Question

Let ˆθ be a statistic that is normally distributed with mean θ and standard error σθˆ. Show that 100(1 − α)% confidence interval for θ is given by: ˆθ ± z α 2 σθˆ

Answer 1

Answer:

$\hat{\theta} \pm z_{\frac{\alpha}{2}}(\hat{\sigma_{\theta}})$              

Step-by-step explanation:

We are given the following in the question:

$\hat{\theta}$ is distributed normally.

Mean = $\hat{\theta}$

Standard error = $\hat{\sigma_{\theta}}$

Significance level = $\alpha$

Thus, the confidence percentage is

$100(1 - \alpha)\%$

Confidence interval:

$\mu \pm z_{critical}\frac{\sigma}{\sqrt{n}}$

$\mu \pm z_{critical}(\text{Standard error})$

$z_{critical}\text{ at}~\alpha = z_{\frac{\alpha}{2}}$

Putting the values, we get,

$\hat{\theta} \pm z_{\frac{\alpha}{2}}(\hat{\sigma_{\theta}})$

which is the required confidence interval.