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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

Silicone implant augmentation rhinoplasty is used to correct congenital nose deformities. The success of the procedure depends on various biomechanical properties of the human nasal periosteum and fascia. The article 鈥淏iomechanics in Augmentation Rhinoplasty鈥� reported that for a sample of 15 (newly deceased) adults, the mean failure strain (%) was 25.0, and the standard deviation was 3.5.

a. Assuming a normal distribution for failure strain, estimate true average strain in a way that conveys information about precision and reliability.

b. Predict the strain for a single adult in a way that conveys information about precision and reliability. How does the prediction compare to the estimate calculated in part (a)?

A state legislator wishes to survey residents of her district to see what proportion of the electorate is aware of her position on using state funds to pay for abortions.

a. What sample size is necessary if the \({\rm{95\% }}\)CI for p is to have a width of at most . \({\rm{10}}\)irrespective of p?

b. If the legislator has strong reason to believe that at least \({\rm{2/3}}\)of the electorate know of her position, how large a sample size would you recommend?

Determine the t critical value for a lower or an upper confidence bound for each of the situations

\(\begin{array}{*{20}{l}}{{\rm{a}}{\rm{. Confidence level = 95\% , df = 10}}}\\{{\rm{b}}{\rm{. Confidence level = 95\% , df = 15}}}\\{{\rm{c}}{\rm{. Confidence level = 99\% , df = 15}}}\\{{\rm{d}}{\rm{. Confidence level = 99\% , n = 5}}}\\{{\rm{\;e}}{\rm{. Confidence level = 98\% , df = 24}}}\\{{\rm{f}}{\rm{. Confidence level = 99\% , n = 38}}}\end{array}\)

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 article 鈥淐oncrete Pressure on Formwork鈥� (Mag. of Concrete Res., 2009: 407鈥�417) gave the following observations on maximum concrete pressure (kN/m2 ):

33.2 41.8 37.3 40.2 36.7 39.1 36.2 41.8

36.0 35.2 36.7 38.9 35.8 35.2 40.1

a.Is it plausible that this sample was selected from a normal population distribution?

b. Calculate an upper confidence bound with confidence level 95% for the population standard deviation of maximum pressure.
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