Question

Let X₁, X₂,…, X_n be a random sample from a continuous probability distribution having median \(\widetilde {\bf{\mu }}\) (so that \({\bf{P(}}{{\bf{X}}_{\bf{i}}} \le \widetilde {\bf{\mu }}{\bf{) = P(}}{{\bf{X}}_{\bf{i}}} \le \widetilde {\bf{\mu }}{\bf{) = }}{\bf{.5}}\)).

a. Show that

\({\bf{P(min(}}{{\bf{X}}_{\bf{i}}}{\bf{) < }}\widetilde {\bf{\mu }}{\bf{ < max(}}{{\bf{X}}_{\bf{i}}}{\bf{)) = 1 - }}{\left( {\frac{{\bf{1}}}{{\bf{2}}}} \right)^{{\bf{n - 1}}}}\)

So that \({\bf{(min(}}{{\bf{x}}_{\bf{i}}}{\bf{),max(}}{{\bf{x}}_{\bf{i}}}{\bf{))}}\)is a \({\bf{100(1 - \alpha )\% }}\) confidence interval for \(\widetilde {\bf{\mu }}\) with (\({\bf{\alpha = 1 - }}{\left( {\frac{{\bf{1}}}{{\bf{2}}}} \right)^{{\bf{n - 1}}}}\).Hint :The complement of the event \(\left\{ {{\bf{min(}}{{\bf{X}}_{\bf{i}}}{\bf{) < \mu < max(}}{{\bf{X}}_{\bf{i}}}{\bf{)}}} \right\}\)is \({\bf{\{ max(}}{{\bf{X}}_{\bf{i}}}{\bf{)}} \le \widetilde {\bf{\mu }}{\bf{\} }} \cup {\bf{\{ min(}}{{\bf{X}}_{\bf{i}}}{\bf{)}} \ge \widetilde {\bf{\mu }}{\bf{\} }}\). But \({\bf{max(}}{{\bf{X}}_{\bf{i}}}{\bf{)}} \le \widetilde {\bf{\mu }}\) iff \({\bf{(}}{{\bf{X}}_{\bf{i}}}{\bf{)}} \le \widetilde {\bf{\mu }}\)for all i.

b.For each of six normal male infants, the amount of the amino acid alanine (mg/100 mL) 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(“The Essential Amino Acid Requirements of Infants,” Amer. J. Of Nutrition, 1964: 322–330).

c.Let x₍₂₎denote the second smallest of the x_i’s and x_(n-1) denote the second largest of the x_i’s. What is the confidence level of the interval(x₍₂₎, x_(n-1)) for \(\widetilde \mu \)?

Short answer

a)Hence, derived the given equality,\(P(\min ({X_i}) < \widetilde \mu < \max ({X_i})) = 1 - {\left( {\frac{1}{2}} \right)^{n - 1}}\).

b) we will be about \(97\% \) confident that the true population median is between the minimum and maximum of the data set:\(1.44 < \widetilde \mu < 3.54\)

c) \(P({X_{(2)}} < \widetilde \mu < {X_{(n - 1)}}) = 1 - {\left( {\frac{1}{2}} \right)^{n - 2}}\).

Step-by-step solution

Component rule. 

Complement rule:

\(P(\overline A ) = 1 - P(A)\)

Addition rule for disjoint or mutually exclusive events:

\(P(A{\rm{or}}B) = P(A) + P(B)\)

Multiplication rule for independent events:

\(P(A{\rm{ and }}B) = P(A) \times P(B)\)

Solution for part a).

Given that \({X_1},{X_2},....,{X_n}\) is a random sample from a continuous probability distribution with median \(\widetilde \mu \)

\(\begin{aligned}P({X_i} \le \widetilde \mu ) & = P({X_i} \le \widetilde \mu )\\ & = 0.5\end{aligned}\)

To prove:

\(P(\min ({X_i}) < \widetilde \mu < \max ({X_i})) = 1 - {\left( {\frac{1}{2}} \right)^{n - 1}}\)

Use the complement and addition rule:

\(\begin{aligned}P(\min ({X_i}) < \widetilde \mu < \max ({X_i})) & = 1 - P(\max ({X_i}) < \widetilde \mu {\rm{ or}}\min ({X_i}) \ge \widetilde \mu )\\ & = 1 - P(\max ({X_i}) < \widetilde \mu ) - P(\min ({X_i}) \ge \widetilde \mu )\\ & = 1 - P({X_1} \le \widetilde \mu ,{X_2} \le \widetilde \mu ,...{X_n} \le \widetilde \mu ) - P({X_1} \ge \widetilde \mu ,{X_2} \ge \widetilde \mu ,...{X_n} \ge \widetilde \mu )\end{aligned}\)

