Chapter 12
Chapter 12
1.
A two-sided or two-tailed hypothesis test is one in which
A. the null hypothesis includes values in either direction from a specific standard.
B. the null hypothesis includes values in one direction from a specific standard.
C. the alternative hypothesis includes values in one direction from a specific standard
D. the alternative hypothesis includes values in either direction from a specific standard
KEY: D
2.
Null and alternative hypotheses are statements about
A. population parameters.
B. sample parameters.
C. sample statistics.
D. it depends - sometimes population parameters and sometimes sample statistics.
KEY: A
3.
Which statement is correct about a p-value?
A. The smaller the p-value the stronger the evidence in favor of the alternative hypothesis.
B. The smaller the p-value the stronger the evidence in favor the null hypothesis
C. Whether a small p-value provides evidence in favor of the null hypothesis depends on whether the test is
one-sided or two-sided.
D. Whether a small p-value provides evidence in favor of the alternative hypothesis depends on whether the
test is one-sided or two-sided.
KEY: A
A hypothesis test gives a p-value of 0.03. If the significance level  = 0.05, the results are said to be
A. not statistically significant because the p-value ≤ .
B. statistically significant because the p-value ≤ .
C. practically significant because the p-value ≤ .
D. not practically significant because the p-value ≤ .
KEY: B
4.
A hypothesis test gives a p-value of 0.050. If the significance level  = 0.05, the results are said to be
A. not statistically significant because the p-value is not smaller than .
B. statistically significant because the p-value ≤ .
C. practically significant because the p-value is the same as .
D. inconclusive because the p-value is not smaller nor larger than .
KEY: B
5.
6.
The likelihood that a statistic would be as extreme or more extreme than what was observed is called a
A. statistically significant result.
B. test statistic.
C. significance level.
D. p-value.
KEY: D
7.
The data summary used to decide between the null hypothesis and the alternative hypothesis is called a
A. statistically significant result.
B. test statistic.
C. significance level.
D. p-value.
KEY: B
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8.
The designated level (typically set at 0.05) to which the p-value is compared to, in order to decide whether the
alternative hypothesis is accepted or not is called a
A. statistically significant result.
B. test statistic.
C. significance level.
D. none of the above.
KEY: C
9.
When the p-value is less than or equal to the designated level of 0.05, the result is called a
A. statistically significant result.
B. test statistic.
C. significance level.
D. none of the above.
KEY: A
10. Which of the following conclusions is not equivalent to rejecting the null hypothesis?
A. The results are statistically significant.
B. The results are not statistically significant.
C. The alternative hypothesis is accepted.
D. The p-value ≤  (the significance level)
KEY: B
11. If the result of a hypothesis test for a proportion is statistically significant, then
A. the null hypothesis is rejected.
B. the alternative hypothesis is rejected.
C. the population proportion must equal the null value.
D. None of the above.
KEY: A
12. The smaller the p-value, the
A. stronger the evidence against the alternative hypothesis.
B. stronger the evidence for the null hypothesis.
C. stronger the evidence against the null hypothesis.
D. None of the above.
KEY: C
13. Which of the following is not a valid conclusion for a hypothesis test?
A. Reject the null hypothesis.
B. Do not reject the null hypothesis.
C. We have proven the null hypothesis is true.
D. We have proven the alternative hypothesis is true.
KEY: C and D
14. In hypothesis testing for one proportion, the "null value" is used in which of the following?
A. The null hypothesis.
B. The alternative hypothesis.
C. The computation of the test statistic.
D. All of the above.
KEY: D
15. A result is called statistically significant whenever
A. the null hypothesis is true.
B. the alternative hypothesis is true.
C. the p-value is less than or equal to the significance level.
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D. the p-value is larger than the significance level.
KEY: C
16. Which one of the following is not true about hypothesis tests?
A. Hypothesis tests are only valid when the sample is representative of the population for the question of
interest.
