Question

The technology underlying hip replacements has changed as these operations have become more popular (over 250,000 in the USA in 2008). 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 and interpret a 99% confidence interval for the true proportion of ceramic hips that develop squeaking.

b Perform a hypothesis test to determine if more than 5% of all ceramic hips squeak.

Answer 1

A) = 10/143 = 0.07

At 99% confidence interval the critical value is z^* = 2.58

The 99% confidence interval for population proportion is

+/- z^* * sqrt((1 - )/n)

= 0.07 +/- 2.58 * Sqrt(0.07 * (1 - 0.07)/143)

= 0.07 +/- 0.055

= 0.015, 0.125

B) H0: P = 0.05

H1: P > 0.05

The test statistic z = ( - p)/sqrt(p(1 - p)/n)

= (0.07 - 0.05)/Sqrt(0.05 * 0.95/143)

= 1.10

P-value = P(Z > 1.10)

= 1 - P(Z < 1.10)

= 1 - 0.8643

= 0.1357

At alpha = 0.05, since the P-value is greater than the significance level (0.1357 > 0.05), we should not reject the null hypothesis.

So there is not sufficient evidence to support the claim that more than 5% of all ceramic hips squeak.