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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

Determine the following:

a. The 95th percentile of the chi-squared distribution with \({\rm{v = 10}}\)

b. The 5th percentile of the chi-squared distribution with\({\rm{v = 10}}\)

\(\begin{array}{l}{\rm{c}}{\rm{.P}}\left( {{\rm{10}}{\rm{.98拢 }}{{\rm{\chi }}^{\rm{2}}}{\rm{拢 36}}{\rm{.78}}} \right){\rm{,where }}{{\rm{\chi }}^{\rm{2}}}{\rm{ is achi - squared rv with \nu = 22}}\\{\rm{d}}{\rm{.P}}\left( {{{\rm{\chi }}^{\rm{2}}}{\rm{ < 14}}{\rm{.611}}} \right.{\rm{ or }}\left. {{{\rm{\chi }}^{\rm{2}}}{\rm{ > 37}}{\rm{.652}}} \right){\rm{,where }}{{\rm{\chi }}^{\rm{2}}}{\rm{ is achi squared rv with \nu = 25}}\end{array}\)

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?)

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?

Determine the confidence level for each of the following large-sample one-sided confidence bounds:

\(\begin{array}{l}{\rm{a}}{\rm{.Upperbound:\bar x + }}{\rm{.84s/}}\sqrt {\rm{n}} \\{\rm{b}}{\rm{.Lowerbound:\bar x - 2}}{\rm{.05s/}}\sqrt {\rm{n}} \\{\rm{c}}{\rm{.Upperbound:\bar x + }}{\rm{.67\;s/}}\sqrt {\rm{n}} \end{array}\)

Let \({\mathbf{0}} \leqslant {\mathbf{\gamma }} \leqslant {\mathbf{\alpha }}\). Then a 100(1 - 伪)% CI for 碌 when n is large is

\(\left( {\overline {\bf{x}} {\bf{ - }}{{\bf{z}}_{\bf{\gamma }}}{\bf{.}}\frac{{\bf{s}}}{{\sqrt {\bf{n}} }}{\bf{,}}\overline {\bf{x}} {\bf{ + }}{{\bf{z}}_{{\bf{\alpha - \gamma }}}}{\bf{ \times }}\frac{{\bf{s}}}{{\sqrt {\bf{n}} }}} \right)\)

The choice 蠏 =伪/2 yields the usual interval derived in Section 7.2; if 蠏 鈮� 伪/2, this interval is not symmetric about \(\overline x \). The width of this interval is \({\mathbf{w = s(}}{{\mathbf{z}}_{\mathbf{\gamma }}}{\mathbf{ + }}{{\mathbf{z}}_{{\mathbf{\alpha - \gamma }}}}{\mathbf{)/}}\sqrt {\mathbf{n}} \) .Show that w is minimized for the choice 蠏 =伪/2, so that the symmetric interval is the shortest. (Hints: (a) By definition of za, 蠒 (za) = 1 - 伪, so that za =蠒-1 (1 -伪 );

(b) the relationship between the derivative of a function y= f(x) and the inverse function x= f-1(y) is (dydy) f-1(y) = 1/f鈥�(x).)
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