Question

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

Mean Weight of Male Statistics Students Data Set 1 “Body Data” in Appendix B includes weights of 153 randomly selected adult males, and those weights have a standard deviation of 17.65 kg. Because it is reasonable to assume that weights of male statistics students have less variation than weights of the population of adult males, let\(\sigma = 17.65\,kg\). How many male statistics students must be weighed in order to estimate the mean weight of all male statistics students? Assume that we want 90% confidence that the sample mean is within 1.5 kg of the population mean. Does it seem reasonable to assume that weights of male statistics students have less variation than weights of the population of adult males?

Short answer

The required sample size is 375.

Yes, the assumption that t the weights of male Statistics students has less variation compared to the weights of the population of adult males seems reasonable.

Step-by-step solution

Given information

Theweights of 153 randomly selected adult males, has a standard deviation of 17.65 kg. Assume thattheweights of the male Statistics students have less variation than the weights of the population of adult males.

It is required to be 90% confident that sample mean is within 1.5 kg of the true mean..

Describe the determination of sample size

The sample size n needs to be determined to estimate the value of the population mean\(\mu \).

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 90% corresponds to\(\alpha = 0.10\,\,\,\,{\rm{and}}\,\,\,\,\frac{\alpha }{2} = 0.05\).

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

Mathematically,

\(\begin{array}{c}P\left( {z < {z_{\frac{\alpha }{2}}}} \right) = 1 - \frac{\alpha }{2}\\ = 0.95\end{array}\)

From the standard normal table, the area of 0.95 is observed corresponding to the row value 1.6 and between column value 0.04 and column value 0.05, which implies\({z_{\frac{\alpha }{2}}}\)is 1.645.

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).

\(\begin{array}{c}{\rm{n}} = {\left( {\frac{{{z_{\frac{\alpha }{2}}} \times \sigma }}{{\rm{E}}}} \right)^2}\\ = {\left( {\frac{{1.645 \times 17.65}}{{1.5}}} \right)^2}\\ = 374.66\\ = 375\,\,\,\,\left( {{\rm{rounded}}\,\,{\rm{off}}} \right)\end{array}\)

Thus, with 375 samples values, you can be 90% confident that your sample mean lies within 1.5kg of the true mean.

Conclude that the assumption seems reasonable

The sample size required to estimate the Mean Weight of Male Statistics Students is 374.

Yes,itdoes seem reasonable to assume that the weights of male Statistics students have less variation than the weights of the population of adult males as adult males tend show more dispersed weight measures as compared to the students.