TOPIC 7 EXERCISES
Topic 7 Exercises
Angi Ware
Grand Canyon University PSY 520
July 27, 2017
TOPIC 7 EXERCISES
19.9 Randomly selected records of 140 convicted criminals reveal that their crimes were
committed on th
...
TOPIC 7 EXERCISES
Topic 7 Exercises
Angi Ware
Grand Canyon University PSY 520
July 27, 2017
TOPIC 7 EXERCISES
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.
Statistical Hypothesis: H0 = Psun = Pmon = Ptues = Pwed = Pthurs = Pfri = Psat =
1
7
H1 : H0 is false
Decision Rule: Reject H0 at the 0.01 level of significance = x2 > 16.81
fe = (expected proportion) (total sample size)
frequency 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 returned at the 0.01 level because the observed x2 of 3.4 is < the
critical x2
of 16.81
The result is that crimes are likely to be committed on any day of the week.
TOPIC 7 EXERCISES
(b) Specify the approximate p -value for this test result.
The null hypothesis H0 is returned at the 1% level of significance, which indicates that p >
0.01.
The x2 table values indicate the value 12.6 at the 0.05 level and degrees of freedom =6. The
null hypothesis can be retained at the 5% level, so p>0.05.
(c) How might this result be reported in the literature?
The evidence states that the crimes would equally likely occur on any day of the week. [x2
(6, n=200) = 3.4, p>0.05]. This indicates that x2
based on 6 degrees of freedom with a sample
size of 140 = 3.4.The test result has an approximate p value >0.05 because the null
hypothesis was retained.
19.10 While playing a coin-tossing game in which you are to guess whether heads or tails will
appear, you observe 30 heads in a string of 50 coin tosses.
(a) Test the null hypothesis that this coin is unbiased, that is, that heads and tails are equally
likely to appear in the long run.
Null hypothesis: H0: Pheads =
1
2 48
Decision rule: Reject H0 at the 0.05 level of significance. If x2>3.84 given the degrees of
freedom to be: c-1=2-1=1
Frequency Heads Tails Total
f0 30 20 50
fe 25 25 50
Null hypothesis:
X
2 = ∑
( f 0−fe) 2
fe
= {
(30−25) 2
25 +
(20−25) 2
25 }
TOPIC 7 EXERCISES
=
25
25
+
25
25
= 1+1
=2
Retain due to x2 of 2 < critical x2 of 3.84
(b) Specify the approximate p -value for this test result.
P value is the smallest level of significance that would lead to rejecting H0. Accepting this
would lead to a greater p value than significance level. Retaining H0 at the 5% level
suggests that p value >a=0.05.
*19.13 In 1912, over 800 passengers perished after the ocean liner Titanic collided with an
iceberg and sank. The table below 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
Decision Rule: We will reject the null hypothesis at 0.05 level of significance if x2 > 3.84,
given the degrees of freedom:
= (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
TOPIC 7 EXERCISES
fe (steerage, survived) =
(712)(485)
1291 =
345320
1291 = 276.48
fe (cabin, not survived) =
(579)(806)
1291 =
466674
1291 = 361.48
fe (steerage, not survived) =
(712)(806)
1291 =
573872
1291 = 444.52
Survived/Accommodation
s
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
TOPIC 7 EXERCISES
A cabin passenger is 3.06 times more likely to survive than a steerage passenger.
19.14 In a classic study, Milgram et al. “lost” stamped envelopes with fictitious addresses
(Medical Research Association, Personal Address, Friends of Communist Party, and Friends of
Nazi Party).* One hundred letters with each address were distributed among four locations
(shops, cars, streets, and phone booths) in New Haven Connecticut, with the following results:
(a) Using the .05 level of significance, test the null
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