I have following problem

Determine the length of the confidence interval for a sample $X\sim N(\mu,\sigma)$ and $H_0:\mu = \mu^*$ is rejected with $H_1:\mu = \mu^{**}>\mu^*$ if $\overline{x}>\mu^*+1$

I know that if the population distribution is normal, the interval $\overline{x}\pm Z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ is a confidence interval for $\mu$ with confidence level $1-\alpha$.

My issue is What is the procedure that I have to follow to solve this problem? Is the same confidence interval than the above without condition of $\overline{x}>\mu^*+1$ ?

  • $\begingroup$ In this case, you would obtain a drastically different and more complicated confidence interval by inverting the statistical test you describe... but you can also obtain a "vanilla" confidence interval by using the usual error of $\mathcal{Z}_{\alpha/2} \hat{\sigma} / \sqrt{n}$. $\endgroup$ – AdamO Nov 21 '16 at 20:26
  • $\begingroup$ @AdamO the meaning of "vanilla" is a proof or solution less hard or less formal, sorry for my ignorance $\endgroup$ – Cyberguille Nov 22 '16 at 17:11
  • $\begingroup$ the asymptotic sampling distribution of the mean is $\sqrt{n} \left( \bar{X} - E(X) \right) \rightarrow_d \mathcal{N} \left(0, \sigma^2 \right)$ regardless of the hypothesis. So a correct 95% CI is always obtained using a normal approximation of $\bar{X} \pm \mathcal{Z}_{\alpha/2} \hat{\sigma}/\sqrt{n}$. This is a basic result of the CLT, so I think a "less formal" approach is adequate. $\endgroup$ – AdamO Nov 22 '16 at 18:26
  • $\begingroup$ @AdamO While I was looking your comment I discovered that I lack a detail in alternative hypothesis $H_1: \mu=\mu{**} > \mu{*}$ $\endgroup$ – Cyberguille Nov 22 '16 at 18:42
  • $\begingroup$ your problem is half baked: you can't have an alternative hypothesis which depends upon the data you've collected. Intuitively that should immediately reek of flawed reasoning. $\endgroup$ – AdamO Nov 22 '16 at 20:23

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