Ch. 11 Review Sheet
The information in this box is used in questions 1, 2, and 3.
All current-carrying wires produce electromagnetic (EM) radiation, including the electrical
wiring running into, through, and out of our homes. High-frequency EM radiation is thought to
be a cause of cancer; the lower frequencies associated with household current are generally
assumed to be harmless. To investigate this, researchers visited the addresses of children in the
Denver area who had died of some form of cancer (leukemia, lymphoma, or some other type)
and classified the wiring configuration outside the building as either a high-current configuration
(HCC) or as a low-current configuration (LCC). Here are some of the results of the study.
HCC
LCC
Leukemia
52
84
Lymphoma
10
21
Other Cancers
17
31
The Minitab output for the above table is given below. The output includes the cell counts, the
expected cell counts, and the chi-square statistic.
Expected counts are printed below observed counts.
C1
C2
C3
1
52
10
17
49.97
11.39
17.64
Total
79
2
84
86.03
21
19.61
31
30.36
136
Total
136
31
48
215
ChiSq = 0.082 + 0.170 + 0.023 + 0.048 + 0.099 + 0.013 = 0.435
1. Using the above data, what are the appropriate degrees of freedom for the chi-square
statistic?
(a) 6
(b) 2
(c) 3
(d) 5
(e) None of the above
2. Using the above data, what is the P-value for the chi-square statistic?
(a) Larger than 0.10
(b) Between 0.05 and 0.10
(c) Between 0.01 and 0.05
(d) Less than 0.01
(e) It is impossible to tell from the information presented.
3. Using the above data, which of the following is the best conclusion?
(a) There is strong evidence of an association between wiring configuration and the chance a
child will develop some form of cancer.
(b) HCC either causes cancer directly or is a major contributing factor to the development of
cancer in children.
(c) There is weak evidence that HCC causes cancer in children.
(d) There is not much evidence of an association between wiring configuration and the type
of cancer children in the study died of.
(e) There is insufficient information provided to reach a conclusion.
4.
a)
b) Find the expected number of females who gave the response “somewhat agree” without
using matrices on your calculator. Show your work.
c) How would the question need to be different so that the chi-square test for homogeneity
of populations would be the appropriate test? What would the null hypothesis be?
5. When performing a
test for association, for what values of
hypothesis if α = 0.01 and there are 6 degrees of freedom?
6. Find the P-value for a test with a
would you reject the null
statistic of 19.8 and 8 degrees of freedom.
7. A random sample of 830 stolen vehicles included 140 white cars, 100 blue cars, 270 red cars,
230 black cars, and 90 cars that were other colors. It is known that 15% of all cars are white,
15% are blue, 35% are red, 30% are black, and 5% are other colors. Test the hypothesis that
the color proportions for stolen cars are identical to the color proportions of all cars. Use a
significance level of α = 0.05.
We are interested in the distribution of
H0:
Ha:
Assumptions: Random -
Independent -
Large Counts -
We will perform a
=
df =
P-value =
Because the P-value is
There
strong evidence that some of the color proportions for stolen cars are
different than the color proportions for all cars.
8. Random samples of 500 male inmates and 500 female inmates are selected, and each inmate
is classified according to type of offense. We would like to know whether male and female
inmates differ with respect to type of offense.
Gender
Type of Crime
Male
Female
Violent
117
66
Property
150
160
Drug
109
168
Public-Order
124
106
a. Is this a test of homogeneity or a test of association/independence? Why?
b. State the hypotheses for this test.
c. Find the expected number of females incarcerated for drug offenses. Show your work.
d. Find the degrees of freedom for the
e. Find the value of the
statistic.
statistic using the
test on your calculator.
Chapter 11 Learning Targets:
__ I can calculate
given the observed and expected counts.
__ I can use the
probability table, the
calculator to find P-values.
__ I can carry out a
cdf command on the calculator, or a
Goodness of Fit Test.
__ I can calculate the expected counts based on a two-way table.
__ I can calculate the degrees of freedom based on a two-way table.
__ I can carry out a
Test for Homogeneity.
__ I can carry out a
Test for Association/Independence.
__ I can identify which
test to use.
Test on the