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

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

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(\sigma)/(\sqrt n)`

Calculating
`\bar x`

`\bar x = \bar x_1 - \bar x_2`

`\bar x = (x_1)/(n) - (x_2)/(n)`

`\bar x = (34)/(200) - (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 = √((\bar x_1)*(1 - \bar x_1) + (\bar x_2)*(1 - \bar x_2) )`

So:

`\sigma = √((0.17)*(1 - 0.17) + (0.27)*(1 - 0.27) )`

`\sigma = √((0.17)*(0.83) + (0.27)*(0.73))`

So:

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

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