Question

Sample Size Advances in technology are dramatically affecting different aspects of our lives. For example, the number of daily print newspapers is decreasing because of easy access to Internet and television news. To help address such issues, we want to estimate the percentage of adults in the United States who use a computer at least once each day. Find the sample size needed to estimate that percentage. Assume that we want 95% confidence that the sample percentage is within two percentage points of the true population percentage.

Short answer

The sample size needed to estimate the percentage of people who use the computer at least once a day is 2401.

Step-by-step solution

Given information

The sample size (n) has to be determined to estimate the percentage of people who use the computer at least once a day.

The confidence level is 95% or 0.95.

Determination of the sample size

The following formula is used to estimate the sample size:

\(n = \frac{{{{\left[ {{z_{\frac{\alpha }{2}}}} \right]}^2} \times 0.25}}{{{E^2}}}\)

Here, E denotes the margin of error and\[\alpha \]denotes the significance level.

Also,

\(\begin{array}{c}100\left( {1 - \alpha } \right) = 95\\1 - \alpha = 0.95\\\alpha = 0.05.\\\end{array}\)

From the standard normal table, the value of Z corresponding to\(\frac{\alpha }{2} = 0.025\)is\[ \pm 1.96\].

It is given thatthe sample percentage should be within two percentage points of the true population percentage, that is, \(E = 0.02\).

Substituting the above values, you will obtain the following sample size:

\(\begin{array}{c}n = \frac{{{{\left[ {{z_{\frac{\alpha }{2}}}} \right]}^2} \times 0.25}}{{{E^2}}}\\ = \frac{{{{\left[ {1.96} \right]}^2} \times 0.25}}{{{{0.02}^2}}}\\ = 2401\end{array}\)

Thus, the estimated sample size is 2401.