Question

A research worker wishes to estimate the mean of a population using a sample large enough that the probability will be \(.95\) that the sample mean will not differ from the population mean by more than 25 percent of the standard deviation. How large a sample should he take?

Short answer

The research worker should take a sample of size 62 to ensure there is a \(.95\) probability that the sample mean will not differ from the population mean by more than \(25\%\) of the standard deviation.

Step-by-step solution

Determine the value of Z

For a \(.95\) probability, we will use a \(95\%\) confidence interval. Since the standard normal distribution is symmetrical, we will look for the critical value (\(Z_{\alpha/2}\)) with an area of \(.975\) to the right of it (i.e., the cut-off point of the middle \(95\%\)). Using the standard normal distribution table or calculator functions, we find that: \[Z_{0.975} \approx 1.96\] This is the critical value needed for our calculation.

Set up the equation

Now that we have found the value of Z, we can set up the equation. Since the allowable error is \(25\%\) of the standard deviation, we can write it as: \[E = 0.25\sigma\] Now, we can use the formula we mentioned above: \[n = \left(\frac{Z \cdot \sigma}{E}\right)^2\] In our case: \[n = \left(\frac{1.96 \cdot \sigma}{0.25\sigma}\right)^2\]

Solve for the sample size

Let us solve for \(n\): \[n = \left(\frac{1.96}{0.25}\right)^2\] \[n = (7.84)^2\] \[n = 61.4656\] Since we cannot have a fraction of a sample, we round up to ensure we meet the desired level of accuracy: \[n = 62\] So, the research worker should take a sample of size 62 to ensure there is a \(.95\) probability that the sample mean will not differ from the population mean by more than \(25\%\) of the standard deviation.