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The following is a homework problem from my textbook, I am totally confused on how to solve this problem. Please help!!!!!

Let $X_1, X_2, ………, X_n$ be a random sample from a continuous probability distribution having median μ~ (so that P(Xi ≤ μ~) = P(Xi ≥ μ~) = 0.50).

(a) Show that P(min (Xi) < μ~ < max(Xi)) = 1 – (1/2)n-1 so that (min (xi), max (xi) is a 100(1 – α)% confidence interval for μ~ with α = 1 – (1/2)n-1. [Hint: The complement of the event {min (Xi) < μ~ < max(Xi)} is {max(Xi) ≤ μ~} U {min(Xi) ≥ μ~}. But max(Xi) ≤ μ~ iff Xi < μ~ for all i.]

(b) For each six normal male infants, the amount of the amino acid alanine (mg/100mL) was determined while the infants were on an isoleucine-free diet, resulting in the following data:

2.84 3.54 2.80 1.44 2.94 2.70

Compute a 97% CI for the true median amount of alanine for infants on such a diet.

(c) Let x(2) denote the second smallest of the xi‘s and x(n-1) denote the second largest of the xi’s. What is the confidence level of the interval (x(2), x(n-1)) for μ~?

user133976
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1 Answers1

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No information has been supplied about what you have tried, so it is not clear whether you know how to start.

(a) Let $\mu$ be a median of our distribution. Then for any $i$ we have $\Pr(X_i\ge \mu)=\frac{1}{2}$. It follows by independence that the probability that all the $X_i$ are $\ge \mu$ is $\left(\frac{1}{2}\right)^n$.

Similarly, the probability that all the $X_i$ are $\le \mu$ is $\left(\frac{1}{2}\right)^n$.

It follows that the probability that all the $X_i$ are either $\ge \mu$ or $\le \mu$ is $\left(\frac{1}{2}\right)^n+\left(\frac{1}{2}\right)^n$, that is, $\frac{1}{2^{n-1}}$.

(b) In the given numerical example, we have $n=6$. Imagine that we make $6$ observations, and record the smallest and the largest. Call these $U$ and $V$. Note that $U$ and $V$ are random variables.

By part (a), we have $\Pr(U\le \mu\le V)=1-\frac{1}{2^5}= 0.96875$. It follows that a roughly $97\%$ confidence interval for the median is $(U, V)$. In our case, that is $(1.44,3.54)$.

(c) Let $S$ be the second smallest observation, and $T$ the second largest. We first calculate $\Pr(T\lt \mu)$. This is the probability that all the observations are $\lt \mu$ or all but one of the observations are $\lt \mu$. By our earlier work, the probability all the observations are $\lt \mu$ is $\frac{1}{2^n}$. The probability that $1$ observation is $\gt \mu$ and the rest are $\lt \mu$ is $\binom{n}{1}\frac{1}{2^n}$. (Note that we casually write $\lt$ and $\le$ interchangeably. This is OK since the underlying distribution is continuous.)

Thus the probability that $T\lt \mu$ is $\frac{n+1}{2^n}$. Similarly, the probability that $S\gt \mu$ is $\frac{n+1}{2^n}$.

Thus the probability that the interval $(S,T)$ does not contain $\mu$ is $\frac{2n+2}{2^n}$. The confidence level is, as a number, $1-\frac{2n+2}{2^n}$. To express as a percentage, multiply by $100$.

André Nicolas
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