Given \(\displaystyle{X}_{{1}},\ldots,{X}_{{{100}}}\) and \(\displaystyle{Y}_{{1}},\ldots,{Y}_{{{50}}}\) which are

navantegipowh

navantegipowh

Answered question

2022-03-23

Given X1,,X100 and Y1,,Y50 which are independent random samples from identical distribution N(μ,1). Each of two statisticians is trying to build confidence interval for μ on a confidence level of 0.8 1α=0.8. You have to find probability that those intervals are disjoint. I know i have to find P(|XY|>11.2850+11.28100) which is 1P(0.309XY0.309)
Question is what should i do next how to calculate this probability?

Answer & Explanation

Drahthaare89c

Drahthaare89c

Beginner2022-03-24Added 19 answers

Step 1
One interval has endpoints X±1.28d1100 and the other
Y±1.28d150.
The intervals are disjoint if X+1.28d1100<Y1.28d150 or Y+1.28d150<X1.28d1100.
That happens if YX<1.28(d1100+d150) or XY>1.28(d1100+d150). In other words
|XY|>1.28(d1100+d150).
This is equivalent to
|XY|d1100+d150>1.28.
You have XYN(0,d1100+d150)=N(0,d3100), so that
XY3100N(0,1).
standard deviation(X¯-Y¯1100+150)=
standard deviation(X¯-Y¯)1100+150
[10pt]={dd3100d1100+d150
=31+2=63
So,
Pr(|XY|d1100+d150>1.28)
=Pr(||XY|d1100+d150|t63>1.2863)
[10pt]={Pr(|Z|>1.2863)  where  ZN(0,1)
If the variance were unknown but the variances of the two populations were equal then we would be working with a t-distribution.

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