I don't know where I'm making a mistake on the solution for the question. The question is as follows:

Suppose that $(x_1,\ldots,x_n)$ is a sample from an $N(\mu, \sigma^2)$ distribution, $\mu$ unknown but $\sigma^2$ is known. Suppose we want to find an interval $C(x_1, \ldots, x_n) = (- \infty, u(x_1, \ldots, x_n))$ that covers the interval $(- \infty, \mu)$ with probability at least $\gamma$. Then we want $u$ such that for every $\mu$, $$P_{\mu}(\mu \leq u(x_1, \ldots, x_n)) \geq \gamma.$$ Then we want to obtain an exact left-sided $\gamma$-confidence interval for $\mu$ using $u(x_1, \ldots, x_n) = \bar{x} + k(\sigma / \sqrt{n})$. In other words, we're tasked with finding $k$.

So, I begin the problem by rewriting the equation to now be $$P_{\mu}(\mu \leq \bar{x} + k(\sigma/\sqrt{n})) = \gamma.$$ We can then rewrite this using algebra to get $$P_{\mu}(\frac{\mu - \bar{x}}{\sigma/\sqrt{n}} \leq k) = \gamma.$$ Note that we let $Z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}}$. Therefore, we want to rearrange the equation so that we have $Z$. In order to do so, we multiply $\frac{\mu-\bar{x}}{\sigma/\sqrt{n}}$ and $k$ by $-1$. Therefore, we have $$P_{\mu}(Z \geq -k) = \gamma. $$ Note that $\Phi$ denote the cdf of the normal distribution. Hence, we have $$1 - \Phi(-k) = \gamma $$ $$ \rightarrow \Phi(-k) = 1- \gamma$$ $$ \rightarrow -k = \Phi^{-1}(1-\gamma).$$ In other words, this means that our one-sided confidence interval is $(- \infty, \bar{x} - \Phi^{-1}(1-\gamma)\cdot (\sigma/\sqrt{n}))$. This does not seem right to me --- shouldn't it be $(-\infty, \bar{x} + \Phi^{-1}(1- \gamma) \cdot (\sigma/\sqrt{n}))$? Where did I go wrong?

  • 2
    $\begingroup$ I don't think you went wrong. For example, if $\gamma = 0.95$ then $\Phi^{-1}(1 - \gamma) = \Phi^{-1}(0.05) = -1.64$ so $\bar{x} - \Phi^{-1}(1 - \gamma)\sigma/\sqrt{n}$ is $\bar{x} + 1.64 \sigma/\sqrt{n}$. $\endgroup$ – markseeto Mar 6 '17 at 1:38
  • $\begingroup$ Ah, I forgot that $\Psi^{-1}(1- \gamma)$ would be negative. Thanks for the help. $\endgroup$ – User203940 Mar 6 '17 at 1:43

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