4
$\begingroup$

I am trying to understand how $$ \mathbb{P}\left(-z_{\alpha / 2} \leqslant \frac{\bar{X}-\mu}{\sigma / \sqrt{n}} \leqslant z_{\alpha / 2}\right) \approx 1-\alpha $$

can be converted to $$ \mathbb{P}\left(\bar{X}-\frac{z_{\alpha / 2}}{\sqrt{n}} \cdot \sigma \leqslant \mu \leqslant \bar{X}+\frac{z_{\alpha / 2}}{\sqrt{n}} \cdot \sigma\right) \approx 1-\alpha $$

In order to make $\mu$ positive I tried multiplying everything by $-1$. But I understand that causes the greater than signs to switch.

$\endgroup$
1
$\begingroup$

You're right about the signs switching. The symmetry of the normal distribution may be confusing you.

If $X \sim \mathsf{Norm}(\mu, \sigma),$ then $Z = \frac{\bar X - \mu}{\sigma/\sqrt{n}} \sim \mathsf{Norm}(0,1).$

In order to make a 95% CI, you can find numbers $L$ and $U$ such that $P(L < Z < U) = 0.95.$

Then the inequality inside the probability statement becomes $$L < \frac{\bar X - \mu}{\sigma/\sqrt{n}} < U$$ or, upon "pivoting," $$\bar X - U\frac{\sigma}{\sqrt{n}} < \mu < \bar X - L\frac{\sigma}{\sqrt{n}}.$$

It is customary to cut the same probability $0.025$ from each tail of the symmetrical normal distribution, so $L = -1.96, U = 1.96$ and plugging those values into the second displayed inequality above becomes

$$\bar X - 1.96\frac{\sigma}{\sqrt{n}} < \mu < \bar X + 1.96\frac{\sigma}{\sqrt{n}}.$$

$\endgroup$
1
$\begingroup$

I'll assume that $X \sim N(\mu, \sigma^2)$, then, $\frac{\bar{X} - \mu}{\sigma / \sqrt{n}} \sim N(0, 1)$, therefore,

$$P( Z_{\alpha / 2} \leq \frac{\bar{X} - \mu}{\sigma / \sqrt{n}} \leq Z_{1 - \alpha/2}) = 1-\alpha $$

Some algebra,

$$P( \frac{\sigma}{\sqrt{n}} Z_{\alpha / 2} \leq \bar{X} - \mu \leq \frac{\sigma}{\sqrt{n}}Z_{1 - \alpha/2}) = 1-\alpha $$

Subtracting the average,

$$P( \frac{\sigma}{\sqrt{n}} Z_{\alpha / 2} - \bar{X} \leq - \mu \leq \frac{\sigma}{\sqrt{n}}Z_{1 - \alpha/2} - \bar{X}) = 1-\alpha $$

Now we multiply by -1, so we need to change the direction of the inequalities,

$$P( - \frac{\sigma}{\sqrt{n}} Z_{\alpha / 2} + \bar{X} \geq \mu \geq -\frac{\sigma}{\sqrt{n}}Z_{1 - \alpha/2} + \bar{X}) = 1-\alpha $$

Rearrange,

$$P( \bar{X} -\frac{\sigma}{\sqrt{n}}Z_{1 - \alpha/2} \leq \mu \leq \bar{X} - \frac{\sigma}{\sqrt{n}} Z_{\alpha / 2} ) = 1-\alpha $$

We now note, that $Z_{\alpha /2} = -Z_{1- \alpha/ } $ due to the symmetry of the standard normal distribution with respect to 0

$$P( \bar{X} -\frac{\sigma}{\sqrt{n}}Z_{1 - \alpha/2} \leq \mu \leq \bar{X} + \frac{\sigma}{\sqrt{n}} Z_{1 - \alpha / 2} ) = 1-\alpha, $$

which means the CI can also be written as $\bar{X} \pm \frac{\sigma}{\sqrt{n}} Z_{1 - \alpha / 2} $.

$\endgroup$
1
  • $\begingroup$ Thank you. However in my question the left hand side starts with $$-z_{\alpha / 2}$$ $\endgroup$
    – Kirsten
    Jul 19 '21 at 0:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.