Question

A random sample of n1 = 272 people who live in a city were selected and 112 identified as a religious person. A random sample of n2 = 105 people who live in a rural area were selected and 53 identified as a religious person. Find the 90% confidence interval for the difference in the proportion of people that live in a city who identify as a religious person and the proportion of people that live in a rural area who identify as a religious person.

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

Answer 1

Explanation:

To find the 90% confidence interval for the difference in proportions, we can use the following formula:

CI = (p1 - p2) ± zα/2 * √( (p1 * (1-p1) / n1) + (p2 * (1-p2) / n2) )

where:

p1 and p2 are the sample proportions from the city and rural area, respectively

n1 and n2 are the sample sizes from the city and rural area, respectively

zα/2 is the critical value from the standard normal distribution for a 90% confidence level, which is 1.645

First, we need to calculate the sample proportions:

p1 = 112 / 272 = 0.4118

p2 = 53 / 105 = 0.5048

Next, we can plug in the values and calculate the confidence interval:

CI = (0.4118 - 0.5048) ± 1.645 * √( (0.4118 * (1-0.4118) / 272) + (0.5048 * (1-0.5048) / 105) )

CI = (-0.146, -0.031)

Therefore, the 90% confidence interval for the difference in proportions is (-0.146, -0.031). This means we are 90% confident that the true difference in proportions of religious people between the city and rural areas lies between -0.146 and -0.031. Since this interval does not include zero, we can conclude that the proportion of religious people is significantly different between the two areas.