Topic 7 Exercises
Kristin Barner
Chapter 19
19.9 Randomly selected records of 140 convicted criminals reveal that their crimes were
committed on the following days of the week:
(a) Using the .01 level of significanc
...
Topic 7 Exercises
Kristin Barner
Chapter 19
19.9 Randomly selected records of 140 convicted criminals reveal that their crimes were
committed on the following days of the week:
(a) Using the .01 level of significance, test the null hypothesis that in the underlying population,
crimes are equally likely to be committed on any day of the week.
H1: H0 is false
H0= Psun = Pmon = Ptues = Pwed = Pthurs = Pfri = Psat = 1/7
We reject H0 at the 0.01 level of significance = X
2
> 16.81
Fe = (expected proportion)(total sample size)
frequenc
y
Mon Tues Wed Thurs Fri Sat Sun Total
f0 17 21 22 18 23 24 15 140
fe 20 20 20 20 20 20 20 140
Null Hypothesis:
X
2 = ∑
f 0−fe
(¿¿ fe)2
¿
= { (17−20) 2
20 +
(21−20) 2
20 +
(22−20) 2
20
+
(18−20)2
20
+
(23−20) 2
20 +
(24−20)2
20
+
(15−20) 2
20 }
=
9
20 +
1
20 +
4
20 +
4
20 +
9
20 +
16
20 +
25
20
=
68
20 = 3.4
The null hypothesis is rejected at the 0.01 significance level because the observed X2 at 3.4 is
less than the critical X2 at 16.81
It can be said that crimes are likely to be committed on any day of the week, not one more than
another.
(b) Specify the approximate p-value for this test result.
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The table indicates the value of 12.6 at the .05 level and degrees of freedom of 6. This means
that the null hypothesis can be retained at the .05 level. p > .05.
(c) How might this result be reported in the literature?
In the literature, it may be reported that crimes are likely to occur on any day of the week.
There is not one day of the week that is more likely to have crimes committed.
19.10 A number of investigators have reported a tendency for more people to die (from natural
causes, such as cancer and strokes) after, rather than before, a major holiday. This post-holiday
death peak has been attributed to a number of factors, including the willful postponement of
death until after the holiday, as well as holiday stress and post-holiday depression. Writing in the
Journal of the American Medical Association (April 11, 1990), Phillips and Smith report that
among a total of 103 elderly California women of Chinese descent who died of natural causes
within one week of the Harvest Moon Festival, only 33 died the week before, while 70 died the
week after.
(a) Using the .05 level of significance, test the null hypothesis that, in the underlying population,
people are equally likely to die either the week before or the week after this holiday.
Within a week before: 33
Week after: 70
N = 103
P1 = P2 = ½
E (X1)2 E (X2) = 103/2 = 51.5
E
2 = (33 – 51.5)+(70 -51.5)2
= 13.291
X = .05 critical = 3.81
TS > critical
Because of the data, we can reject the null hypothesis. We can then conclude that there is
significance data to show that more people die after the holiday than before the holiday.
(b) Specify the approximate p-value for this test result.
p = .05
p > .05
19.13 Students are classified according to religious preference (Buddhist, Jewish, Protestant,
Roman Catholic, or Other) and political affiliation (Democrat, Republican, Independent, or
Other).
(a) Is anything suspicious about these observed frequencies?
Yes, there is something suspicious about these observed frequencies. In this problem, all of the
frequencies end in multiples of 10. This makes me think that the observed frequencies may not
be real and may be being used to simplify the data and calculations.
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(b) Using the .05 level of significance, test the null hypothesis that these two variables are
independent.
Statistical hypothesis:
H0: Political affiliation and religious preferences are independent factors
H1: H0 is not true
X
2 = {
(30−20) 2
20 +
(30−20) 2
20 +
(30−20) 2
20
+
(18−20)2
20
+
(100−40)2
40
}
= { (30−20) 2
20 +
(30−20) 2
20 +
(40−20)2
20
+
(30−20) 2
20
+
(20−20)2
20 +
(20−20) 2
20
+
(10−20) 2
20 }
= 220
X
2 = 220 is larger than 21.03. This means that there is a relationship between the religious
preferences of students and their political affiliation.
(c) If appropriate, estimate the effect size.
Ø
2
c =
220
500 (4−1)
= 0.15
Effect size: Medium
19.14 In 1912 over 800 passengers perished after the ocean liner Titanic collided with an
iceberg and sank. The following table compares the survival frequencies of cabin and steerage
passengers.
(a) Using the .05 level of significance, test the null hypothesis that survival rates are
independent of the passengers’ accommodations (cabin or steerage).
Statistical hypothesis:
H0 : type of accommodations and survival rates are independent
H1 : H0 is false
= (c-1) (r-1)
= (2-1) (2-1)
= 1
fe
(column total)(rowtotal)
grand total
fe (cabin, survived) = (579)(485)
1291
=
280815
1291 = 217.52
fe (steerage, survived) = (712)(485)
1291
=
345320
1291 = 276.48
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fe (cabin, not survived) = (579)(806)
1291
=
466674
1291 = 361.48
fe (steerage, not survived) = (712)(806)
1291
=
573872
1291 = 444.52
Survived/Accommodations Cabin Steerage Total
Yes f0
fe
299
217.52
186
276.48
485
No f0
fe
280
361.48
526
444.52
806
Total 579 712 1291
X
2
=
(299−217.52) 2
217.52 +
(186−267.48) 2
267.48 +
(280−231.48) 2
361.48 +
(516−444.52)2
444.52
=
6638.99
217.52 +
6638.99
267.48 +
6638.99
361.48 +
6638.99
444.52
= 30.52 + 24.82 + 18.37+ 14.94 = 88.65
The null hypothesis will be rejected because x2
of 88.65 > critical x2
of 3.84.
There seems to be a correlation between the type of accommodation and the survival rate.
(b) Assuming a significant c2, estimate the strength of the relationship.
Ø
2
c =
88.65
1291(2−1)
=
88.65
1291 = 0.07
Strength of relationship = medium
(c) To more fully appreciate the importance of this relationship, calculate an odds ratio to
determine how much more likely a cabin passenger is to have survived than a steerage
passenger.
OR =
299/280
186 /526 =
1.07
0.35 = 3.06
A cabin passenger is 3.06 times more likely to survive than a steerage passenger.
19.16 Test the null hypothesis at the .01 level of significance that the distribution of blood types
for college students complies with the proportions described in the blood bank bulletin, namely,
.44 for O, .41 for A, .10 for B, and .05 for AB. Now, however, assume that the results are
available for a random sample of only 60 students. The results are as follows: 27 for O, 26 for A,
4 for B, and 3 for AB.
Since the p value in this problem is 0.7642 is greater than .01 ( 0.7642 > .01) the test is not
considered significant. Because of this, we cannot reject the null hypothesis.
Chapter 20
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