Question

Weekly Salaries. In the following table, we repeat the salary data in Data Set II from Example 3.1.

(a) Use Definitions 3.4 and 3.6 on pages 99 and 108, respectively, to obtain the sample mean and sample standard deviation of this (ungrouped) data set.

(b) A frequency distribution for Data Set II, using single-value grouping, is presented in the first two columns of the following table. The third column of the table is for thexf- values, that is, classmark or midpoint (which here is the same as the class) times class frequency. Complete the missing entries in the table and then use the grouped-data formula to obtain the sample mean.

(c) Compare the answers that you obtained for the sample mean in parts (a) and (b). Explain why the grouped-data formula always yields the actual sample mean when the data are grouped by using single-value grouping. (Hint: What does xf represent for each class?)

(d) Construct a table similar to the one in part (b) but with columns for $x,f,x-\overline{x},{\left(x-\overline{x}\right)}^2,and{\left(x-\overline{x}\right)}^{2}f$.Use the table and the grouped-data formula to obtain the sample standard deviation.

(e) Compare your answers for the sample standard deviation in parts (a) and (d). Explain why the grouped-data formula always yields the actual sample standard deviation when the data are grouped by using single-value grouping.

Short answer

Part (a) 474 and 281.2

Part (b) 474

Part (c) Same answers.

Part (d) 281.23

Part (e) Same answers.

Step-by-step solution

Part (a) Step 1. Given information.

The given data is:

Part (a) Step 2. Find the sample mean and sample standard deviation.

$$\begin{gathered} \overline{x}=\frac{300+300+940+.....+300+1050+300}{10} \\ =\frac{4740}{10} \\ =474 \\ \sigma =\sqrt{\frac{{\left(300-474\right)}^2+{\left(300-474\right)}^2+.......+{\left(1050-474\right)}^2+{\left(300-474\right)}^2}{10-1}} \\ \approx 281.2 \end{gathered}$$

Part (b) Step 1. Complete the missing entries in the table and obtain the sample mean.

Salary (x)	Frequency (f)	xf
300	5	1500
400	2	800
450	1	450
940	1	940
1050	1	1050
Total	10	4740

$$\begin{gathered} \overline{x}=\frac{\sum x_i{f}_i}{n} \\ =\frac{4740}{10} \\ =474 \end{gathered}$$

Part (c) Step 1. Compare the answers.

We can see that in the numerical component of part (a), we added each sample ntimes, and in the numerical part of part (b), the sum is multiplied by the number of counts (f) in the given data.

When data is grouped into classes based on a single value, the grouped data formula always returns the actual sample mean.

Part (d) Find sample standard deviation by completing the table.

x	f	$x-\overline{x}$	${\left(x-\overline{x}\right)}^2$	${\left(x-\overline{x}\right)}^{2}f$
300	5	-174	30276	151380
400	2	-74	5476	10952
450	1	-24	576	576
940	1	466	217156	217156
1050	1	576	331776	331776
Total	10			711840

Sample standard deviation:

$$\begin{gathered} s=\sqrt{\frac{\sum {\left(x-\overline{x}\right)}^2{f}_i}{n-1}} \\ =\sqrt{\frac{711840}{9}} \\ =281.23 \end{gathered}$$

Part (e) Step 1. Compare answers in parts (a) and (d). 

Parts (a) and (d) have the same values when comparing the responses we got for the sample standard deviation.

We can see that the numerical portion from part (a) sums squared deviations, whereas the numerical part from part (d) sums squared deviations multiplied by the number of counts (f) in the given data.

When data is grouped into classes based on a single value, the grouped data formula always returns the actual sample standard deviation.