Probability and One-sample Z tests
Due: 11/01. This assignment is to be done by hand. Please show your work.
1. Lisa wants to conduct a survey about people's attitudes towards the death penalty. She
posts a questionnaire on the internet asking people to respond to her survey. After one
month, she collects her data and presents her results as a representation of people's
attitudes towards the death penalty. Can Lisa's data be considered representative of the
population? Be sure to provide a justification for your answer. What alternative
selection method could Lisa use?
Lisa’s data is not representative on many levels:
1. Internet users make up a minority of the American population and have a
higher SES than the general population.
2. Lisa’s respondents have self-selected and numerous studies have shown that
self-selectors are not representative of the population.
3. Lisa’s self-selecting respondents are also only a subset of those who somehow
managed to find her website where the questionnaire was posted.
A better way to collect her sample would be to identify the target population in
which she is interested and randomly sample.
2. Does aspirin reduce the risk of a subsequent heart attack? Here are the data for
subjects who had survived one heart attack and who then began to take a placebo or
aspirin each day.
Placebo
Attack
239
Aspirin
139
No Attack
10795
10898
Using this data; calculate the probability of having a heart attack when 1) taking a
placebo, and 2) taking aspirin.
1) 239/11034 = .022 or a 2.2% chance.
2) 139/11037 = .012 or a 1.2% chance.
3. The Internal Revenue Service needs to hire 150 claim reviewers. According to
previous research, applicants who score at or above the 80th percentile on the PDQ
mathematics test complete successfully the IRS training program and are then certified as
claim reviewers. The PDQ mathematics has a mean of 50 and a standard deviation of 12.
After the job announcement was advertised, 1637 people applied to take the test. Use
this information to complete the following.
a. How many of the people taking the test do you predict will achieve a score high
enough to be considered for the job of claim reviewer?
.20*1637 = 327.4 or 328 people.
b. What score is necessary to be considered for this job?
1. 80th percentile is a z-score of .85.
2. A z-score of .85 converts to a raw score on PDQ of 60.2.
t PDQ  z ( s)  X  .85(12)  50  10.2  50  60.2
c. Thomas scores a 70 on the test. What is his percentile ranking?
1. A score of 70 is a z-score of 1.67.
z
X  X 70  50 20


 1.67
s
12
12
2. A z-score of 1.67 is at the 95.25 percentile level.
d. Of the applicants taking the test, how many do you estimate will
1. have a score less than 45?
a. 45 is a z-score of -.42
z
X  X 45  50
5

   .42
s
12
12
b. A z-score of -.42 is at the 33.72 percentile level.
c. 1637 * .3372 = 552
2. have a score between 38 and 72?
a. 38 is a z-score of -1.00
z
X  X 38  50
12

   1.00
s
12
12
b. A z-score of -1.00 is at the 15.87 percentile level.
c. 72 is a z-score of 1.83
z
X  X 72  50 22


 1.83
s
12
12
d. A z-score of 1.83 is at the 96.64 percentile level.
e. 96.64 – 15.87 = 80.77
f. 1637 * .8077 = 1,322
3. have a score greater than 72 or a score less than 38?
a.
From the previous exercise, 1,322 can be expected to have a
score between 38 and 72 so 315 (1637-1322 = 315) can be
expected to have a score greater than 72 or a score less than 38.
e. What is the probability of randomly selecting 3 people whose score is 50 or greater?
50 is the mean so there is a 50% chance of selecting 1 person at random whose
score is 50 or greater. Using the multiplication rule for independent events
would give:
.5*.5*.5=.125 or a 12.5% chance.
f. What is the probability of randomly selecting 3 people whose score is 65 or greater?
1. A score of 65 is a z-score of 1.25
z
X  X 65  50 15


 1.25
s
12
12
2. A z-score of 1.25 is at the 89.44 percentile level leaving a 10.56% chance of
selecting 1 person at random whose score is 65 or greater. Using the
multiplication rule for independent events would give:
.10*.10*.10=.001 or a .1% chance.
4. The table below shows the result of survey by a car dealer about the relationship
between income level and type of car they buy. A total of 250 participants were randomly
sampled.
a. Calculate the following:
a. P(h1) = 95/250 = .38
b. P(D1) = 40/250 = .16
c. P(h1 ∩ D1) = 25/250 = .10
d. P(h1 U D1) = P(h1) + P(D1) - P(h1 ∩ D1) = .38 + .16 -.10 = .44
e. P(h2 U D3) = P(h2) + P(D3) – P(h2 ∩ D3) = .62 + .46 - .3 = .78
f. P(h1|D3) = 40/115 = .34
g. P(D1|h2) = 15/155 = .09
5. The scores of a physical performance test for boys of junior high age have a mean of
175 and a standard deviation of 12 for the general population. In a large city school
system, a random sample of 225 junior high school boys is tested. The sample mean is
173.6.
a) Find the standard error of the mean. s.e. = 12/225 = .8
a. Explain what the standard error of the mean represents.
The standard error is the standard deviation of a sampling distribution of sample
means of size n.
b) Using the hypothesis testing format provided in class test the hypothesis that the
mean test score for the population is 175 vs. the alternative that it is not 175 using
α = .05 (i.e., H0: µ = 175).
Step 1: We wish to determine whether our sample of students mean score is different than
that of the national norms on a physical performance test.
Step 2: Ho: µ = 175
H1: µ ≠ 175
Step 3: Test Statistic
Z 
X   hyp
X
Step 4: Assumptions underlying test statistic
See handout
Step 5: Formulation of α and decision rule
With α = .05, if z exceed 1.96 or -1.96 we will reject the null hypothesis.
Step 6: Collect the data, calculate the summary statistics and test statistic.
Z 
173.6  175
 1.75
.8
Step 7: Since a z of –1.75 does not exceed –1.96 or 1.96 we do not reject the null
hypothesis
Step8: Verbal Conclusion
Our sample likely comes from a population with a µ = 175