B. Hypotheses are statements about the population represented by the samples.
C. Hypotheses are statements about the sample (or samples) from the population.
D. Conclusions are statements about the population represented by the samples.
KEY: C
17. In a hypothesis test which of the following can (and should) be determined before collecting data?
A. The null and alternative hypotheses.
B. The value of the test statistic.
C. The p-value.
D. Whether the test statistic will be positive or negative.
KEY: A
18. The level of significance (usually .05) associated with a significance test is the probability
A. that the null hypothesis is true.
B. that the alternative hypothesis is true.
C. of not rejecting a true null hypothesis.
D. of rejecting a true null hypothesis.
KEY: D
Questions 19 to 22: Suppose the significance level for a hypothesis test is  = 0.05.
19. If the p-value is 0.001, the conclusion is to
A. reject the null hypothesis.
B. accept the null hypothesis.
C. not reject the null hypothesis.
D. None of the above.
KEY: A
20. If the p-value is 0.049, the conclusion is to
A. reject the null hypothesis.
B. accept the null hypothesis.
C. not reject the null hypothesis.
D. None of the above.
KEY: A
21. If the p-value is 0.05, the conclusion is to
A. reject the null hypothesis.
B. accept the alternative hypothesis.
C. not reject the null hypothesis.
D. None of the above.
KEY: C
22. If the p-value is 0.999, the conclusion is to
A. reject the null hypothesis.
B. accept the alternative hypothesis.
C. not reject the null hypothesis.
D. None of the above.
KEY: C
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23. In hypothesis testing, a Type 1 error occurs when
A. the null hypothesis is not rejected when the null hypothesis is true.
B. the null hypothesis is rejected when the null hypothesis is true.
C. the null hypothesis is not rejected when the alternative hypothesis is true.
D. the null hypothesis is rejected when the alternative hypothesis is true.
KEY: B
24. In hypothesis testing, a Type 2 error occurs when
A. the null hypothesis is not rejected when the null hypothesis is true.
B. the null hypothesis is rejected when the null hypothesis is true.
C. the null hypothesis is not rejected when the alternative hypothesis is true.
D. the null hypothesis is rejected when the alternative hypothesis is true.
KEY: C
25. In a hypothesis test the decision was made to not reject the null hypothesis. Which type of mistake could have
been made?
A. Type 1.
B. Type 2.
C. Type 1 if it's a one-sided test and Type 2 if it's a two-sided test.
D. Type 2 if it's a one-sided test and Type 1 if it's a two-sided test.
KEY: B
26. If, in a hypothesis test, the null hypothesis is actually true, which type of mistake can be made?
A. Type 1.
B. Type 2.
C. Type 1 if it's a one-sided test and Type 2 if it's a two-sided test.
D. Type 2 if it's a one-sided test and Type 1 if it's a two-sided test.
KEY: A
27. In an American criminal trial, the null hypothesis is that the defendant is innocent and the alternative
hypothesis is that the defendant is guilty. Which of the following describes a Type 2 error for a criminal
trial?
A. A guilty verdict for a person who is innocent.
B. A guilty verdict for a person who is not innocent.
C. A not guilty verdict for a person who is guilty
D. A not guilty verdict for a person who is innocent
KEY: C
28. A Washington Post poll shows that concerns about housing payments have spiked despite some
improvements in the overall economy. In all, 53 percent of the 900 American adults surveyed said they are
"very concerned" or "somewhat concerned" about having the money to make their monthly payment. Let p
represent the population proportion of all American adults who are "very concerned" or "somewhat
concerned" about having the money to make their monthly payment. Which are the appropriate hypotheses to
assess if a majority of American adults are worried about making their mortgage or rent payments?
A. H0: p = 0.50 versus Ha: p > 0.50
B. H0: p ≥ 0.50 versus Ha: p < 0.50
C. H0: p = 0.53 versus Ha: p > 0.53
D. H0: p = 0.50 versus Ha: p = 0.53
KEY: A
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Questions 29 to 32: A hypothesis test for a population proportion p is given below:
H0: p = 0.10
Ha: p ≠ 0.10
For each sample size n and sample proportion p̂ compute the value of the z-statistic.
