Question

One study 145 individuals who received ceramic hips between 2003 and 2005, 20 hips developed squeaking. Using a=.02

1. Calculate the confidence interval for the true proportion of such hips that develop squeaking

2. Assume the population proportion of such hips that develop squeaking is p=0.15, what is the probability that between 8% and 21% of the sampled hips develop squeaking

3. Decide whether the proportion of such hips that develop squeaking is greater than 0.1. What is the rejection region? Calculate p-value. State appropriate conclusion.

Answer 1

a) = 20/145 = 0.1379

Confidence level = 1 - 0.02 = 0.98

At 98% confidence level, the critical value is z_0.01 = 2.33

The 98% confidence interval is

b) n = 145

    p = 0.15

= p = 0.15

     

     

P(0.08 < < 0.21)

= P(-2.36 < Z < 2.02)

= P(Z < 2.02) - P(Z < -2.36)

= 0.9783 - 0.0091

= 0.9692

c) H₀: p = 0.1

   H₁: p > 0.1

The test statistic is

   

  

At alpha = 0.02, the critical value is z_0.02 = 2.05

Reject H₀, if z > 2.05

Since the test statistic value is not greater than the critical value(1.52 < 2.05), so we should not reject the null hypothesis.

P-value = P(Z > 1.52)

             = 1 - P(Z < 1.52)

            = 1 - 0.9357

            = 0.0643

At 0.02 significance level, there is not sufficient evidence to conclude that the proportion of such hips that develop squeaking is greater than 0.1.