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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

1 Answer

4 votes

Answer:


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

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)(\sigma)/(√(n))


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


z_(critical)\text{ at}~\alpha = z_{(\alpha)/(2)}

Putting the values, we get,


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

which is the required confidence interval.

User Kevin Fang
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