Question

A random sample of n1 = 252 people who live in a city were selected and 104 identified as a cat person. A random sample of n2 = 107 people who live in a rural area were selected and 55 identified as a cat person. Find the 99% confidence interval for the difference in the proportion of people that live in a city who identify as a cat person and the proportion of people that live in a rural area who identify as a cat person.

Round answers to 2 decimal places, use confidence interval notation :

Answer 1

Confidence interval = -0.1013 ± 0.1991

Confidence interval = (-0.30, 0.10)

Explanation:

We can use the following formula to find the confidence interval for the difference in two population proportions:

Confidence interval = (p1 - p2) ± z*sqrt[p1(1-p1)/n1 + p2(1-p2)/n2]

where p1 and p2 are the sample proportions, n1 and n2 are the sample sizes, and z is the z-score corresponding to the level of confidence.

First, we need to calculate the sample proportions:

p1 = 104/252 = 0.4127

p2 = 55/107 = 0.5140

Next, we need to find the value of z for a 99% confidence level. We can look up this value in a standard normal distribution table or use a calculator, which gives us a z-score of 2.576.

Then, we can plug in the values:

Confidence interval = (0.4127 - 0.5140) ± 2.576*sqrt[0.4127(1-0.4127)/252 + 0.5140(1-0.5140)/107]

Simplifying this expression, we get:

Confidence interval = -0.1013 ± 0.1991

Rounding to two decimal places and using interval notation, we get:

Confidence interval = (-0.30, 0.10)