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Natasha_Volkova [10]
3 years ago
7

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

tistics and confidence level. ​95% confidence; nequals=​2388, xequals=1672
Mathematics
1 answer:
seropon [69]3 years ago
6 0

Answer:

The margin of error is of 0.0184 = 1.84%.

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}}

95% confidence level

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

In this question:

\pi = \frac{1672}{2388} = 0.7, n = 2388

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

M = 1.96\sqrt{\frac{0.7*0.3}{2388}}

M = 0.0184

The margin of error is of 0.0184 = 1.84%.

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