Since \({X_1},{X_2},....,{X_n}\) are random sample, we can assume that \({X_1},{X_2},....,{X_n}\) are independent and thus we can use the multiplication rule for independent events:

\(\begin{aligned} &= 1 - P({X_1} \le \widetilde \mu ,{X_2} \le \widetilde \mu ,...{X_n} \le \widetilde \mu ) - P({X_1} \ge \widetilde \mu ,{X_2} \ge \widetilde \mu ,...{X_n} \ge \widetilde \mu )\\ & = 1 - P({X_1} \le \widetilde \mu )P({X_2} \le \widetilde {\mu )},...P({X_n} \le \widetilde \mu ) - P({X_1} \ge \widetilde \mu )P({X_2} \ge \widetilde \mu ),...P({X_n} \ge \widetilde \mu )\\ & = 1 - {(P({X_1} \le \widetilde \mu ))^n} - {(P({X_1} \ge \widetilde \mu ))^n}\end{aligned}\)

\(\begin{aligned}{} & = 1 - 2{(P({X_1} \le \widetilde \mu ))^n}\\ & = 1 - 2{(0.5)^n}\\ & = 1 - {(0.5)^{n - 1}}\\ & = 1 - {\left( {\frac{1}{2}} \right)^{n - 1}}\end{aligned}\)

Thus, we have derived the given equality.

Solution for part b).

\(\begin{aligned}{}c & = 97\% \\n & = 6\end{aligned}\)

The minimum of the data set is \(1.44\) and the maximum is \(3.54\).

By part a) we can determine the confidence lever for \(n = 6\):

\(\begin{aligned}{}c & = 1 - {\left( {\frac{1}{2}} \right)^{n - 1}}\\c & = 1 - {\left( {\frac{1}{2}} \right)^{6 - 1}}\\c & = \frac{{31}}{{32}}\end{aligned}\)

\(\begin{aligned}{}c & = 0.96875\\c \approx 0.97\\c & = 97\% \end{aligned}\)

Thus we will be about \(97\% \) confident that the true population median is between the minimum and maximum of the data set:

\(1.44 < \widetilde \mu < 3.54\)

Solution for part c).

Given that \({X_1},{X_2},....,{X_n}\) is a random sample from a continuous probability distribution with median \(\widetilde \mu \)

\(\begin{aligned}P({X_i} \le \widetilde \mu ) & = P({X_i} \le \widetilde \mu )\\ & = 0.5\end{aligned}\)

To prove:

Use the complement and addition rule (we will assume that \({X_1}\) is the smallest values) :

\(\begin{aligned}P({X_{(2)}} < \widetilde \mu < {X_{(n - 1)}}) & = 1 - P({X_{(n - 1)}} < \widetilde \mu or{X_{(2)}} \ge \widetilde \mu )\\ & = 1 - P({X_{(n - 1)}} < \widetilde \mu ) - P({X_{(2)}} \ge \widetilde \mu )\\ & = 1 - P({X_1} \le \widetilde \mu ,{X_2} \le \widetilde \mu ,...{X_{n - 1}} \le \widetilde \mu ) - P({X_2} \ge \widetilde \mu ,{X_3} \ge \widetilde \mu ,...{X_n} \ge \widetilde \mu )\end{aligned}\)

Since \({X_1},{X_2},....,{X_n}\) are random sample, we can assume that \({X_1},{X_2},....,{X_n}\) are independent and thus we can use the multiplication rule for independent events:

\(\begin{aligned}{} & = 1 - P({X_1} \le \widetilde \mu ,{X_2} \le \widetilde \mu ,...{X_{n - 1}} \le \widetilde \mu ) - P({X_2} \ge \widetilde \mu ,{X_3} \ge \widetilde \mu ,...{X_n} \ge \widetilde \mu )\\ & = 1 - P({X_1} \le \widetilde \mu )P({X_2} \le \widetilde {\mu )},...P({X_{n - 1}} \le \widetilde \mu ) - P({X_2} \ge \widetilde \mu )P({X_3} \ge \widetilde \mu ),...P({X_n} \ge \widetilde \mu )\\ & = 1 - {(P({X_1} \le \widetilde \mu ))^{n - 1}} - {(P({X_1} \ge \widetilde \mu ))^{n - 1}}\end{aligned}\)

\(\begin{aligned}{} & = 1 - 2{(P({X_1} \le \widetilde \mu ))^{n - 1}}\\ & = 1 - 2{(0.5)^{n - 1}}\\& = 1 - {(0.5)^{n - 2}}\\ & = 1 - {\left( {\frac{1}{2}} \right)^{n - 2}}\end{aligned}\)

Hence, \(P({X_{(2)}} < \widetilde \mu < {X_{(n - 1)}}) = 1 - {\left( {\frac{1}{2}} \right)^{n - 2}}\).