# calculating confidence interval for critical value multiplying by negative 1

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.

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

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

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 $$\bar{X} \pm \frac{\sigma}{\sqrt{n}} Z_{1 - \alpha / 2}$$.

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