Regression Problems:
1. A researcher wants to know if there is a relationship between the number of shopping
centers in a state and the retail sales (in billions $) of that state. A random sample of 8
states is listed below. After determining, via a scatter-plot, that the data followed a
linear pattern, the regression line was found. Using the given data and the given
regression output answer the following questions.
State
1
2
3
4
5
6
7
8
a.
b.
c.
d.
e.
f.
Num
630
370
616
700
430
568
1200
2976
Sales
15.5
7.5
13.9
18.7
8.2
13.2
23.0
87.3
Output
a=
b=
r=
-4.930
0.030
0.991
What is the equation of the regression line?
Interpret the slope in the words of the problem.
Find r2 and interpret its meaning in the words of the problem.
Is there a significant linear relationship between Num and Sales.
Find the error for predicting the sales of a state with 1200 stores.
Use the regression line to predict the sales for a state with 100 stores
a. y-hat = -4.930 + 0.030x
b. Slope = 0.030 means that for every increase of 1 shopping center retail sales increases
0.030 billion dollars, on average.
c. r2 = .982 means that there is a 98.2% reduction in error in predicting retail sales by
using number of shopping centers over the sample mean.
d. The test statistic was 18.30 with a p-value 1.7 * 10-6 so there is sufficient evidence that
the true slope is different from 0, meaning there is a significant linear relationship
between num and sales.
Or, the 95% confidence interval for the true slope was 0.026 and 0.034, which does not
contain 0, so there is sufficient evidence that the true slope is different from 0, meaning
there is a significant linear relationship between num and sales.
e. x = 1200, y = 23.0 y-hat = 31.07 so error = 23.0 – 31.07 = -8.07
f. This is an example of extrapolation so you should not do it.
2. A pharmaceutical company is investigating the relationship between advertising
expenditures and the sales of some over-the-counter (OTC) drugs. The following data
represents a sample of 10 common OTC drugs. Find the equation of the regression line,
using Advertising dollars as the independent variable and Sales as the response variable.
Interpret the slope of the line in the words of the problem. Find r 2 and interpret it in the
words of the problem. Use the line to predict the Sales if Advertising dollars = $50
million. Note that AD = Advertising dollars in millions and S = Sales in millions $.
AD
22
25
29
35
38
42
46
52
65
88
S
64
74
82
90
100
120
120
142
180
230
Calculator Output
a = 6.629,
b = 2.569,
r = .996
y-hat = 6.629 + 2.569x
The slope = 2.569 means that for every increase of $1 million Advertising, sales
increases $2.569 million on average.
r2 = .993 so there is a 99.3% reduction in error for predicting Sales using advertising
dollars.
y-hat = 6.629 + 2.596(50) = 135.079
3. A chemical company wants to study the effect of extraction time on the efficiency of an
extraction process. They obtained a random sample of extraction times and the
corresponding efficiency scores. The output from Excel is given below. What is the
regression line? Interpret the slope and R2 in the words of the problem. Use the
regression line to estimate the efficiency for an extraction time of 20. You can assume
20 is in the range of the x’s.
Regression Statistics
Multiple R
0.864
R Square
0.746
Std Error
5.139
Obs
15
Coefficients
Intercept
39.022
Time
0.764
Std Error
4.173079
0.123639
t Stat
9.350943
6.178365
P-value
3.9E-07
3.33E-05
Lower 95%
30.00684
0.496782
Upper 95%
48.03761
1.030995
y-hat = 39.022 + .764x
The slope = .764 means that for every increase of 1 unit in extraction time efficiency score
increases .764 units on average.
r2 = .746 means that there is a 74.6% reduction is error for predicting efficiency using extraction
time.
y-hat = 39.022 + .764 (20) = 54.302
The model is useful because the true slope is significantly different from 0, because the 95% CI
for the true slops is 0.497 to 1.031 and the is r2 = .746 reasonably high.
4. The following is output from Excel for regression analysis. The researcher wanted to
predict the total cholesterol (mg/100ml) using weight (kg) as the predictor variable.
Using the output, please answer the following questions?
a. Use ŷ to predict the total cholesterol for a subject who weighs 70kg.
b. Find the coefficient of determination and explain what this means in the words
of the problem?
c. Find and interpret 95% Confidence interval for B.
d. Do you think weight is a good predictor total cholesterol, Explain?
SUMMARY OUTPUT
Regression
Multiple R
R Square
Standard Error
Observations
Intercept
Weight
Statistics
0.265293
0.070381
76.65431
25
Coeff Std Err t Stat
199.30 85.82 2.322
1.62 1.229 1.320
ANOVA
Source
Regress
Residual
Total
df
1
23
24
P-value
0.0294
0.1999
Lower 95%
21.77
-0.921
SS
MS
F
10231 10231 1.741
135145 5875.8
145377
Upper 95%
376.825
4.1656
y-hat = 199.30 + 1.62x  for x = 70 y-hat = 312.7
r2 = .070 means that there is a 7% reduction in error for predicting total cholesterol using
weight.
95% CI for B (-.921, 4.1656), means that we 95% sure that the true mean slope is between 0.921 and 4.166.
This is not a good model because r2 is low and B could be 0, because 0 is in the 95% CI for B.