Question

K
Consider a study conducted in 2018 to estimate the percentage of people from a certain region who do not use the
Internet. Complete parts (a) through (c) below.
a. If a 99% confidence level is used, how many people should be included in the survey if the researchers wanted to
have a margin of error of 7%?
There should be people included in the survey.
(Round up to the nearest person as needed.)
rred

Answer 1

The number of people who must be included in the survey, that is, the sample size should be 340.

We assume the sample size to be n.

The confidence interval required is 99%.

The Z-score corresponding to the 99% confidence interval is (Z) 2.58.

The true proportion (p) of the sample is not given, so we assume it to be 50% or 0.5.

The margin of error for the sample (E) is given to be 7% or 0.07.

By the formula of margin of error, we know that:

E = Z√[{p(1 - p)}/n].

Putting in the values, we have, we get:

0.07 = 2.58√[{0.5*0.5}/n],

or, (0.07/2.58)² = 0.25/n,

or, n = 0.25/{(0.07/2.58)²} = 339.6122449 ≈ 340.

Thus, the number of people who must be included in the survey, that is, the sample size should be 340.

Learn more about the sample size at

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