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 :

Answer:

[tex]\hat{\theta} \pm z_{\frac{\alpha}{2}}(\hat{\sigma_{\theta}})[/tex]              

Step-by-step explanation:

We are given the following in the question:

[tex]\hat{\theta}[/tex] is distributed normally.

Mean = [tex]\hat{\theta}[/tex]

Standard error = [tex]\hat{\sigma_{\theta}}[/tex]

Significance level = [tex]\alpha[/tex]

Thus, the confidence percentage is

[tex]100(1 - \alpha)\%[/tex]

Confidence interval:

[tex]\mu \pm z_{critical}\frac{\sigma}{\sqrt{n}}[/tex]

[tex]\mu \pm z_{critical}(\text{Standard error})[/tex]

[tex]z_{critical}\text{ at}~\alpha = z_{\frac{\alpha}{2}}[/tex]

Putting the values, we get,

[tex]\hat{\theta} \pm z_{\frac{\alpha}{2}}(\hat{\sigma_{\theta}})[/tex]

which is the required confidence interval.

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