Chapter 14 Homework
13. (p.575)
(1) Use the least squares method to develop the estimated regression equation with
adjusted gross income as the independent variable.
(2) Estimate a reasonable level of total itemized deductions for a taxpayer with an
adjusted gross income of $52,500. If this taxpayer has claimed total itemized
deductions of $20,400, would the IRS agent’s request for an audit appear justified?
Explain.
(3) Provide an interpretation for the slope of the estimated regression equation.
(4) Compute the coefficient of determination.
(5) Interpret coefficient of determination.
(6) Compute the standard error of estimate.
(7) Compute the correlation coefficient.
(8) Is there a significant relationship between the total itemized deductions and the
adjusted gross income? (a=0.05)
(9) Set up a 95% confidence interval estimate of the mean total itemized deduction for all
taxpayers whose adjusted gross income is $70,000.
(10) Set up a 95% prediction interval estimate of the total itemized deduction for an
individual taxpayer whose adjusted gross income is $70,000.
Answer:
Prepare sample data:
Adjusted Gross
Total Itemized
Income ($1000s) Deductions ($1000s)
22
9.6
27
9.6
32
10.1
48
11.1
65
13.5
85
17.7
120
25.5
∑
X 
(1)
399
399
 57
7
b1 
Y 
97.1
X2
484
729
1024
2304
4225
7225
14400
Y2
92.16
92.16
102.01
123.21
182.25
313.29
650.25
XY
211.2
259.2
323.2
532.8
877.5
1504.5
3060
30391
1555.33
6768.4
97.1
 13.8714 3
7
n xi yi  ( xi )( yi )
n x  ( xi )
2
i
2

(7)(6768.4)  (399)(97.1)
 0.16131
(7)(30391)  (399) 2
b0  y  b1 x  13.87143  (0.16131)(57)  4.67675
Yˆ  4.67675  0.16131X
(2) Given X = 52.5 ($52,500), Yˆ  4.67675  (0.16131)(52.5)  13.14553 ($13,145.53).
If this taxpayer has claimed total itemized deductions of $20,400, the agent’s request for
an audit appears to be justified.
(3) For each increase of $1,000 adjusted gross income (one unit of X), the expected total
itemized deduction (Y) is estimated to increase by an average of $161.31 (0.16131 units
of Y).
[ xi y i  ( xi  y i ) / n] 2 [6768.4  (399)(97.1) / 7] 2
SSR


 199.0083
(4)
2
2
2
x

(
x
)
/
n
30391

(
399
)
/
7
 i  i
SST   y 
2
i
r2 
( y i ) 2
n
 1555.33 
(97.1) 2
 208.4143
7
SSR 199.0083

 0.9549
SST 208.4143
(5) The 95.49% of variation in total itemized deductions can be explained by this model or
by the adjusted gross income.
SSE

n2
SST  SSR
208.4143  199.0083

 1.3716
n2
72
(6)
s
(7)
r  (Sign of b1 ) r 2  0.9549  0.9772
(8) 1. H0: β1 = 0
Ha: β1 ≠ 0
2. α = 0.05
3. Test Statistic Value:
SSR
199.0083
F
(n  2) 
(7  2)  105.7880
SSE
208.4143  199.0083
4. p-Value: df1 = 1, df2 = n-2 = 5, F = 105.7880, from the F table: p-value < 0.01.
5. Reject H0
6. There is a significant relationship between the total itemized deductions and the
adjusted gross income.
*: Alternatively, you can use the t-test to test the same hypotheses (see Clemson Apartment
Example online).
(9) Confidence Interval for E(yp):

1. y p  4.67675  (0.16131)(70)  15.96845
($15,968.45)
2. tα/2: df = n-2 = 5, the upper tail = α/2 = 0.025. From the t table, t α/2 = 2.571
3. Margin of error:
s = 1.3716 (from question (6)),
hp 
1

n
(x p  x)2
 xi2 
( xi ) 2

1

7
n
(70  57) 2
 0.16495
(399) 2
30391 
7
MOE  t / 2 S yˆ p  t / 2 S h p  (2.571)(1.3716) 0.16495  1.43221
4. Confidence interval:
UCL = 15.96845 + 1.43221 = 17.40066
LCL = 15.96845 – 1.43221 = 14.53624
The 95% confidence interval for the mean total itemized deduction of all taxpayers
who have $70,000 of adjusted gross income (E( y p ) ) is [$14,536.24, $17,400.66].
(10) Prediction Interval for yp:

1. y p  15.96845
($15,968.45)
2. tn-2 = 2.571
3. Margin of error:
s = 1.3716 (from question (6)),
h p  0.16495
MOE  t / 2 S ind  t / 2 S 1  h p  (2.571)(1.3716) 1  0.16495  3.8061
4. Prediction interval:
UCL = 15.96845 + 3.8061 = 19.77455
LCL = 15.96845 – 3.8061 = 12.16235
The 95% prediction interval for the total itemized deduction of an individual taxpayer
who has $70,000 of adjusted gross income ( y p ) is [$12,162.35, $19,774.55].