The technology underlying hip replacements has changed as these operations have become more popular (over 250,000 in United States in 2008). Starting in 2003, highly durable ceramic hips were marketed. Unfortunately, for too many patients the increased durability has been couunterbalanced 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 95% confidence level for the true proportion of such hips that develop squeaking.

b. Interpret the 95% confidence level used in (a).

#### Solution

The R code to compute the confidence interval for true proportion of hip that develop squeaking is as follows:

```
X<-10
n<-143
pbar<-round(X/n,4)
alpha<-0.05
SE<-sqrt(pbar*(1-pbar)/n)
E<-qnorm(1-alpha/2)*SE
left<-round(pbar-E,3)
right<-round(pbar+E,3)
left
```

`[1] 0.028`

`right`

`[1] 0.112`

(a) The $95$ percent confidence interval for true proportion of hip that develop squeaking is

`$$ \begin{aligned} & \big(\overline{p}-z_{\alpha/2}\times \sqrt{\frac{\overline{p}*(1-\overline{p})}{n}},\overline{p}+z_{\alpha/2}\times \sqrt{\frac{\overline{p}*(1-\overline{p})}{n}}\big)\\ & = \big(0.0699-1.96\times \sqrt{\frac{0.0699*(1-0.0699)}{143}},0.0699+1.96\times \sqrt{\frac{0.0699*(1-0.0699)}{143}}\big)\\ &= \big(0.028,0.112\big) \end{aligned} $$`

Thus an lower confience bound for true proportion of hip that develop squeaking using a confidence level of $95$ % is $0.028$.

(b) The $95$ percent confidence interval for true proportion of hip that develop squeaking is $(0.028, 0.112)$. We are 95 % confident that the true proportion of hip that develop squeaking lies in the interval ($0.028, 0.112)$.