29. Sample size n = 100 and sample proportion p̂ = 0.10. z-statistic = ?
A. –1.00
B. 0.00
C. 0.10
D. 1.00
KEY: B
30. Sample size n = 100 and sample proportion p̂ = 0.15. z-statistic = ?
A. –1.12
B. 0.04
C. 1.12
D. 1.67
KEY: D
31. Sample size n = 500 and sample proportion p̂ = 0.04. z-statistic = ?
A. –6.84
B. –4.47
C. 4.47
D. 6.84
KEY: B
32. Sample size n = 500 and sample proportion p̂ = 0.20. z-statistic = ?
A. –7.45
B. –5.59
C. 5.59
D. 7.45
KEY: D
Questions 33 to 37: A sample of n = 200 college students is asked if they believe in extraterrestrial life and 120 of
these students say that they do. The data are used to test H0: p = 0.5 versus Ha: p > 0.5, where p is the population
proportion of college students who say they believe in extraterrestrial life. The following Minitab output was obtained:
Sample
1
X
N
120
Sample p
200 0.600000
95.0 % CI
(0.532105, 0.667895)
Z-Value
2.83
P-Value
0.002
33. What is the correct description of the area that equals the p-value for this problem?
A. The area to the right of 0.60 under a standard normal curve.
B. The area to the right of 2.83 under a standard normal curve.
C. The area to the right of 2.83 under the standard normal curve.
D. The area between 0.532105 and 0.667895 under a standard normal curve.
KEY: B
34. Suppose that the alternative hypothesis had been Ha: p ≠ 0.5. What would have been the p-value of the test?
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A.
B.
C.
D.
KEY: C
0.002
0.001
0.004
0.5
35. Using a 5% significance level, what is the correct decision for this significance test?
A. Fail to reject the null hypothesis because the p-value is greater than 0.05.
B. Fail to reject the null hypothesis because the p-value is less than 0.05.
C. Reject the null hypothesis because the p-value is greater than 0.05.
D. Reject the null hypothesis because the p-value is less than 0.05.
KEY: D
36. Using a 5% significance level, what is the correct conclusion for this significance test?
A. The proportion of college students who say they believe in extraterrestrial life is equal to 50%.
B. The proportion of college students who say they believe in extraterrestrial life is not equal to 50%.
C. The proportion of college students who say they believe in extraterrestrial life seems to be greater than 50%.
D. The proportion of college students who say they believe in extraterrestrial life seems to be equal to 60%.
KEY: C
37. Based on the decision made in question 70, what mistake could have been made?
A. Type 1.
B. Type 2.
C. Neither one; the p-value is so small that no mistake could have been made.
KEY: A
38. About 90% of the general population is right-handed. A researcher speculates that artists are less likely to be
right-handed than the general population. In a random sample of 100 artists, 83 are right-handed. Which of
the following best describes the p-value for this situation?
A. The probability that the population proportion of artists who are right-handed is 0.90.
B. The probability that the population proportion of artists who are right-handed is 0.83.
C. The probability the sample proportion would be as small as 0.83, or even smaller, if the population
proportion of artists who are right-handed is actually 0.90.
D. The probability that the population proportion of artists who are right-handed is less than 0.90, given that
the sample proportion is 0.83.
KEY: C
39. Consider testing the alternative hypothesis that the proportion of adult Canadians opposed to same-sex
marriage in Canada is less than 0.5. The test was conducted based on a poll of n = 1003 adults and it had a pvalue of 0.102. Which of the following describes the probability represented by the p-value for this test?
A. It is the probability that fewer than half of all adults in Canada that year were opposed to same-sex
marriage.
B. It is the probability that more than half of all adult Canadians that year were opposed to same-sex marriage.
C. It is the probability that a sample of 1003 adults in Canada that year would result in 48% or fewer saying
they are opposed to same-sex marriage, given that a majority (over 50%) of Canadian adults actually were
opposed that year.
D. It is the probability that a sample of 1003 adults in Canada that year would result in 48% or fewer saying
they are opposed to same-sex marriage, given that 50% of Canadian adults actually were opposed that year.
KEY: D
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Questions 40 to 44: An airport official wants to prove that the p1 = proportion of delayed flights after a storm for
Airline 1 was different from p2 = the proportion of delayed flights for Airline 2. Random samples from the two
airlines after a storm showed that 50 out of 100 of Airline 1’s flights were delayed, and 70 out of 200 of Airline 2’s
flights were delayed.
