Question

Problem 6 Let X , X,X,+1 be a sample from normal distribution N(hC). Let X-ΣΧ . Find c such that X belongs to the interval (R-c, R+c) with probability 0.95. Calculate that interval if X = 1, n = 8, σ

Answer 1

Given $X_1,X_2,...,X_n,X_{n+1}$ be a sample from normal distribution $N\left ( \mu ,\sigma ^2 \right )$ . Let $\overline{X}=\frac{1}{n}\sum_{i=1}^{n}X_i$ .

Now, we know $\overline{X}=\frac{1}{n}\sum_{i=1}^{n}X_i\sim N\left ( \mu ,\sigma ^2/n \right )$ . Also $Y=X_{n+1}-\overline{X}\sim N\left ( 0 ,\sigma ^2\left ( n+1 \right )/n \right )$

The probability

$P\left ( \overline{X}-c<X_{n+1}<\overline{X}+c \right )= P\left ( -c<X_{n+1}-\overline{X}<c \right )\\ P\left ( \overline{X}-c<X_{n+1}<\overline{X}+c \right )= P\left ( -c<Y<c \right )\\ P\left ( \overline{X}-c<X_{n+1}<\overline{X}+c \right )= P\left ( -c/\sqrt{\sigma ^2\left ( n+1 \right )/n }<\left ( Y-0 \right )/\sqrt{\sigma ^2\left ( n+1 \right )/n }<c/\sqrt{\sigma ^2\left ( n+1 \right )/n } \right )\\ P\left ( \overline{X}-c<X_{n+1}<\overline{X}+c \right )= 1-2\Phi \left ( -c/\sqrt{\sigma ^2\left ( n+1 \right )/n } \right )\\$
We need,

$1-2\Phi \left ( -c/\sqrt{\sigma ^2\left ( n+1 \right )/n } \right )=0.95\\ \Phi \left ( -c/\sqrt{\sigma ^2\left ( n+1 \right )/n } \right )=0.025\\ -c/\sqrt{\sigma ^2\left ( n+1 \right )/n}=\Phi ^{-1}\left ( 0.025 \right )\\ -c/\sqrt{\sigma ^2\left ( n+1 \right )/n}=-1.959964\\ c=1.959964\times \sqrt{\sigma ^2\left ( n+1 \right )/n}\\ {\color{Blue} c=2.079}$