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% confidence interval for the difference in population proportions (City C minus City K) who subscribe to the streaming service?
A. (0.17 – 0.27) + or - 1.65 underroot 0.17/200 + 0.27/200.
B. 0.17 – 0.27) + or -1.96 underroot (0.17)(0.83) + (0.27)(0.73)/400
C. 0.17 – 0.27) + or - 1.65 underroot (0.17)(0.83) + (0.27)(0.73)/400
D. (0.17 – 0.27) + or - 1.96 underroot (0.17)(0.83) + (0.27)(0.73)/200
E. (0.17 – 0.27) + or - 1.65 underroot (0.17)(0.83) + 0.27)(0.73)/200

Answer 1

Answer:

$(0.17 - 0.27) \pm 1.65\sqrt{\frac{0.17*0.83 + 0.27*0.73}{200}}$, that is, option C

Step-by-step explanation:

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.

This means that the proportions are:

$p_C = \frac{34}{200} = 0.17$

$p_K = \frac{54}{200} = 0.27$

Subtraction of proportions:

In the confidence interval, we subtract the proportions. So:

$p = p_C - p_K = 0.17 - 0.27$

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence level of $1-\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

Standard error:

For a subtraction, as the standard deviation of the distribution is the square root of the sum of the variances, we have that:

$\sqrt{\frac{\pi(1-\pi)}{n}} = \sqrt{\frac{0.17*0.83 + 0.27*0.73}{200}}$

90% confidence level

So $\alpha = 0.1$, z is the value of Z that has a pvalue of $1 - \frac{0.1}{2} = 0.95$, so $Z = 1.645$.

So the confidence interval is:

$(0.17 - 0.27) \pm 1.65\sqrt{\frac{0.17*0.83 + 0.27*0.73}{200}}$, that is, option C