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

Question

Let

X_{1}, \dots X_{n}

be iid

N (θ, 1)

. A

95 %

confidence interval for

θ

is

\overset{―}{X} \pm \frac{1.96}{\sqrt{n}}

. Let p denote the probability that an additional independent observation,

X_{n + 1}

, will fall in this interval. Is p greater than, less than, equal to

0.95

? Prove your answer.

Answer 1

Step 1
Perhaps a little more than a hint but here goes
The question appears to be asking if
$p = P {\overset{―}{X} - 1.96 \sqrt{\frac{1}{n}} \leq X_{n + 1} \leq \overset{―}{X} + 1.96 \sqrt{\frac{1}{n}}}$
$= P {- 1.96 \sqrt{\frac{1}{n}} \leq X_{n + 1} - \overset{―}{X} \leq 1.96 \sqrt{\frac{1}{n}}}$
is less than, equal to, or greater than 0.95.
Now, it turns out $X_{n + 1} - \overset{―}{X}$ (itself a linear combination of normal random variables) is also normal, with mean 0 and variance $1 + \frac{1}{n}$ (due to independence of $X_{n + 1}$ from the other $X_{i}$ and hence from $\overset{―}{X}$ ).
Then we have that
$P {- 1.96 \sqrt{1 + \frac{1}{n}} \leq X_{n + 1} - \overset{―}{X} \leq 1.96 \sqrt{1 + \frac{1}{n}}} = 0.95$ as well.
Now take a look at the interval above and notice that
$[- 1.96 \sqrt{1 + \frac{1}{n}}, 1.96 \sqrt{1 + \frac{1}{n}}]$
$= [- 1.96 \sqrt{1 + \frac{1}{n}}, - 1.96 \sqrt{\frac{1}{n}})$
$⋃ \begin{array}{cc} - 1.96 \sqrt{\frac{1}{n}} & 1.96 \sqrt{\frac{1}{n}} \end{array}$
$⋃ (1.96 \sqrt{\frac{1}{n}}, 1.96 \sqrt{1 + \frac{1}{n}}]$
Can you deduce where p stands relative to $0.95$ ?

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

Answers (1)

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