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Chapter 7: Q22E (page 294)

The technology underlying hip replacements has changed as these operations have become more popular . Starting in 2003, highly durable ceramic hips were marketed. Unfortunately, for too many patients the increased durability has been counterbalanced by an increased incidence of squeaking. The May 11, 2008, issue of the New York Times reported that in one study of 143 individuals who received ceramic hips between 2003 and 2005, 10 of the hips developed squeaking.

a. Calculate a lower confidence bound at the \({\rm{95\% }}\)confidence level for the true proportion of such hips that develop squeaking.

b. Interpret the \({\rm{95\% }}\)confidence level used in (a).

Short Answer

Expert verified

a) The lower confidence bound is \(0.0423\)

b) There is \(95\% \)confidence that the true proportion is larger than \(0.0423\)

Step by step solution

01

To calculate a lower confidence bound

(a):

Denote with

\(\begin{array}{l}{\rm{\bar p - }}\left( {\frac{{{\rm{\hat p + z}}_{{\rm{\alpha /2}}}^{\rm{2}}}}{{{\rm{2n}}}}} \right){\rm{/}}\left( {{\rm{1 + }}\frac{{{\rm{z}}_{{\rm{\alpha /2}}}^{\rm{2}}}}{{\rm{n}}}} \right)\\{\rm{\hat q - 1 - \hat p}}{\rm{.}}\end{array}\)

A confidence interval for a

population proportion

p with confidence level approximately \({\rm{100(1 - \alpha )}}\)is

\(\left( {{\rm{\tilde p - }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\sqrt {{\rm{\hat p\hat q/n + z}}_{{\rm{\alpha /2}}}^{\rm{2}}{\rm{/4}}{{\rm{n}}^{\rm{2}}}} }}{{{\rm{1 + z}}_{{\rm{\alpha /2}}}^{\rm{2}}{\rm{/n}}}}{\rm{,\bar p + }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\sqrt {{\rm{\hat p\hat q/n + z}}_{{\rm{\alpha /2}}}^{\rm{2}}{\rm{/4}}{{\rm{n}}^{\rm{2}}}} }}{{{\rm{1 + z}}_{{\rm{\alpha /2}}}^{\rm{2}}{\rm{/n}}}}} \right)\)

This is also called a score \({\rm{Cl for p}}\)

By replacing \({{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{with }}{{\rm{z}}_{\rm{\alpha }}}{\rm{ and in \pm }}\)only 鈥� or + stand, the large-sample upper confidence bound for \({\rm{\mu }}\)and the large-sample lower confidence bound for \({\rm{\mu }}\)are obtained.

The lower confidence bound at the \(95\% \)confidence level for the true proportion can be computed using

\({\rm{\bar p + }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{,}}\frac{{\sqrt {{\rm{\hat p\hat q/n + z}}_{{\rm{\alpha /2}}}^{\rm{2}}{\rm{/4}}{{\rm{n}}^{\rm{2}}}} }}{{{\rm{1 + z}}_{{\rm{\alpha /2}}}^{\rm{2}}{\rm{/n}}}}{\rm{.}}\)

02

To solve the equation

The values in the formula are

\(\begin{array}{l}{\rm{\bar p - }}\frac{{{\rm{0}}{\rm{.07 + 1}}{\rm{.64}}{{\rm{5}}^{\rm{2}}}{\rm{/2143}}}}{{{\rm{1 + 1}}{\rm{.64}}{{\rm{5}}^{\rm{2}}}{\rm{/143}}}}{\rm{ = 0}}{\rm{.078,}}\\{\rm{\hat p - }}\frac{{10}}{{{\rm{143}}}}{\rm{ - 0}}{\rm{.07,}}\\{\rm{\hat q - 1 - \hat p - 0}}{\rm{.93,}}\end{array}\)

where

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

and

\({{\rm{z}}_{\rm{\alpha }}}{\rm{ - }}{{\rm{z}}_{{\rm{0}}{\rm{.05}}}}\mathop {\rm{ - }}\limits^{{\rm{(1)}}} {\rm{1}}{\rm{.645}}\)

(1) : this is obtained from

\({\rm{P}}\left( {{\rm{Z > }}{{\rm{z}}_{{\rm{0}}{\rm{.05}}}}} \right){\rm{ - 0}}{\rm{.05}}\)

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

\(\begin{array}{l}{\rm{\bar p - }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\sqrt {{\rm{\hat p\hat q/n + z}}_{{\rm{\alpha /2}}}^{\rm{2}}{\rm{/4}}{{\rm{n}}^{\rm{2}}}} }}{{{\rm{1 + z}}_{{\rm{\alpha /2}}}^{\rm{2}}{\rm{/n}}}}{\rm{ - 0}}{\rm{.078 - }}\frac{{{\rm{1}}{\rm{.645 \times }}\sqrt {{\rm{0}}{\rm{.07 \times 0}}{\rm{.93/143 + 1}}{\rm{.64}}{{\rm{5}}^{\rm{2}}}{\rm{/}}\left( {{\rm{4 \times 14}}{{\rm{3}}^{\rm{2}}}} \right)} }}{{{\rm{1 + 1}}{\rm{.64}}{{\rm{5}}^{\rm{2}}}{\rm{/143}}}}\\{\rm{ - 0}}{\rm{.078 - 0}}{\rm{.0357 - 0}}{\rm{.0423}}{\rm{.}}\end{array}\)

