Question

3. Use the data table below to answer the following questions: (a) The least absolute deviation line for the data in the table is y - 1.5x +17. What is the sum of the absolute deviations? (b) Find the median-median regression line for the data in the table. (c) Using the equations for the least absolute deviation line and the median-median regression line determine which model does a better job of predicting the y-value for x 11.* 21 20 38 7.5 30 25 10.5 32 16 41

Answer 1

a)

The least absolute deviation line is,

y = 1.5x+ 17

Now, the sum of absolute deviation is obtained using the formula,

sum of absolute deviation-Xly-y

Where, $\widehat{y}$ is the predicted value of y using the least absolute deviation line

x	y	y-hat	y-y-hat
2	21	1.5*2+17=20	1
3	20	21.5	1.5
11	38	33.5	4.5
7.5	30	28.25	1.75
5	25	24.5	0.5
10.5	32	32.75	0.75
16	41	41	0
		SUM	10

sum of absolute deviation-Σ y-yi-10

b)

The median median line is obtained in following steps,

Step 1:Sort both x and y with respect to x data points from smallest to largest.

x	y
2	21
3	20
5	25
7.5	30
10.5	32
11	38
16	41

Step 2: Divide the data into three groups (since data points are odd put one data point in middle group)

x	y
2	21
3	20
5	25
7.5	30
10.5	32
11	38
16	41

Step 3: Calculate median for x and y for each group,

x	y	Median x	Median y
2	21	2.5	20.5
3	20
5	25	7.5	30
7.5	30
10.5	32
11	38	13.5	39.5
16	41

Step 4: The slope of the median median line is obtained by taking the gradient of upper and lower group median values,

f- y 20.5-39.51.727 = 1.727 r-2.5-13.5

Step 5: The median median equation is obtained from middle group median point,

y-30-1.7272(x-7.5)

y = 1.72722、+17.04545

c)

Actual y value for x = 11 is,

$y=38$

For Least absolute deviation line,

y = 1.5x + 17 × 11 = 33.5

For median median line

y = 1.7272 × 11 + 17.04545 = 36.04545

Since the predicted value of y for median median line is closer compare to Least absolute deviation line hence median median line done a better job.