Question

Canine hip dysplasia is a degenerative disease that causes pain in many dogs. Sometimes advanced warning signs appear in puppies as young as 6 months. A veterinarian checked 42 puppies whose owners brought them to a vaccination clinic, and she found 5 with early hip dysplasia. She considers this group to be a random sample of all puppies. Construct a "plus-four" confidence interval and interpret it in this context.

Answer 1

Answer with explanation:

Formula for plus-four confidence interval :

$\hat{p}\pm z^* \sqrt\dfrac{\hat{p}(1-\hat{p})}{n+4}}$

, where n= Sample size.

$\hat{p}$  = Sample proportion and $\hat{p}=\dfrac{x+2}{n+4}$

z* = Critical z-value.

Let p be the proportion of puppies area found with early hip dysplasia.

As per given , we have

n= 42

$\hat{p}=\dfrac{5+2}{42+4}=0.1522$

Since confidence interval is not given , so we assume it as 95% .

z-critical value for 95% confidence is 1.96.

Then, the required confidence interval will become :

$0.1522\pm (1.96)\sqrt{\dfrac{0.1522(1-0.1522)}{42+4}}\\\\ 0.152\pm (1.96)\sqrt{0.002805112173}\\\\ 0.152\pm 0.1038\\\\ =(0.1522- 0.1038 ,\ 0.152+ 0.1038)\approx(0.0484,\ 0.256)$

Hence, the plus four confidence interval for p = (0.0484, 0.256)

Interpretation: We are 95% sure that the true proportion of puppies area found with early hip dysplasia lies in  (0.0484, 0.256).