Q 8.84. - Elementary Statistics

Question

Body Fat. J. McWhorter et al. of the College of Health Sciences at the University of Nevada, Las Vegas, studied physical therapy students during their graduate-school years. The researchers were interested in the fact that, although graduate physical therapy students are taught the principles of fitness, some have difficulty finding the time to implement those principles. In the study, published as "An Evaluation of Physical Fitness Parameters for Graduate Students" (Journal of American College Health, Vol. $51$, No.$1$ pp. $32-37$), a sample of $27$ female graduate physical therapy students had a mean of $22.46$ percent body fat.

a. Assuming that the percent body fat of female graduate physical therapy students is normally distributed with standard deviation $4. 10$ percent body fat, determine a $95\%$ confidence interval for the mean percent body fat of all female graduate physical-therapy students.

b. Obtain the margin of error, $E$ for the confidence interval you found in part (a).

c. Explain the meaning of $E$ in this context in terms of the accuracy of the estimate.

d. Determine the sample size required to have a margin of error of $1.55$ percent body fat with a $99\%$ confidence level.

Step-by-Step Solution

Answer

Part (a) $(20.91, 24.01)$

Part (b) $1.55$

Part (c) Approximately $95\%$ of the sample means (i.e. $95\%$ of the drawn samples) are projected to differ from $\mu$ by nearly $1.55$ units.

Part (d) Sample size is 47

Part (a) Step 1: Given information

The average body fat percentage of $27$ female graduate physical therapy students was $22.46$

Part (a) Step 2: Concept

The formula used: $E=z_{\frac{a}{2}}\times \frac{\sigma }{\sqrt{n}}$

Part (a) Step 3: Calculation

Let $\mu$ be the population s.d. and $\sigma =4.10$ be the population mean % body fat.

We have to determine $95\%$ confidence interval of $\mu$

$$\begin{gathered} \therefore \text{ Confidence level }=0.95\times 100\% \\ \therefore 1-\alpha =0.95 \\ \alpha =0.05 \\ \therefore z_{\frac{a}{2}}=z_{\frac{0.05}{2}}=z_{0.025}=1.96 \end{gathered}$$

$95\%$ Confidence interval of $\mu$ is given by

$\left(\overline{x}-z_{\frac{a}{2}}\times \frac{\sigma }{\sqrt{n}},\overline{x}+z_{\frac{a}{2}}\times \frac{\sigma }{\sqrt{n}}\right)=(\overline{x}-E,\overline{x}+E)$

Given that sample mean $\overline{x}=22.46$

Sample size $n=27$

$$\begin{gathered} \therefore E=z_{\frac{\alpha }{2}}\times \frac{\sigma }{\sqrt{n}} \\ =1.96\times \frac{4.10}{\sqrt{27}} \\ =1.55 \end{gathered}$$

$95\%$ Confidence interval of $\mu$

$$\begin{gathered} =(\overline{x}-E,\overline{x}+E) \\ =(22.46-1.55,22.46+1.55) \\ =(20.91,24.01) \end{gathered}$$

i.e. we have a $95\%$ confidence level that all-female graduate physical therapy students have a mean $\%$ body fat of between $20.91$ and $24.01$

Part (b) Step 1: Calculation

Margin of error$E=z_{\frac{\alpha }{2}}\times \frac{\sigma }{\sqrt{n}}$

$=1.96\times \frac{4.10}{\sqrt{27}}$

Part (c) Step 1: Calculation

For a $95\%$confidence interval. We are $95\%$ certain that our mistake in predicting the population mean $\mu$ by sample mean $\overline{x}$ is no more than $$1.55\%$

In other words, if we took several simple random samples of size $n=27$ from a population with a mean of $\mu$ we may assert that. Approximately $95\%$ of the sample means (i.e. $95\%$ of the drawn samples) are projected to differ from $\mu$ by nearly $\overline{x}$ units.

Part (d) Step 1: Calculation

We know that the formula for calculating the required sample size for a specific margin of error with a confidence level of $100(1-\alpha )\%$ is given by

$n={\left\{\frac{z_{\alpha /2}\times \sigma }{E}\right\}}^2$

Here confidence level $=99\%=100\times 0.99\%$

$$\begin{gathered} \therefore 1-\alpha =0.99 \\ \Rightarrow \alpha =0.01 \\ \Rightarrow \frac{\alpha }{2}=0.05 \\ {z}_{\frac{\alpha }{2}}=z_{0.005}=2.57583 \end{gathered}$$

Required $E=1.55$

Population s.d., $\sigma =4.10$

Thus $n={\left\{\frac{z_{0\cos }\times \sigma }{E}\right\}}^2$

$$\begin{gathered} ={\left\{\frac{2.57583\times 4.10}{1.55}\right\}}^2 \\ =46.42 \end{gathered}$$

[ 47

$\therefore$ required sample size is $47$