Question

The article “Application of Surgical Navigation to To-
tal Hip Arthroplasty” (T. Ecker and S. Murphy, Journal

of Engineering in Medicine, 2007:699–712) reports
that in a sample of 113 people undergoing a certain
type of hip replacement surgery on one hip, 65 of them
had surgery on their right hip. Can you conclude that
frequency of this type of surgery differs between right
and left hips?

(a) State the null and alternative hypotheses.

(b) Use a two-sided 95% Agresti confidence interval for the population proportion to conduct this hypoth-

esis test. After constructing your confidence interval, state the conclusion in the context of the study

and the conclusion of the hypothesis test.

(c) Go ahead and formally conduct the hypothesis test using a = 0.05. Calculate the test statistic and

P-value appropriately. Does your conclusion to the hypothesis test agree with part (b)?

Answer 1

Part a

Null hypothesis: H₀: The frequency of hip replacement surgery do not differs between right and left hips.

Alternative hypothesis: H_a: The frequency of hip replacement surgery differs between right and left hips.

H₀: p₁ = p₂ vs. H_a: p₁ ≠ p₂

Part b

Here, we have to find 95% confidence interval for difference between two population proportions.

Confidence interval = (P1 – P2) ± Z*sqrt[(P1*(1 – P1)/N1) + (P2*(1 – P2)/N2)]

We are given

X1=65, n1 = 113, X2=113-65=48, n2 = 113

P1 = X1/n1 = 65/113 = 0.575221239

P2 = X2/n2 = 48/113 = 0.424778761

Z = 1.96 (by using z-table)

Confidence interval = (P1 – P2) ± Z*sqrt[(P1*(1 – P1)/N1) + (P2*(1 – P2)/N2)]

Confidence interval = (0.575221239 – 0.424778761) ± 1.96*sqrt[(0.575221239*(1 – 0.575221239)/113) + (0.424778761*(1 – 0.424778761)/113)]

Confidence interval = (0.575221239 – 0.424778761) ± 1.96* 0.0658

Confidence interval = 0.150442478 ± 1.96* 0.0658

Lower limit = 0.150442478 - 1.96* 0.0658 = 0.021474478

Upper limit = 0.150442478 + 1.96* 0.0658 = 0.279410478

The value of zero is not lies within this confidence interval, so we reject the null hypothesis H₀.

So, we conclude that the frequency of hip replacement surgery differs between right and left hips.

Part c

Here, we have to use two sample z test for population proportions.

H₀: p₁ = p₂ vs. H_a: p₁ ≠ p₂

Z = (P1 – P2) / sqrt[(P1*(1 – P1)/N1) + (P2*(1 – P2)/N2)]

Z = (0.575221239 – 0.424778761) / sqrt[(0.575221239*(1 – 0.575221239)/113) + (0.424778761*(1 – 0.424778761)/113)]

Z = 0.150442478 / 0.0658

Z = 2.286359848

P-value = 0.0222

(by using z-table)

P-value < α = 0.05

So, we reject the null hypothesis H₀

There is sufficient evidence to conclude that frequency of hip replacement surgery differs between right and left hips.