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Katen [24]
3 years ago
10

Assume that a sample is used to estimate a population proportion p. Find the margin of error M.E. that corresponds to a sample o

f size 329 with 49 successes at a confidence level of 99.5%.
Mathematics
1 answer:
Naddika [18.5K]3 years ago
3 0

Answer:

The magin of error is 0.0552 = 5.52 percentage points.

Step-by-step 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{\frac{\pi(1-\pi)}{n}}

In which

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

The margin of error is:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

For this problem, we have that:

n = 329, \pi = \frac{49}{329} = 0.1489

99.5% confidence level

So \alpha = 0.005, z is the value of Z that has a pvalue of 1 - \frac{0.005}{2} = 0.9975, so z = 2.81.

M.E:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

M = 2.81\sqrt{\frac{0.1489*(0.8511)}{329}} = 0.0552

The magin of error is 0.0552 = 5.52 percentage points.

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