Review for 1 sample HT
Name___________________________________
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
1) A statistics professor wants to see if more than 80% of her students enjoyed taking her class. At the end of the
term, she takes a random sample of students from her large class and asks, in an anonymous survey, if the
students enjoyed taking her class. Which set of hypotheses should she test?
E) H0 : p > 0.80
A) H0 : p < 0.80
B) H0 : p = 0.80
C) H0 : p = 0.80
D) H0 : p < 0.80
HA : p > 0.80
HA : p > 0.80
HA : p < 0.80
HA : p ≠ 0.80
HA : p = 0.80
SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.
A report on health care in the US said that 28% of Americans have experienced times when they havenʹt been able to afford
medical care. A news organization randomly sampled 801 black Americans, of whom 38% reported that there had been
times in the last year when they had not been able to afford medical care. Does this indicate that this problem is more severe
among black Americans?
2) Test an appropriate hypothesis and state your conclusion. (Make sure to check any necessary conditions and to
state a conclusion in the context of the problem.)
3) Was your test one-tail upper tail, one-tail lower tail, or two-tail? Explain why you chose that kind of test in
this situation.
4) Explain what your P-value means in this context.
According to the 2010 U.S. Census, 11.7% of the people in the state of Oregon were Hispanic or Latino. A political party
wants to know how much impact the Hispanic and Latino vote will have, so they wonder if that percentage has changed
since then. They take a random sample of 853 adults in Oregon and ask, among other things, their race. 113 of the people
surveyed were Hispanic or Latino. Can the political party conclude that the Hispanic proportion of the population has
changed since 2010?
5) Test an appropriate hypothesis and state your conclusion.
6) Was your test one-tail upper tail, lower tail, or two-tail? Explain why you chose that kind of test in this
situation.
7) Explain what your P-value means in this context.
A professor at a large university believes that students take an average of 15 credit hours per term. A random sample of
24 students in her class of 250 students reported the following number of credit hours that they were taking:
8) Does this sample indicate that students are taking more credit hours than the professor believes? Test an
appropriate hypothesis and state your conclusion.
The average American sees 3.9 movies at the theater each year. A curious student polls 30 friends and family over the
course of a week. He finds that his friends have seen an average of 4.5 movies with a standard deviation of 1.2 movies.
9) Does this sample provide evidence that people are attending the movies more often? Provide a complete
significance test to support your answer.
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10) The student insists to his media teacher at school that movie attendance is on the rise. Do you agree with this
conclusion?
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
11) Two agronomists analyzed the same data, testing the same null hypothesis about the proportion of tomato
plants suffering from blight. One rejected the hypothesis but the other did not. Assuming neither made a
mistake in calculations, which of these possible explanations could account for this apparent discrepancy?
I. One agronomist wrote a one-tailed alternative hypothesis, but the other used 2 tails.
II. They wrote identical hypotheses, but the one who rejected the null used a higher α-level.
III. They wrote identical hypotheses, but the one who rejected the null used a lower α -level.
A) I or II
B) II only
C) I or III
D) I only
E) III only
SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.
12) Sleep Do more than 50% of U.S. adults feel they get enough sleep? According to Gallupʹs December 2004
Lifestyle poll, 55% of U.S. adults said that that they get enough sleep. The poll was based on a random sample
of 1003 U.S. adults. Test an appropriate hypothesis and state your conclusion in the context of the problem.
13) Game Wizard A friend of yours is constantly bragging that he performs exceptionally well at a popular video
game since he scores an average of 1400 points. You think he is not all that hot. Having taken a statistics class,
you know that you can actually test to see whether or not your friend scores higher than other players of the
game. You collect a random sample of scores from other players at the local arcade and get the following data:
1375 1420 1250 1310 1450 1380 1390 1410 1290 1360
Is your friend right? That is, is there evidence that other players average less than 1400 points per game?
Conduct a hypothesis test, making sure to state your conclusion in context of the problem.
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Answer Key
Testname: UNTITLED1
1) B
2) Hypotheses: H0 : p = 0.28. The proportion of all black Americans that were unable to afford medical care in the last
year is 28%.
HA : p > 0.28. The proportion of all black Americans that were unable to afford medical care in the last
year is greater than 28%.
Model: Okay to use the Normal model because the sample is random, these 801 black Americans are less than 10% of
all black Americans, and
np = (801)(0.28) = 224.28 ≥ 10 and nq = (801)(0.72)= 576.72 ≥ 10.
We will do a one-proportion z-test.
^
Mechanics: n = 801, p = 0.38
0.38 - 0.28
z = (0.28)(0.72)
801
^
P(p > 0.38) = P z > 6.29 ≈ 0
Conclusion: With a P-value so small (just about zero), I reject the null hypothesis. There is enough evidence to
suggest that the proportion of black Americans who were not able to afford medical care in the past year is more than
28%.
3) One-tail, upper tail test. We are concerned that the proportion of people who are not able to afford medical care is
higher among black Americans.
4) If the proportion of black Americans was 28%, we would almost never expect to find at least 38% of 801 randomly
selected black Americans responding ʺyesʺ.
