Q58SE Let X1, X2,鈥�, Xn be a random ... [FREE SOLUTION]

91影视

Probability And Statistics For Engineering And Sciences

Jay L. Devore

$Math Studyset 91影视 Explanations$ Math

9th Edition 路 714 pages

ISBN: 9781305251809

Chapter 7: Q58SE (page 308)

Let X₁, X₂,鈥�, 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(鈥淭he Essential Amino Acid Requirements of Infants,鈥� Amer. J. Of Nutrition, 1964: 322鈥�330).
c.Let x₍₂₎denote the second smallest of the x_i鈥檚 and x_(n-1) denote the second largest of the x_i鈥檚. What is the confidence level of the interval(x₍₂₎, x_(n-1)) for $\widetilde \mu $?

Short Answer

Expert verified

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

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Short Answer

Step by step solution

Component rule.

Solution for part a).

Solution for part b).

Solution for part c).

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