Question

Sample Size. In Exercises 29–36, find the sample size required to estimate the population mean.

Mean Age of Female Statistics Students Data Set 1 “Body Data” in Appendix B includes ages of 147 randomly selected adult females, and those ages have a standard deviation of 17.7 years. Assume that ages of female statistics students have less variation than ages of females in the general population, so let\(\sigma = 17.7\)years for the sample size calculation. How many female statistics student ages must be obtained in order to estimate the mean age of all female statistics students? Assume that we want 95% confidence that the sample mean is within one-half year of the population mean. Does it seem reasonable to assume that ages of female statistics students have less variation than ages of females in the general population?

Short answer

The required sample size is 4815.

Yes, the assumption about the ages of female Statistics students having less variation than the ages of females in the general population seems reasonable.

Step-by-step solution

Given information

Assuming that the variation in the ages of the population of female students is lesser than the general population, let\(\sigma = 17.7\;{\rm{years}}\).

A 95% level of confidence is required that the sample mean lies within one-half year of the true mean.

Describe the determination of the sample size

The sample size n can be determined by using the following formula.

\(n = {\left( {\frac{{{z_{\frac{\alpha }{2}}} \times \sigma }}{E}} \right)^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)\)

Here, E is the margin of error.

Find the critical value \[{z_{\frac{{_\alpha }}{2}}}\]

The\({z_{\frac{\alpha }{2}}}\)is a z-score that separates an area of\(\frac{\alpha }{2}\)in the right tail of the standard normal distribution.

The confidence level of 95% corresponds to\(\alpha = 0.05\,\,\,\,{\rm{and}}\,\,\,\,\frac{\alpha }{2} = 0.025\).

The value\({z_{\frac{\alpha }{2}}}\)has cumulative area\(1 - \frac{\alpha }{2}\)to its left .

Mathematically,

\(\begin{array}{c}P\left( {z < {z_{\frac{\alpha }{2}}}} \right) = 1 - \frac{\alpha }{2}\\ = 0.975\end{array}\)

From the standard normal table, the area of 0.975 is observed corresponding intersection of the row value 1.9 and column value 0.06, which implies\({z_{\frac{\alpha }{2}}}\)is 1.96.

Find the required sample size

The sample size is calculated by substituting the values of\({z_{\frac{\alpha }{2}}}\),\(\sigma \),and E in equation (1).

\(\begin{array}{c}{\rm{n}} = {\left( {\frac{{{z_{\frac{\alpha }{2}}} \times \sigma }}{{\rm{E}}}} \right)^2}\\ = {\left( {\frac{{1.96 \times 17.7}}{{0.5}}} \right)^2}\\ = 4814.14\\ = 4815\,\,\,\,\left( {{\rm{rounded}}\,\,{\rm{off}}} \right)\end{array}\)

So, with 4815 samples values, you can be 95% confident that your sample mean lies within a one-half year of the true mean.

Conclude that the assumption seems reasonable

The sample size required to estimate the Mean Age of Female Statistics Studentsis 4815.

The assumption seems reasonable as the female Statistics students are expected to show less variation than the general population of females.