Answer Key
Psychology 235B Spring, 2000 Midterm Exam #2
-
University:
Amherst College -
Course:
PSYC 221 | Personality Psychology Academic year:
2000
-
Views:
485
Pages:
6
Author:
ntacyozvx
Psychology 235B
Spring, 2000
Midterm Exam #2
Name_________________________________________
TA________ Time________
1) The average freight train that runs down the tracks next to Neil Street is 50 cars long with a
standard deviation of 10 cars. Which of the following possibilities is more likely (pretend that
the number of boxcars can be modeled as a continuous random variable)? Explain.
a) The next freight train that passes will be 45 cars long.
b) The average length of the next 10 trains will be 48 cars. (5 pts)
A) Z = (45-50) / 10
= -.5
B) Z = (48-50) / 10 / SQRT(10)
= -2 / 3.16
= -.63
Because the z-score in part a) is closer to 0, the probability of event A is greater than the
probability of Event b. Therefore, we are much more likely to observe one train that is 45 cars
long, than for the average of the next 10 trains to be 48.
2) Although in most regards, you and Biff have a stable and happy friendship, you are
occasionally quite competitive. For example, the two of you are independently chasing the
wealth, fame and power, that will be granted to the person that derives a point estimator that is
better than the mean. You two agree to test your estimators using the same population of data
for which = 72.387. The sampling distribution of Biff's point estimator, which he has
modestly labeled the "Biff", has a mean of 72.387, but a variance of 26. The sampling
distribution of your point estimator, which you have modestly named the "", has a mean of 72.395 and a variance of 1. Which is a better estimator and why? (2 pts)
You would have to take your hat off to Biff. Although his point estimator is much more variable
than yours, the “Biff” is unbiased in that the mean of the sampling distribution is equal to .
Even though your estimator has very low variability, it is biased and our statistical tools will not
function properly with a biased estimator.
1) You have a firm rule not to date anyone who is a member of because anyone who joins a
Stats Honor Fraternity must be a geek! Describe each of the following situations as either a Type I Error, a Type II Error or no error in particular. Be sure to maintain the same null
hypothesis for each part (i.e., H0: = geek), and explain your answer.
a) An attractive member of asks you out, but you say no. Your friend goes out with this
person and s/he turns out to be the exact opposite of a geek. Not only is this person not a geek,
s/he turns out to be the only heir to the Solo Cup Company fortune, so your friend lives happily
ever after in a big mansion.
Type I Error. You failed to reject the null hypothesis; that is, you concluded that the
member was a geek. This was a mistake in this case because the member was not a geek.
b) After witnessing your friend's good fortune, you decide to go out with the next person
who asks you out. Unfortunately, this person asks you to help collect specimens for an
experiment to determine whether earthworms can be used as a cheap alternative to breakfast
cereal.
Type II error. You rejected the null – concluded that the member was not a geek. This was
a mistake because this person quite obviously was a geek.
c) Just to be mean, you give the "Earthworm Experimenter" your younger sibling's phone
number and explain how much your sib likes digging around in the mud. But, your sib rejects
the "Earthwork Experimenter's" request for a date because of the "EE's" affiliation with .
This represents no error. Your sibling failed to reject the null; s/he assumed that the
member was a geek. This decision was correct.
4) A researcher is interested in whether moderate consumption of caffeine by pregnant women
affects the birth weight of their children. She samples 100 women who drank 2 cups of coffee
per day and finds that the average birth weight of their children was 7.5 pounds.
a) Construct the 60 and 95% CIs for the true average birth weight of mothers who consumed
moderate amounts of caffeine if σ = 3. (3 pts each)
60% CI = mean +/- zcrit (s / sqrt(n))
= 7.5 +/- .84 (3 / sqrt(100))
= 7.5 +/- .252
= [7.248 – 7.752]
95% CI = mean +/- zcrit (s / sqrt(n))
= 7.5 +/- 1.96 (3 / sqrt(100))
= 7.5 +/- .588
= [6.912 - 8.088]
b) Does the 95% CI lead you to conclude that moderate caffeine consumption affects birth
weight if the population parameter equals 8 pounds? Explain. (2 pts) The CI leads me to conclude that caffeine consumption does not affect birth weight. Because the
CI includes the value 8, I cannot be sure that the population parameter for women who have
consumed moderate amounts of caffeine is different from the expected value of 8.
d) What percentage of newborns would you expect to fall within the 95% CI? (2 pts)
To answer this question, you need to calculate the z-scores that correspond to the upper and
lower boundaries of the CI based on S, instead of SE. The reason: because now we are dealing
with the probability of individual observations, not the probability of observing means from a
sample.
