= 0.05, we fail to reject
H
0
.
We do not have convincing evidence that the distribution of drink
brand preference is different for the professional and support staffs.
(b)
The responses of these 6
nonresponders might change our conclusion.
In the most extreme case
—
if all 6 of them chose
Cocacola
—
the chisquare statistic for the test would change to 7.449, yielding a
P
value of
0.0063, which would provide strong evidence that there is a difference in the distribution of
drink brand preference between the two groups.
Expected Counts for part (a)
Professional
Staff
Support
Staff
Cocacola
10.2
33.8
Pepsi
11.8
39.2
© 2011 BFW Publishers
The Practice of Statistics, 4/e Chapter 11
541
Test 11C
Part I
1.
d
We are looking for a difference in the proportion cars choosing each of the three lanes, so
the alternative hypothesis is that the population distribution of the categorical variable is
not equal for at least one of the three values (the alternative hypothesis includes the
possibility that the two proportions are equal to each other, but different from the third).
2.
c
3
2
1
2
Observed
Expected
Expected
i
3.
d
Choice a is about tests of means, choices b and c make reference to observed counts,
which are not involved in conditions required for chisquare procedures.
4.
c
For a chisquare test of homogeneity, the null hypothesis is that the distribution of a
categorical variable is the same for two or more different populations.
There are three
random samples in this setting from three distinct populations.
5.
d
Degrees of freedom for any chisquare test involving a 2way table is
1
1
number of rows
number of columns
.
6.
e
Expected counts in a twoway table
row total
column total
grand total
.
7.
a
Since the component for midnight/within specifications is a small fraction of the total
chisquare statistic (0.090), the observed and expected counts must be close to equal.
8.
e
Each component of a chisquare statistic is
2
Observed
Expected
Expected
for the given cell.
9.
b
The 2 x 3 table produces a chisquare statistic with 2 degrees of freedom, thus the
P

value is much less that 0.05, providing strong evidence against the null hypothesis that
accident type and age are independent.
10. a
A chisquare test on a 2 x 2 table is equivalent to a twosided twoproportion
z
test, and
the
2
distribution with
df =
1 is the square of the standard normal distribution.
Thus if
2
2
3.07,
3.07
9.42
z
, and the resulting
P
value is the same.
Part II
11.
State: We are testing the hypothesis
H
o
: The proportion of allergy sufferers born during each
season of the year matches the proportion of the general population born during each season,
against
H
a
: The proportion of allergy sufferers born during each season of the year does not
match the proportion of the general population born during each season.
We will use a
significance level of
= 0.05.
Plan:
The procedure is a chisquare goodnessoffit test.
Conditions:
Random
: the data come from a simple random sample of 500 people who are
allergic to dust mites.
Large sample size
: Expected counts are: Winter: 150; Spring: 110;
Summer: 120; Fall: 120
—
all are at least 5.
Independent
: Randomlyselected allergy sufferers
should be independent; and there are surely at least 10 x 500 people who suffer from dust mite
allergies.
Do:
2
2
2
2
2
117
150
105
110
145
120
133
120
14.10
150
110
120
120
,
df
= 3;
P
value = 0.0028.