Chapter 11, Problem 1P

Hematology

The data in Table 11.17 are given for 9 patients with aplastic anemia [11].

Fit a regression line relating the percentage of reticulocytes (x) to the number of lymphocytes (y).

TABLE 11.17 Hematologic data for patients with aplastic anemia

To determine

Compute the regression line relating percentage of reticulocytes and number of lymphocytes.

Answer to Problem 1P

The regression line is$y=1894.8+112.1x$.

Explanation of Solution

Calculation:

The sum of squares and products can be calculated as follows:

$x$	$y$	$x^2$	$y^2$	$xy$
3.6	1700	12.96	2890000	6120
2.0	3078	4.00	9474084	6156
0.3	1820	0.09	3312400	546
0.3	2706	0.09	7322436	811.8
0.2	2086	0.04	4351396	417.2
3.0	2299	9.00	5285401	6897
0.0	676	0.00	456976	0
1.0	2088	1.00	4359744	2088
2.2	2013	4.84	4052169	4428.6
${\displaystyle \sum x=12.6}$	${\displaystyle \sum y}=18,466$	${\displaystyle \sum x^2}=32.02$	${\displaystyle \sum y^2}=41,504,606$	${\displaystyle \sum xy}=27,464.6$

$\begin{array}{c}{L}_{xx}={\displaystyle \sum x^2}-\frac{{\left({\displaystyle \sum x}\right)}^2}{n}\\ =32.02-\frac{{12.6}^2}{9}\\ =14.38\end{array}$

$\begin{array}{c}{L}_{yy}={\displaystyle \sum y^2}-\frac{{\left({\displaystyle \sum y}\right)}^2}{n}\\ =41,504,606-\frac{18,{466}^2}{9}\\ =3,616,478\end{array}$

$\begin{array}{c}{L}_{xy}={\displaystyle \sum xy}-\frac{\left({\displaystyle \sum x}\right)\left({\displaystyle \sum y}\right)}{n}\\ =27,464.6-\frac{\left(12.6\times 18,466\right)}{9}\\ =1,612.2\end{array}$

The slope can be calculated as follows:

$\begin{array}{c}b=\frac{L_{xy}}{L_{xx}}\\ =\frac{1612.2}{14.38}\\ =112.1\end{array}$

The intercept can be calculated as follows:

$\begin{array}{c}a=\frac{{\displaystyle \sum y}-b{\displaystyle \sum x}}{n}\\ =\frac{18,466-\left(112.1\times 12.6\right)}{9}\\ =1894.8\end{array}$

The regression line is given as follows:

$\begin{array}{c}y=a+bx\\ =1894.8+112.1x\end{array}$

Thus, the regression line is$y=1894.8+112.1x$.