Let X1,…Xn be iid N(θ, 1). A 95% confidence interval for θ is X―±1.96n. Let p denote the probability that an additional independent observation, Xn+1, will fall in this interval. Is p greater than, less than, equal to 0.95? Prove your answer.

Question

Let X1,…Xn be iid N(θ, 1). A 95% confidence interval for θ is X―±1.96n. Let p denote the probability that an additional independent observation, Xn+1, will fall in this interval. Is p greater than, less than, equal to 0.95? Prove your answer.

Jaslyn Allison · Accepted Answer

Step 1Perhaps a little more than a hint but here goesThe question appears to be asking ifp=P{X―−1.961n≤Xn+1≤X―+1.961n}=P{−1.961n≤Xn+1−X―≤1.961n}is less than, equal to, or greater than 0.95.Now, it turns out&amp;nbsp;Xn+1−X―&amp;nbsp;(itself a linear combination of normal random variables) is also normal, with mean 0 and variance&amp;nbsp;1+1n&amp;nbsp;(due to independence of&amp;nbsp;Xn+1&amp;nbsp;from the other&amp;nbsp;Xi&amp;nbsp;and hence from&amp;nbsp;X―).Then we have thatP{−1.961+1n≤Xn+1−X―≤1.961+1n}=0.95&amp;nbsp;as well.Now take a look at the interval above and notice that[−1.961+1n,&amp;nbsp;&amp;nbsp;1.961+1n&amp;nbsp;]=[−1.961+1n,&amp;nbsp;&amp;nbsp;−1.961n&amp;nbsp;)⋃−1.961n&amp;nbsp;&amp;nbsp;1.961n&amp;nbsp;⋃(1.961n,&amp;nbsp;&amp;nbsp;1.961+1n&amp;nbsp;]Can you deduce where p stands relative to&amp;nbsp;0.95?

Let \(\displaystyle{X}_{{1}},\ldots{X}_{{n}}\) be iid \(\displaystyle{N}{\left(\theta,\ {1}\right)}\).

Answered question

Answer & Explanation

New Questions in College Statistics