5) Hypotheses: H0 : p = 0.117. 11.7% of people in Oregon are Hispanic or Latino.
HA : p ≠ 0.117. The percentage of Hispanic and Latinos in Oregon is no longer 11.7%.
Model: OK to use a Normal model because trials are independent (people are randomly selected), and np = 0.117·853
= 99.8, nq = 0.883·853 = 753.2, which are both greater than 10. Do a 1 proportion z-test.
^
Mechanics: p = 0.132 , z = 1.406, P = 0.16
Conclusion: With a P-value so high I fail to reject the null hypothesis. This test does not provide convincing evidence
that the proportion of Oregonʹs population that is Hispanic and Latino is different from 11.7%.
6) Two-tail. The question was whether the proportion had changed. No direction was anticipated.
7) If 11.7% of Oregonʹs population is Hispanic and Latino, there is a 16% chance of getting a sample proportion as far
from 11.7% as 0132, in either direction.
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Answer Key
Testname: UNTITLED1
8) H0 : μ = 15. The mean number of credit hours taken by all of the professorʹs students is 15 credit hours.
HA: μ > 15. The mean number of credit hours taken by all of the professorʹs students is greater than 15 credit hours.
Conditions:
* Randomization condition: Students from the class were randomly sampled.
* 10% condition: The sample is less than 10% of the class population.
* Nearly Normal condition: The histogram of credit hours is unimodal and reasonably symmetric.
Under these conditions, the sampling distribution of the mean can be modeled by Studentʹs t with df = n - 1 = 24 - 1 =
23. Use a one-sample t-test for the mean.
We know: n = 24, y = 16.6 , and s = 2.22 .
2.22
= 0.453.
So, SE(y) = 24
t = y - μ0
SE(y)
= 16.6 - 15
= 3.532.
0.453
The P-value is P(t23 > 3.532) = 0.0009.
The P-value of 0.0009 is low, so reject the null hypothesis. There is strong evidence that the professorʹs students are
taking more than 15 credit hours, on average.
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Answer Key
Testname: UNTITLED1
9) H0 : μ = 3.9 The mean number of movies seen per year is 3.9
HA: μ > 3.9 The mean number of movies seen per year is greater than 3.9
*Randomization Condition: This is a convenience sample. That is of concern. Our conclusion will be narrow.
*10% condition: 30 people is certainly less than 10% of all movie goers.
*Nearly Normal Condition: Our sample is fairly large (30), so the CLT guarantees the sampling distribution is
approximately normal.
Since the conditions are satisfied, we will to use a one-tail, one-sample t-test.
x = 4.5 s = 1.2
n = 30 df = 29
4.5 - 3.9
= 2.74
t = 1.2
30
P(t > 2.74) = 0.0052
α = 0.05
With a P-value of 0.0052 < 0.05, we reject H0 . We have statistically significant evidence that this studentʹs friends
watch more movies than average.
10) Given that this student collected a convenience sample, we are very skeptical about any attempts to generalize it to a
larger population.
11) A
12) Hypothesis: H0 : p = 0.50
HA : p > 0.50
Plan: Okay to use the Normal model because the trials are independent (random sample of U.S. adults), these 1003
U.S. adults are less than 10% of all U.S. adults, and np0 = (1003)(0.50) = 501.5 ≥ 10 and nq0 = (1003)(0.50) = 501.5 ≥ 10.
We will do a one-proportion z-test.
Mechanics: SD(p0 ) = ^
P(p > 0.55) = P(z > p0 q0
^
(0.50)(0.50)
= = 0.0158; sample proportion: p = 0.55
1003
n
0.55 - 0.50
) = P(z > 3.16) = 0.0008
0.0158
With a P-value of 0.0008, I reject the null hypothesis. There is strong evidence that the proportion of U.S. adults who
feel they get enough sleep is more than 50%.
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Answer Key
Testname: UNTITLED1
13) H0 : μ = 1400 points HA: μ < 1400 points
Conditions:
* Randomization condition: Player’s scores were randomly sampled from the arcade.
* 10% condition: The sample is less than 10% of all players at the high arcade.
* Nearly Normal condition: The histogram of game points is unimodal and reasonably symmetric. This is close
enough to Normal for our purposes. (A histogram must be shown.) Under these conditions, the sampling distribution
of the mean can be modeled by Student’s t with degrees of freedom: df = n – 1 = 10 – 1 = 9.
We will use a one-sample t-test for the mean.
62.452
= 19.749.
We know: n = 10, y = 1363.5, and s = 62.452. So SE(y) = 10
t = y - μ 0
SE(y)
= 1363.5 - 1400
= -1.848 19.749
The P-value is P(t9 < -1.848) = 0.048.
At the 5% significance level, the P-value of 0.048 is statistically significant and you would reject the null hypothesis
that the mean player score is 1400 in favor of the alternative that it is lower; this data would provide evidence, at the
5% level, that your friend outperforms, on average, other players of this game. However, using a 1% significance level,
you would fail to reject the null hypothesis and conclude that you have insufficient evidence to support your friend’s
claim of superior play at this game.
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