Solutions Practice Problems: Relationships between Two Variables 1. A carcinogenicity study was conducted to examine the tumor potential of a drug product
scheduled for an initial testing in humans. A total of 300 rats (150 males and 150 females) were
studied for a six-month period. At the beginning of the study, 100 rats (50 males and 50 females)
were randomly assigned to the control group, 100 to the low dose group, and the remaining 100
(50 males and 50 females) to the high-dose group. On each day of the six-month period, the rats
in the control group received an injection of the solution plus drug. The sample data are shown in
the accounting table (drug_test.txt):
Number of Tumors
Rat group
One or more
None
Control
10
90
Low-dose
14
86
High-dose
19
81
a. Give the percentage of rats with one or more tumors for each of the three treatment groups.
Control: 10/100 = 0.1 or 10%
Low-Dose: 14/100 = 0.14 or 14%
High-Dose: 19/100 = 0.19 of 19%
b. Calculate, by hand, the expected counts table.
Number of Tumors
Rat group
One or more
None
Total
Control
(100)(43)/300 = 14.3
(100)(257)/300 = 85.7
100
Low-dose
(100)(43)/300 = 14.3
(100)(257)/300 = 85.7
100
High-dose
(100)(43)/300 = 14.3
(100)(257)/300 = 85.7
100
43
257
Total
300
NOTE: This example use the common row totals of 100 to better represent what we
discussed in the lesson how the expected table would show the same cell proportions across
each row.
c. Based on a visual comparison of the observed and expected tables, does there appear to be a
drug-related problem regarding tumors for this drug product? That is, as the dose is increased,
does there appear to be an increase in the proportion of rats with tumors?
1 In comparing the two tables, the expected counts for the low-dose treatment are the same
(within rounding), while the control group experiences slightly fewer cases of tumors than
expected and the high-dose group slightly more than expected.
d. Using software, conduct a test of whether there is a significant difference in the proportion of
rats having one or more tumors for the three treatment groups with α of 0.05
Ho: There is no relationship between the treatments and development of tumors
Ha: There is a relationship between the treatments and development of tumors.
From the output we have a Chi-square test statistics of 3.312 with 2 degrees of freedom.
This results in a p-value of 0.191 which is above our state 0.05 level of significance. We fail
to reject the null hypothesis. We do not have statistical evidence to conclude that there is a
relationship between the three treatment groups and potential for tumors among the
population of rats.
e. What is the relative risk of the high-dose group experiencing one or more tumors compared to
the control group? Provide an interpretation of this relative risk.
2 The relative risk is asking to compare the risk of the high-dose group to the control group.
This makes the control group the baseline risk since it is the group being “compared to”.
Therefore,
RR = Tumor Risk High-Dose / Tumor Risk Control = 0.19/0.10 = 1.9
Rats receiving the high-dose treatment are about twice as likely to develop one or more
tumors as the rats in the control group.
2. You are interested in determining the relationship between age and sales price of used cars. In
checking local advertisements of used cars in your area you find six with the following age and
price (used cars.txt):
Age Price
6 5500
5 5100
7 5000
8 4500
6 5800
10 3800
a. Prior to calculating the correlation, if a linear relationship exists between Age and Price of
used cars, what direction would you expect the direction to take?
Given that price of cars tend to decrease as a car’s age increases, we would expect a
negative correlation.
b. Calculate by hand the correlation and the test statistic for the correlation test.
The mean price is 4950 and the mean age is 7
Price (Y)
5500
5100
5000
4500
5800
3800
𝑦−𝑦
5500-4950 = 550
5100-4950 = 150
5000-4950 = 50
4500-4950 = -450
5800-4950 = 850
3800-4950 = -1150
Age (X)
6
5
7
8
6
10
𝑥−𝑥
6-7 = -1
5-7 = -2
7-7 = 0
8-7 = 1
6-7 = -1
10-7 = 3
𝑦−𝑦 ∗ 𝑥−𝑥
-550
-300
0
-450
-850
-3450
𝑦 − 𝑦 ∗ 𝑥 − 𝑥 = − 5600
𝑦−𝑦
!
= 2575000
3 𝑥−𝑥
𝑟=
!
= 16
−5600
16 2575000
0.87245
c. Use Minitab to verify you by-hand calculations and to conduct a test of significance for the
correlation at a 5% level of significance. Include your hypotheses statements, p-value, degrees
of freedom, and conclusion.
Ho: ρ = 0 versus Ha: ρ < 0 DF = n -­‐2 = 6 – 2 = 4 From Minitab, the correlation value is 0.872 with a 0.023 p-­‐value. With this p-­‐value being for the two-­‐
sided test of “not equal”, the p-­‐value for our one-­‐sided alternative would be about 0.0115 (half the 0.023). With this p-­‐value being less than our 0.05 level of significance, we would reject the null hypothesis. We conclude the there is a significant negative linear relationship between age and price of cars. 4