Z = (6.912 – 7.5) / 3
Z = (8.088 – 7.5) / 3
= -.196
= .196
Area (Body -.20) – Area (Tail .20) = .5793 - .4207 = .1586
5) Holiday Inn wants to know what proportion of its guests steal hotel towels. To conduct this
study, they must send a person into each room after check-out to count the number of towels left
in a wet, stinky pile in the middle of the bathroom floor. It's gross work, so they want to limit
the number of wet, stinky piles that must be sampled. If Holiday Inn estimates that 20% of its
patrons steal towels, how many room will have to be sampled to estimate the population
parameter within 5%? Assume that alpha = .05. (4 pts)
n = p(1-p) * (z/E)2
= (.2)(.8) * (1.96/.05)2
= (.16) * (39.2)2
= (.16) * 1536.64
= 245.8624
Round up to 246. 6) Mike Huckabee, the current Republican governor of Arkansas is considering using "Our Trout
are Larger than Average" as the state's slogan for the big summer tourist season. To avoid the
stain and scandal of previous administrations, he decides to commission a study to determine
whether this claim is accurate. He has Officers at the State Fish & Wildlife Commission catch
and measure one trout from each of Arkansas' 750 streams, rivers and lakes. The mean of the
sample was 5.3 pounds; we know from previous work that σ = 3 pounds. Governor Huckabee
wants you to conduct a hypothesis test to determine whether these data provide enough evidence
to conclude that the tourism slogan is accurate if we know that (the mean weight for all fullgrown adult trout) = 5 pounds. He is very concerned about avoiding any appearance of
impropriety, so he asks you to conduct a two-tailed test with = .01. Your answer should
include a statement of the null and alternative hypotheses, the critical value for your test statistic,
the observed value of your test statistic, your decision regarding the null, and a statement of what
this means about the size of Arkansas trout, and the appropriateness of the tourism slogan. (12
pts)
Ho: = 5
Ha: 5
Zcrit = 2.57
Zobs = (mean – ) / s (sqrt(n)
= (5.3 – 5) / 3 (sqrt(750)
= .3 / .1095
= 2.74
Zobs > Zcrit. Therefore, I would REJECT the null. This would lead me to conclude that tourism
slogan is accurate. Arkansas’ trout are indeed larger than average.
b) What is the p-value for your test? (2 pts)
p-value = 2 * (Tail 2.74)
= 2 * .0031
= .0062 7) Moike, a start-up pharmaceutical company, has developed a new heartburn medication called
Dyspyron. They run clinical trials to determine how well the drug works relative to Zantac. So,
the Moike scientists give 10 people Zantac and 10 people Dyspyron, and measure how long it
takes for the drug to eliminate heartburn symptoms. The data for each subject appears below.
Zantac
Dispyron
2
1
3
2
5
3
1
2
4
4
7
5
7
6
3
2
5
4
3
1
Mean
4
3
SD
2.0
1.7
Conduct a hypothesis test to determine whether these data provide enough evidence to conclude
that the two drugs differ in terms of effectiveness ( = .05). Your answer should include a
statement of the null and alternative hypotheses, the critical value for your test statistic, the
observed value of your test statistic, your decision regarding the null, and a statement of what
this means about the relative effectiveness of the two drugs. (14 pts)
Ho: z = d
Ha: z d
Tcrit = 2.101
Tobs = (meanz – meand – Do) / SE
SE = sqrt [(S2p/n1) + (S2p/n2)]
S2p = [(n1-1)s21 + (n2-1)s22)] / (n1 + n2 – 2)
= [(10-1)1.72 + (10-1)2.0s2)] / (10 + 10 – 2)
= [9(2.89) + 9(4)] / 18
= (26.01 + 36) / 18
= 62.01 /18 =
3.445
= sqrt[(3.445/10 + 3.445/10)]
= sqrt(.3445 + .3445)
= sqrt(.689) =
.83
Tobs = 4-3-0 / .83
= 1 / .83
=
1.20
Tobs < Tcrit. Therefore, I would FAIL TO REJECT the null. This would lead me to conclude
that the two drugs are equally effective. 8) Let's now assume that Moike conducted its clinical trial in the following way. 10 men were
asked to consume a chili dog. After the inevitable heartburn started, they were given a dose of
Zantac and the time that it took for the drug to take effect was measured. A week later, the same
10 men came back and repeated the procedure, only this time, they took Dyspyron. The data in
table below, are exactly the same as the data for question 10.
Zantac
Dyspyron
Man1
2
1
1
Man2
3
2
1
Man3
5
3
2
Man4
1
2
-1
Man5
4
4
0
Man6
7
5
2
Man7
7
6
1
Man8
3
2
1
Man9
5
4
1
Man10
3
1
2
Conduct a hypothesis test to determine whether these data provide enough evidence to conclude
that the two drugs differ in terms of effectiveness ( = .05). Your answer should include a
statement of the null and alternative hypotheses, the critical value for your test statistic, the
observed value of your test statistic, your decision regarding the null, and a statement of what
this means about the relative effectiveness of the two drugs. (14 pts)
(x) = 10
(x2) = 12 + 12 + 22 + -12 + 02 + 22 + 12 + 12 + 12 + 22 = 18
Mean Diff = 10 / 10 = 1
S2 = [(x2) - ((x)2 / n)] / n-1
= [18 – (102 / 10)] / 9
= (18 – 10) / 9
=8/9
= .89
s = sqrt(.89) = .94
Ho: d = 0
Ha: d 0
Tcrit = 2.262
Tobs = Mean Diff – Do / (s / sqrt (n)
= 1 – 0 / (.94 / (sqrt (10))
= 1 / (.94 / 3.16)
= 1 / .297
=
3.36
Because Tobs > Tcrit, I would reject the null.
This suggests that Dyspyron is more effective than Zantac; that is, relief came more quickly with
the former, than with the latter.
Psychology 235B Spring, 2000 Midterm Exam #2
of 6
Report
Tell us what’s wrong with it:
Thanks, got it!
We will moderate it soon!
Free up your schedule!
Our EduBirdie Experts Are Here for You 24/7! Just fill out a form and let us know how we can assist you.
Take 5 seconds to unlock
Enter your email below and get instant access to your document