume that n = 100 is large enough for the approximations using Gaussian.
I don't know if the ? is equal to 45.2/100 since n=100 I think it sould be (?(from i=1 to 100)_yi )/100 but how can I get from 45.2 to this?
My answer:
Since n is large, Y~G(100?,?(100?(1-?) )). Then ( ?) ?=Y/n ~ G(?,?(?(1-?)/100))
P(-1.96? (? ?-?)/?(?(1-?)/100)?1.96)= 0.95
p(-1.96*?(?(1-?)/100) ? ? ?-??1.96*?(?(1-?)/100)) = 0.95
p(? ?-1.96*?((? ?(1-? ? ))/100) ? ??? ?+1.96*?((? ?(1-? ? ))/100))= 0.95
Since 100/20 = 5; if we draw the sample 100 times, we can divide it into 5 groups of 20.
so,? ?=(?(from i=1 to 100)_yi )/100=45.2/100=0.452
So the confidence interval for ? is approximately (0.35445, 0.54954).
Compute numerically, showing all your working, the 95% confidence interval for ?, where you may assume that n = 100 is large enough for the approximations using Gaussian.
I don't know if the ? is equal to 45.2/100 since n=100 I think it sould be (?(from i=1 to 100)_yi )/100 but how can I get from 45.2 to this?
My answer:
Since n is large, Y~G(100?,?(100?(1-?) )). Then ( ?) ?=Y/n ~ G(?,?(?(1-?)/100))
P(-1.96? (? ?-?)/?(?(1-?)/100)?1.96)= 0.95
p(-1.96*?(?(1-?)/100) ? ? ?-??1.96*?(?(1-?)/100)) = 0.95
p(? ?-1.96*?((? ?(1-? ? ))/100) ? ??? ?+1.96*?((? ?(1-? ? ))/100))= 0.95
Since 100/20 = 5; if we draw the sample 100 times, we can divide it into 5 groups of 20.
so,? ?=(?(from i=1 to 100)_yi )/100=45.2/100=0.452
So the confidence interval for ? is approximately (0.35445, 0.54954).
