Question

Researchers are studying the distribution of subscribers to a certain streaming service in different populations. From a random sample of 200 people in City C, 34 were found to subscribe to the streaming service. From a random sample of 200 people in City K, 54 were found to subscribe to the streaming service. Assuming all conditions for inference are met, which of the following is a 90 percent confidence interval for the difference in population proportions (City C minus City K) who subscribe to the streaming service?
A. (0.17 – 0.27) + 1.65, 0.27 0.17 V 200
B. (0.17 – 0.27) 1.96 V (0.17)(0.83)+(0.27)(0.73) 400
C. (0.17 – 0.27) + 1.657 (0.17)(0.83)+(0.27)(0.73) 400
D. (0.17 – 0.27) + 1.96V (0.17)(0.83)+(0.27)(0.73) 200
E. (0.17 – 0.27) + 1.657 (0.17)(0.83)+0.27)(0.73) 200

Answer 1

Answer:

$CI = (0.17 - 0.27)\± 1.65\sqrt{\frac{(0.17)*(0.83) + (0.27)*(0.73)}{200}}$

Step-by-step explanation:

Given

$n = 200$

$x_1 = 34$ -- City C

$x_2 = 54$ --- City K

Required

Determine the 90% confidence interval

This is calculated using:

$CI = \bar x \± z\frac{\sigma}{\sqrt n}$

Calculating $\bar x$

$\bar x = \bar x_1 - \bar x_2$

$\bar x = \frac{x_1}{n} - \frac{x_2}{n}$

$\bar x = \frac{34}{200} - \frac{54}{200}$

$\bar x = 0.17 - 0.27$

For a 90% confidence level, the ​z-score is 1.65.  So:

$z = 1.65$

Calculating the standard deviation $\sigma$

$\sigma = \sqrt{(\bar x_1)*(1 - \bar x_1) + (\bar x_2)*(1 - \bar x_2) }$

So:

$\sigma = \sqrt{(0.17)*(1 - 0.17) + (0.27)*(1 - 0.27) }$

$\sigma = \sqrt{(0.17)*(0.83) + (0.27)*(0.73)}$

So:

$CI = (0.17 - 0.27)\± 1.65\frac{\sqrt{(0.17)*(0.83) + (0.27)*(0.73)}}{\sqrt {200}}$

$CI = (0.17 - 0.27)\± 1.65\sqrt{\frac{(0.17)*(0.83) + (0.27)*(0.73)}{200}}$