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A national survey of companies included a question that asked whether the customers like the new flavor of a cola from company A. The sample results of 1000 customers, and 850 of them indicated that they liked the new flavor. The 98% confidence interval on the population proportion of people who like the new flavor is _______________.

User Mastisa
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1 Answer

3 votes

Answer:

The 98% confidence interval on the population proportion of people who like the new flavor is (0.8237, 0.8763).

Explanation:

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{(\pi(1-\pi))/(n)}

In which

z is the zscore that has a pvalue of
1 - (\alpha)/(2).

For this problem, we have that:


n = 1000, \pi = (850)/(1000) = 0.85

98% confidence level

So
\alpha = 0.02, z is the value of Z that has a pvalue of
1 - (0.02)/(2) = 0.99, so
Z = 2.327.

The lower limit of this interval is:


\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.85 - 2.327\sqrt{(0.85*0.15)/(1000)} = 0.8237

The upper limit of this interval is:


\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.85 + 2.327\sqrt{(0.85*0.15)/(1000)} = 0.8763

The 98% confidence interval on the population proportion of people who like the new flavor is (0.8237, 0.8763).

User Pierce
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