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Reconsider the CI for p, and focus on a confidence level of 95%. Show that the confidence limits agree quite well with those of the traditional interval once two successes and two failures have been appended to the sample.

Short Answer

Expert verified

The limits agree quite well.

Step by step solution

01

To find the confidence interval

For better understand see Agresti and Coull article mentioned in the exercise.

Remember that the traditional interval is

\({\rm{\hat p \pm }}{{\rm{z}}_{{\rm{\alpha /2}}}}\sqrt {\frac{{{\rm{\hat p\hat q}}}}{{\rm{n}}}} \)

For the \(95\% \)confidence interval the z score is

\(\begin{array}{*{20}{r}}{{\rm{100(1 - \alpha ) - 95}}}\\{{\rm{\alpha - 0}}{\rm{.05}}}\end{array}\)

and

\({{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ - }}{{\rm{z}}_{{\rm{0}}{\rm{.05/2}}}}{\rm{ - }}{{\rm{z}}_{{\rm{0}}{\rm{.025}}}}\mathop {}\limits^{{\rm{(1)}}} {\rm{ - 1}}{\rm{.96}}\)

(1) : this is obtained from

\({\rm{P}}\left( {{\rm{Z > }}{{\rm{z}}_{{\rm{0}}{\rm{.025}}}}} \right){\rm{ - 0}}{\rm{.025}}\)

and from the normal probability table in the appendix. The probability can also be computed with a software.

02

To find the mid point

The midpoint of the new interval is roughly

\({\rm{\hat p - }}\frac{{{\rm{x + 2}}}}{{{\rm{n + 4}}}}\)

for \({{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ - 1}}{\rm{.96\gg 2}}\)

where the midpoint is actually given with

\(\frac{{{\rm{x + }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{/2}}}}{{{\rm{n + z}}_{{\rm{\alpha /2}}}^{\rm{2}}}}\)

The variance is given with

\(\frac{{{\rm{npq}}}}{{{{\left( {{\rm{n + z}}_{{\rm{\alpha /2}}}^{\rm{2}}} \right)}^{\rm{2}}}}}\)

which is, for \({{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ - 1}}{\rm{.96\gg 2}}\), approximately

\(\frac{{{\rm{pq}}}}{{{\rm{n + 4}}}}\)

and the standard deviation is given with

\(\sqrt {\frac{{{\rm{pq}}}}{{{\rm{n + 4}}}}} \)

This indicates that the \(95\% \)confidence interval becomes

\({\rm{\hat p \pm }}{{\rm{z}}_{{\rm{a/2}}}}{\rm{ \times }}\sqrt {\frac{{{\rm{\hat p\hat q}}}}{{{\rm{n + 4}}}}} \)

which is the same as the mentioned traditional interval. The CL agree because for values

\(\begin{array}{l}{\rm{\bar n - n + 4}}\\{\rm{\bar x - x + 2}}\end{array}\)

03

To find the coincidence interval

the confidence interval becomes

\({\rm{\bar p \pm }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\sqrt {\frac{{{\rm{\bar p\bar q}}}}{{{\rm{\bar n}}}}} \)

Where

where

\(\begin{array}{l}{\rm{\bar p - \hat p - }}\frac{{{\rm{\bar x}}}}{{{\rm{\bar n}}}}{\rm{,}}\\{\rm{\bar q - 1 - \bar p}}{\rm{.n}}\end{array}\)

Hence The limits agree quite well.