Hence the lower confidence bound is \(0.0423\)

(b)

There is \(95\% \)confidence that the true proportion is larger than \(0.0423\)

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Most popular questions from this chapter

The article 鈥淢easuring and Understanding the Aging of Kraft Insulating Paper in Power Transformers鈥� contained the following observations on degree of polymerization for paper specimens for which viscosity tim\({\rm{418 421 421 422 425 427 431 434 437 439 446 447 448 453 454 463 465}}\)es concentration fell in a certain middle range:

a. Construct a boxplot of the data and comment on any interesting features.

b. Is it plausible that the given sample observations were selected from a normal distribution?

c. Calculate a two-sided \({\rm{95\% }}\)confidence interval for true average degree of polymerization (as did the authors of the article). Does the interval suggest that \({\rm{440}}\) is a plausible value for true average degree of polymerization? What about \({\rm{450}}\)?

Consider a normal population distribution with the value of \({\rm{\sigma }}\) known. a. What is the confidence level for the interval \({\rm{\bar x \pm 2}}{\rm{.81\sigma /}}\sqrt {\rm{n}} \)? b. What is the confidence level for the interval \({\rm{\bar x \pm 1}}{\rm{.44\sigma /}}\sqrt {\rm{n}} \)? c. What value of \({{\rm{z}}_{{\rm{\alpha /2}}}}\) in the CI formula (\({\rm{7}}{\rm{.5}}\)) results in a confidence level of \({\rm{99}}{\rm{.7\% }}\)? d. Answer the question posed in part (c) for a confidence level of \({\rm{75\% }}\).

Suppose that a random sample of \({\rm{50}}\) bottles of a particular brand of cough syrup is selected and the alcohol content of each bottle is determined. Let \({\rm{\mu }}\) denote the average alcohol content for the population of all bottles of the brand under study. Suppose that the resulting \({\rm{95\% }}\)confidence interval is \({\rm{(7}}{\rm{.8,9}}{\rm{.4)}}\). a. Would a \({\rm{90\% }}\) confidence interval calculated from this same sample have been narrower or wider than the given interval? Explain your reasoning. b. Consider the following statement: There is a \({\rm{95\% }}\) chance that \({\rm{\mu }}\) is between \({\rm{7}}{\rm{.8}}\) and \({\rm{9}}{\rm{.4}}\). Is this statement correct? Why or why not? c. Consider the following statement: We can be highly confident that \({\rm{95\% }}\) of all bottles of this type of cough syrup have an alcohol content that is between \({\rm{7}}{\rm{.8}}\) and \({\rm{9}}{\rm{.4}}\). Is this statement correct? Why or why not? d. Consider the following statement: If the process of selecting a sample of size \({\rm{50}}\) and then computing the corresponding \({\rm{95\% }}\) interval is repeated \({\rm{100}}\)times, \({\rm{95}}\) of the resulting intervals will include \({\rm{\mu }}\). Is this statement correct? Why or why not?

A study of the ability of individuals to walk in a straight line reported the accompanying data on cadence (strides per second) for a sample of n =\({\rm{20}}\) randomly selected healthy men.

\({\rm{.95 }}{\rm{.85 }}{\rm{.92 }}{\rm{.95 }}{\rm{.93 }}{\rm{.86 1}}{\rm{.00 }}{\rm{.92 }}{\rm{.85 }}{\rm{.81 }}{\rm{.78 }}{\rm{.93 }}{\rm{.93 1}}{\rm{.05 }}{\rm{.93 1}}{\rm{.06 1}}{\rm{.06 }}{\rm{.96 }}{\rm{.81 }}{\rm{.96}}\)

A normal probability plot gives substantial support to the assumption that the population distribution of cadence is approximately normal. A descriptive summary of the data from Minitab follows:

Variable N Mean Median TrMean StDev SEMean cadence

\({\rm{20 0}}{\rm{.9255 0}}{\rm{.9300 0}}{\rm{.9261 0}}{\rm{.0809 0}}{\rm{.0181}}\)

Variable Min Max Q1 Q3 cadence

\({\rm{0}}{\rm{.7800 1}}{\rm{.0600 0}}{\rm{.8525 0}}{\rm{.9600}}\)

a. Calculate and interpret a \({\rm{95\% }}\) confidence interval for population mean cadence.

b.Calculate and interpret a \({\rm{95\% }}\)prediction interval for the cadence of a single individual randomly selected from this population.

c. Calculate an interval that includes at least \({\rm{99\% }}\)of the cadences in the population distribution using a confidence level of \({\rm{95\% }}\)

Consider the next \({\rm{1000}}\) 95% CIs for m that a statistical consultant will obtain for various clients. Suppose the data sets on which the intervals are based are selected independently of one another. How many of these\({\rm{1000}}\)intervals do you expect to capture the corresponding value of m? What is the probability that between \({\rm{940 and 960}}\)Do these intervals contain the corresponding value of m? (Hint: Let Y = the number among the \({\rm{1000}}\) intervals that contain m. What kind of random variable is Y?)
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