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

Mean Pulse Rate of Males Data Set 1 “Body Data” in Appendix B includes pulse rates of 153 randomly selected adult males, and those pulse rates vary from a low of 40 bpm to a high of 104 bpm. Find the minimum sample size required to estimate the mean pulse rate of adult males. Assume that we want 99% confidence that the sample mean is within 2 bpm of the population mean.

a. Find the sample size using the range rule of thumb to estimate $\sigma$.

b. Assume that $\sigma =11.3 bpm$, based on the value of $s=12.5 bpm$for the sample of 153 male pulse rates.

c. Compare the results from parts (a) and (b). Which result is likely to be better?

The pulse rates vary from a low 40 bpm to a high 104 bpmfor adult males.

The required confidence level is 99%, and the sample mean is within 2 bpm of the true mean.

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.

The range rule of thumb is a simple tool for understanding and interpreting the standard deviation.It is used to estimate the standard deviation roughly from the collection of sample data.

The formula for the range rule of thumb is as follows.

$\sigma \approx \frac{\text{range}}{4} ...\left(2\right)$

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

The confidence level 99% corresponds to $\alpha =0.01 \text{and} \frac{\alpha }{2}=0.005$.

The value$z_{\frac{\alpha }{2}}$, has the cumulative are$1-\frac{\alpha }{2}$to its left.

Mathematically,

$$\begin{gathered} P\left(z<z_{\frac{\alpha }{2}}\right)=1-\frac{\alpha }{2} \\ =0.995 \end{gathered}$$

From the standard normal table, the area of 0.995 is observed corresponding to the row value 2.5, between the column value 0.07, and the column value 0.08, which implies that localid="1648020139074" $z_{\frac{\alpha }{2}}$is 2.575.

a.

Themean pulse rate of males varies from a low 40 bpm to a high 104 bpm.

Therefore, the range of pulse rate is as follows.

$$\begin{gathered} \text{Range} =104-40 \\ =64 \end{gathered}$$

The estimate of $\sigma$is obtained by substituting the value of range in equation (2). So,

$$\begin{gathered} \sigma \approx \frac{\text{range}}{4} \\ =\frac{64}{4} \\ =16 \end{gathered}$$

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

$$\begin{gathered} \text{n}={\left[\frac{z_{\frac{\alpha }{2}}\times \sigma }{\text{E}}\right]}^2 \\ ={\left[\frac{2.575\times 16}{2}\right]}^2 \\ =424.36 \\ =425 \left(\text{rounded} \text{off}\right) \end{gathered}$$

Thus, with 425 samples values, we can be 99% confident that the sample mean is within 2 bpm of the true mean.

b.

Assume that $\sigma =11.3 \text{bpm}$is based on the value of $s=11.3 \text{bpm}$for the sample of 147 female pulse rates.

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

$$\begin{gathered} \text{n}={\left[\frac{z_{\frac{\alpha }{2}}\times \sigma }{\text{E}}\right]}^2 \\ ={\left[\frac{2.575\times 11.3}{2}\right]}^2 \\ =211.66 \\ =212 \left(\text{rounded} \text{off}\right) \end{gathered}$$

Thus, with 212 samples values, we can be99% confident that the sample mean is within 2 bpm of the true mean.

c.

The sample size required to estimate the mean pulse rate of males by using the range rule estimate of is 425.

The sample size required to estimate the mean pulse rate of males by using instead of is 212.

The result obtained in part (a) is larger than the result obtained in part (b).

The result from part (b) is better than results from part (a) because it uses instead of the estimated obtained from the range rule of thumb.