Question

A veterinarian wants to know if pit bulls or golden retrievers have a higher incidence of tooth decay at the age of three. The vet surveys 120 three-year-old pit bulls and finds 30 of them have tooth decay. The vet then surveys 160 three-year-old golden retrievers and finds 32 of them have tooth decay. Number the population of pit bulls and golden retrievers by 1 and 2, respectively. Calculate a 90% confidence interval for the difference in the population proportion of pit bulls and golden retrievers that have tooth decay. Which of the following is correct?
A) [-0.0492, 0.1492]
B) [-0.0332, 0.1332]
C) [0.0199. 0.0801]
D) [0.0428, 0.0572]

Answer 1

Answer: B) [-0.0332,0.1332]

Step-by-step explanation: Confidence Interval is an interval where we can be a percentage sure the true mean is.

The confidence interval for a difference in population proportion is calculated following these steps:

First, let's find population proportion for each population:

$p_{1}=\frac{30}{120}=0.25$

$p_{2}=\frac{32}{160}=0.2$

Second, calculate standard deviation for each proportion:

$\sigma_{1}=\sqrt{\frac{0.25(0.75)}{120} } = 0.0395$

$\sigma_{2}=\sqrt{\frac{0.2(0.8)}{160} } = 0.0316$

Now, we calculate standard error for difference:

$SE=\sqrt{\sigma_{1}^{2}+\sigma_{2}^{2}}$

$SE=\sqrt{0.0395^{2}+0.0316^{2}}$

SE = 0.0505

The z-score for a 90% CI is 1.645.

Then, confidence interval is

$p_{1}-p_{2}$ ± z-score.SE

$0.25-0.2$ ± $1.645(0.0505)$

0.05 ± 0.0831

The limits of this interval are:

inferior: 0.05 - 0.0831 = $-0.0332$

superior: 0.05 + 0.0831 = 0.1332

The 90% confidence interval for the difference in the population proportion of pit pulls and golden retrievers is $[-0.0332,0.1332]$.