Question

4. (15 pts) Hip dysplasia is a hip socket abnormality that affects many large breed dogs. A review of medical records of dogs seen at 27 veterinary medical teaching hospitals found that hip dysplasia was more common in Golden retrievers than in Border Collies; the data are shown in the following table. Golden retriever 3,995 42,946 46,941 Border collie 221 5,007 5,228 Hip dysplasia Yes No Total (a) Calculate the relative risk of hip dysplasia for Golden retriever compared to border collie. (b) Calculate the sample value of odds ratio. (c) Construct a 95% confidence interval for the population value of the odds ratio. (d) Interpret the confidence interval from the part (b) in the context of this setting.

Answer 1

(a)

From table:

P(Hip dysplasia in golden retriever) = 3995/ 46941

P(Hip dysplasia in border collie) = 221/ 5228

The relative risk is:

RR =P(Hip dysplasia in golden retriever) / P(Hip dysplasia in border collie) = ( 3995/ 46941) / ( 221/ 5228) = 2.01

Since RR is greater than 1 so it shows that risk of hip dysplasia is higher in golden retriever group.

(b)

Odds for golden retriever group:

$O_{1}=3995: 42946$

Odds for border collie group:

$O_{2}=221:5007$

Odds ratio:

$OR=\frac{O_{1}}{O_{2}}=\frac{3999/42946}{221/5007}=2.11$

(c)

The 95% confidence interval for odds ratio

$ln(OR)\pm 1.96\sqrt{\frac{1}{3995}+\frac{1}{221}+\frac{1}{42946}+\frac{1}{5007}}=ln(2.11)\pm 1.96\sqrt{\frac{1}{3995}+\frac{1}{221}+\frac{1}{42946}+\frac{1}{5007}}$

$=0.7467\pm 0.1386=(0.6081,0.8853)$

So required 95% confidence interval for odds ratio is

$\left (e^{0.6081},e^{0.8853} \right )=(1.84,2.42)$

(d)

Since confidence interval does not contain 1 so we cannot conclude that risk of hip dysplasia is same in both group.