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.
Attack No Attack
Placebo 239 10795
Aspirin 139 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.
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.
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
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
b. A z-score of -1.00 is at the 15.87 percentile level.
c. 72 is a z-score of 1.83
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
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:
- P(h1) = 95/250 = .38
- P(D1) = 40/250 = .16
- P(h1 ∩ D1) = 25/250 = .10
- P(h1 U D1) = P(h1) + P(D1) - P(h1 ∩ D1) = .38 + .16 -.10 = .44
- P(h2 U D3) = P(h2) + P(D3) – P(h2 ∩ D3) = .62 + .46 - .3 = .78
- P(h1|D3) = 40/115 = .34
- 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
- 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
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.
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