Chapter 8, Problem 17CR

Expand Your Knowledge: Two Confidence Intervals What happens if we want several confidence intervals to hold at the same time (concurrently)? Do we still have the same level of confidence we had for each individual interval?

(a) Suppose we have two independent random variables x₁ and x₂ with

respective population means µ₁ and µ₂. Let us say that we use sample data to construct two 80% confidence intervals.

Confidence Interval	Confidence Level
$A_1<{\mu }_1<B_1$	0.80
$A_2<{\mu }_2<B_2$	0 80

Now. what is the probability that both intervals hold at the same time? Use methods of Section 5.2 to show that

$P\left(A_1<{\mu }_1<B_{1}and A_2<{\mu }_2<B_2\right)=0.64$

Hint: You are combining independent events. If the confidence is 64% that both intervals hold concurrently, explain why the risk that at least one interval does not hold (i.e., fails) must be 36%.

(b) Suppose we want both intervals to hold with 90% confidence (i.e., only 10%risk level). How much confidence c should each interval have to achieve this combined level of confidence? (Assume that each interval has the same confidence level c.)

Hint: $P(A_1<{\mu }_1<B_{1}and{A}_2< {\mu }_2<B_2)=0.90$

$$ $P(A_1<{\mu }_1<B_1\text{ ) }\times P(A_2< {\mu }_2<B_2)=0.90$

$c\times c=0.90$

Now solve for c.

(c) If we want both intervals to hold at the 90% level of confidence, then the individual intervals must hold at a higher level of confidence. Write a brief but detailed explanation of how this could be of importance in a large, complex engineering design such as a rocket booster or a spacecraft.

To determine

To find: The probability that both interval hold at the same time and also explain the risk that at least one interval does not hold must be 36%.

Answer to Problem 17CR

Solution: The probability that both interval hold at the same time is 0.64.

Explanation of Solution

Calculation:

$P(A_1<{\mu }_1<B_1\cap A_2<{\mu }_2<B_2)=P(A_1<{\mu }_1<B_1)\cdot P(A_2<{\mu }_2<B_2)$

Since, $x_1\text{and}{x}_2$ are independent random variables.

$\begin{array}{l}P(A_1<{\mu }_1<B_1)=0.80\\ P(A_2<{\mu }_2<B_2)=0.80\end{array}$

Probability that both interval together is

$\begin{array}{l}P(A_1<{\mu }_1<B_1\text{and}{A}_2<{\mu }_2<B_2)=P(A_1<{\mu }_1<B_1)\cdot P(A_2<{\mu }_2<B_2)\\ P(A_1<{\mu }_1<B_1\cap A_2<{\mu }_2<B_2)=P(A_1<{\mu }_1<B_1)\cdot P(A_2<{\mu }_2<B_2)\\ P(A_1<{\mu }_1<B_1\cap A_2<{\mu }_2<B_2)=0.80(0.80)\\ P(A_1<{\mu }_1<B_1\cap A_2<{\mu }_2<B_2)=0.64\end{array}$

Thus the probability that both interval holds at the same time is 0.64.

$\begin{array}{l}P(\text{at least one interval fails to hold μ})=1-P(\text{both interval capture their μ})\\ P(\text{at least one interval fails to hold μ})=1-0.64\\ P(\text{at least one interval fails to hold μ})=0.36\end{array}$

Thus, the probability that at least one of item does not contain $\mu$ is 0.36 or 36%.

To determine

To find: The confidence level cthat each interval should achieve to have 90% combined level of confidence.

Answer to Problem 17CR

Solution: The value of confidence c to achieve 90% combined level of confidence is 0.949.

Explanation of Solution

$\begin{array}{l}P(A_1<{\mu }_1<B_1\text{ and }{A}_2<{\mu }_2<B_2)=0.90\\ P(A_1<{\mu }_1<B_1\cap A_2<{\mu }_2<B_2)=P(A_1<{\mu }_1<B_1)\cdot P(A_2<{\mu }_2<B_2)\\ P(A_1<{\mu }_1<B_1)\cdot P(A_2<{\mu }_2<B_2)=0.90\\ c\cdot c=0.90\\ c^2=0.90\\ c=\pm \sqrt{0.90}\\ c=0.949\end{array}$

The value of c is 0.949.

To determine

To explain: The importance of having individual intervals at a higher level of confidence, if we both intervals to hold at 90% level of confidence and its application in complex engineering designs.

Explanation of Solution

A higher probability of success for individual components is needed, when both components must succeed in order for a particular system to be functional.

For example:

Consider an aircraft having two engines; if both engines in an aircraft must operate properly for the functionality of aircraft, then each engine must have a higher probability than the combined pair of engines. If the probability of success for the first engine is 0.975 and the probability of success for the second engine is also 0.975, then the probability of success for both of the engines is given as:

$\begin{array}{l}P(A_1<p_1<B_1\cap A_2<p_2<B_2)=P(A_1<{\mu }_1<B_1)\cdot P(A_2<{\mu }_2<B_2)\\ P(A_1<p_1<B_1\cap A_2<p_2<B_2)=(0.975)(0.975)\\ P(A_1<p_1<B_1\cap A_2<p_2<B_2)=0.9506\\ P(A_1<p_1<B_1\cap A_2<p_2<B_2)\approx 0.95\end{array}$

Hence we can see that for getting the 95% success probability for both of the engines, we have to use 97.5% probability of success for the individual engines.