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Most popular questions from this chapter

A CI is desired for the true average stray-load loss \({\rm{\mu }}\) (watts) for a certain type of induction motor when the line current is held at \({\rm{10 amps}}\) for a speed of \({\rm{1500 rpm}}\). Assume that stray-load loss is normally distributed with \({\rm{\sigma = 3}}{\rm{.0}}\). a. Compute a \({\rm{95\% }}\) CI for \({\rm{\mu }}\) when \({\rm{n = 25}}\) and \({\rm{\bar x = 58}}{\rm{.3}}\). b. Compute a \({\rm{95\% }}\) CI for \({\rm{\mu }}\) when \({\rm{n = 100}}\) and \({\rm{\bar x = 58}}{\rm{.3}}\). c. Compute a \({\rm{99\% }}\) CI for \({\rm{\mu }}\) when \({\rm{n = 100}}\) and \({\rm{\bar x = 58}}{\rm{.3}}\). d. Compute an \({\rm{82\% }}\) CI for \({\rm{\mu }}\) when \({\rm{n = 100}}\) and \({\rm{\bar x = 58}}{\rm{.3}}\). e. How large must n be if the width of the \({\rm{99\% }}\) interval for \({\rm{\mu }}\) is to be \({\rm{1}}{\rm{.0}}\)?

Example 1.11 introduced the accompanying observations on bond strength.

11.5 12.1 9.9 9.3 7.8 6.2 6.6 7.0

13.4 17.1 9.3 5.6 5.7 5.4 5.2 5.1

4.9 10.7 15.2 8.5 4.2 4.0 3.9 3.8

3.6 3.4 20.6 25.5 13.8 12.6 13.1 8.9

8.2 10.7 14.2 7.6 5.2 5.5 5.1 5.0

5.2 4.8 4.1 3.8 3.7 3.6 3.6 3.6

a.Estimate true average bond strength in a way that conveys information about precision and reliability.

(Hint: \(\sum {{{\bf{x}}_{\bf{i}}}} {\bf{ = 387}}{\bf{.8}}\) and \(\sum {{{\bf{x}}^{\bf{2}}}_{\bf{i}}} {\bf{ = 4247}}{\bf{.08}}\).)

b. Calculate a 95% CI for the proportion of all such bonds whose strength values would exceed 10.

The following observations are lifetimes (days) subsequent to diagnosis for individuals suffering from blood cancer (鈥淎 Goodness of Fit Approach to the Class of Life Distributions with Unknown Age,鈥 Quality and Reliability Engr. Intl., \({\rm{2012: 761--766):}}\)

\(\begin{array}{*{20}{l}}{{\rm{115 181 255 418 441 461 516 739 743 789 807}}}\\{{\rm{865 924 983 1025 1062 1063 1165 1191 1222 1222 1251}}}\\{{\rm{1277 1290 1357 1369 1408 1455 1478 1519 1578 1578 1599}}}\\{{\rm{1603 1605 1696 1735 1799 1815 1852 1899 1925 1965}}}\end{array}\)

a. Can a confidence interval for true average lifetime be calculated without assuming anything about the nature of the lifetime distribution? Explain your reasoning. (Note: A normal probability plot of the data exhibits a reasonably linear pattern.)

b. Calculate and interpret a confidence interval with a \({\rm{99\% }}\)confidence level for true average lifetime.

A random sample of n=\({\rm{15}}\)heat pumps of a certain type yielded the following observations on lifetime (in years):

\(\begin{array}{*{20}{l}}{{\rm{2}}{\rm{.0 1}}{\rm{.3 6}}{\rm{.0 1}}{\rm{.9 5}}{\rm{.1 }}{\rm{.4 1}}{\rm{.0 5}}{\rm{.3}}}\\{{\rm{15}}{\rm{.7 }}{\rm{.7 4}}{\rm{.8 }}{\rm{.9 12}}{\rm{.2 5}}{\rm{.3 }}{\rm{.6}}}\end{array}\)

a. Assume that the lifetime distribution is exponential and use an argument parallel to that to obtain a \({\rm{95\% }}\) CI for expected (true average) lifetime.

b. How should the interval of part (a) be altered to achieve a confidence level of \({\rm{99\% }}\)?

c. What is a \({\rm{95\% }}\)CI for the standard deviation of the lifetime distribution? (Hint: What is the standard deviation of an exponential random variable?)

Let X1, X2,鈥, Xn 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 xi鈥檚 and x(n-1) denote the second largest of the xi鈥檚. What is the confidence level of the interval(x(2), x(n-1)) for \(\widetilde \mu \)?

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