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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

Sagot :

Answer:

[tex](0.17 - 0.27) \pm 1.65\sqrt{\frac{0.17*0.83 + 0.27*0.73}{200}}[/tex], 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:

[tex]p_C = \frac{34}{200} = 0.17[/tex]

[tex]p_K = \frac{54}{200} = 0.27[/tex]

Subtraction of proportions:

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

[tex]p = p_C - p_K = 0.17 - 0.27[/tex]

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

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].

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:

[tex]\sqrt{\frac{\pi(1-\pi)}{n}} = \sqrt{\frac{0.17*0.83 + 0.27*0.73}{200}}[/tex]

90% confidence level

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

So the confidence interval is:

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

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