Deriving a confidence interval for an unknown population mean

Question

The times that a cashier spends processing each person’s transaction are independent and identically distributed random variables with a mean of µ and a variance of $σ^2$. Thus, if $X_i$ is the processing time for each transaction, E($X_i$) = µ and Var($X_i$) = $σ^2$. Let Y be the total processing time for 100 orders: Y = $X_1$ + $X_2$ + · · · + $X_{100}$

(a) What is the approximate probability distribution of Y, the total processing time of 100 orders?

(b) Suppose for Z ∼ N(0, 1), a standard normal random variable: P(a < Z < b) = 100(1 − α)%. Using your distribution from part (a), show that an approximate 100(1 − α)% confidence interval for the unknown population mean µ is: $($$\frac{Y − 10bσ}{100}$$)$< µ <$($$\frac{Y − 10aσ}{100}$$)$.

I have done part (a) and ended up with Y ∼ N(100µ,$σ^2$) using the central limit theorem, but I am unsure of where to start in part (b).

Answer 1

The goal of the exercise to illustrate that you may derive a confidence interval that might not be the standard symmetric one (the usual conventional one that uses $\bar{x}$ as a pivot).

So starting with some $a$ and $b$ such that $P(a < Z < b) = (1-\alpha)$, we can write

\begin{align} (1-\alpha) & = P(a < Z < b) \\ & = P\left(a < \dfrac{Y-100 \mu}{10 \sigma} < b\right) \\ & = P(10a\sigma < Y-100 \mu< 10 b \sigma) \\ & = P(-Y+10a\sigma < -100 \mu< -Y+10 b \sigma) \\ & = P\left(\dfrac{Y-10a\sigma}{100} > \mu > \dfrac{Y-10 b \sigma}{100}\right) \\ & = P\left(\dfrac{Y-10 b \sigma}{100} < \mu < \dfrac{Y-10a\sigma}{100} \right) \end{align}

As stated at the beginning, this is a general confidence interval not necessarily symmetric. IF one takes $a=\Phi^{-1}(\alpha/2)$ and $b=\Phi^{-1}(1-\alpha/2)$, one gets the standard/conventional symmetric confidence interval. But the point is, if the only requirement is the interval having probability $(1-\alpha)$ without requiring symmetry, then there are infinitely many such intervals possible.

Moral: There is no such thing as "the" confidence interval unless one puts requirements (such as symmetry).

Answer 2

If $X_1, X_1, \dots, X_{100}$ are a random sample from a distribution with mean $\mu$ and variance $\sigma^2$, then $T = \sum_{1=1}^{100}$ has $E(T) = 100\mu$ as you say, and (by independence) $V(T) = 100\sigma^2,$ not $\sigma^2.$ Hence, $SD(\bar X) = \sigma/\sqrt{100},$ where $\bar X = T/n = T/100.$

By the Central Limit Theorem, $Z = \frac{\bar X - \mu}{\sigma/\sqrt{100}}$ should have nearly a standard normal distribution, as you say. Thus, one has $$ 0.95 \approx P(-1.96 < Z < 1.96) = P\left(-1.96 < \frac{\bar X - \mu}{\sigma/\sqrt{100}} < 1.96\right) = P(\bar X - 1.96\sigma/\sqrt{100}, < \mu < \bar X + 1.96\sigma/\sqrt{100}),$$

where the last inequality is found my manipulating the event in the penultimate term to isolate $\mu$ as in the last term. The last term is the basis for saying that $\bar X \pm 1.96\sigma/\sqrt{100}$ is an approximate 95% confidence interval for $\mu.$ [I believe you may have mis-copied the last formula in part(b), and I hope the specific case for $\alpha = 0.05$ helps you put it right and to see how to derive the correct CI.]