40. What are the appropriate null and alternative hypotheses?
A. H0: p1 – p2 = 0 and Ha: p1 – p2 ≠ 0
B. H0: p1 – p2 ≠ 0 and Ha: p1 – p2 = 0
C. H0: p1 – p2 = 0 and Ha: p1 – p2 < 0
D. H0: p1 – p2 = 0 and Ha: p1 – p2 > 0
KEY: A
41. What is the value of the test statistic?
A. 0.79
B. 2.00
C. 2.50
D. None of the above
KEY: C
42. What is the p-value? (computer or calculator for normal distribution required)
A. p-value = 0.0062
B. p-value = 0.0124
C. p-value = 0.0456
D. p-value = 0.2148
KEY: B
43. For a significance level of  = 0.05, are the results statistically significant?
A. No, the results are not statistically significant because the p-value < 0.05.
B. Yes, the results are statistically significant because the p-value < 0.05.
C. No, the results are not statistically significant because the p-value > 0.05
D. Yes, the results are statistically significant because the p-value > 0.05.
KEY: B
44. Report your conclusion in terms of the two airlines.
A. The proportion of delayed flights after a storm for Airline 1 seems to be different than the proportion of
delayed flights after a storm for Airline 2.
B. The results are not statistically significant: there is not enough evidence to conclude there is a difference
between the two proportions.
C. The difference in proportions of delayed flights is at least 15%.
D. The proportion of delayed flights after a storm for Airline 1 seems to be the same as the proportion of
delayed flights after a storm for Airline 2.
KEY: A
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Questions 45 to 49: An airport official wants to prove that the p1 = proportion of delayed flights after a storm for
Airline A is less than p2 = the proportion of delayed flights for Airline B. Random samples from two airlines after a
storm showed that 51 out of 200 of Airline A’s flights were delayed, and 60 out of 200 of Airline B’s flights were
delayed.
45. What are the appropriate null and alternative hypotheses?
A. H0: p1 – p2 = 0 and Ha: p1 – p2 ≠ 0
B. H0: p1 – p2 ≠ 0 and Ha: p1 – p2 = 0
C. H0: p1 – p2 = 0 and Ha: p1 – p2 < 0
D. H0: p1 – p2 = 0 and Ha: p1 – p2 > 0
KEY: C
46. What is the value of the test statistic?
A. 1.00
B. 1.50
C. 2.00
D. None of the above
KEY: A
47. What is the p-value? (computer or calculator for normal distribution required)
A. p-value = 0.0228
B. p-value = 0.0668
C. p-value = 0.1587
D. p-value = 0.3174
KEY: C
48. For a significance level of  = 0.05, are the results statistically significant?
A. No, the results are not statistically significant because the p-value < 0.05.
B. Yes, the results are statistically significant because the p-value < 0.05.
C. No, the results are not statistically significant because the p-value > 0.05
D. Yes, the results are statistically significant because the p-value > 0.05.
KEY: C
49. Report your conclusion in terms of the two airlines.
A. The proportion of delayed flights for Airline A seems to be less than the proportion of delayed flights for
Airline B.
B. The results are not statistically significant: there is not enough evidence to conclude that the proportion of
delayed flights for Airline A is less than the proportion for Airline B.
C. The difference in proportions of delayed flights is at least 4%.
D. The proportion of delayed flights after a storm for Airline A seems to be greater than the proportion of
delayed flights after a storm for Airline B.
KEY: B
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Chapter 13
50. Which statement is not true about hypothesis tests?
A. Hypothesis tests are only valid when the sample is representative of the population for the
question of interest.
B. Hypotheses are statements about the population represented by the samples.
C. Hypotheses are statements about the sample (or samples) from the population.
D. Conclusions are statements about the population represented by the samples.
KEY: C
51. The primary purpose of a significance test is to
A. estimate the p-value of a sample.
B. estimate the p-value of a population.
C. decide whether there is enough evidence to support a research hypothesis about a sample.
D. decide whether there is enough evidence to support a research hypothesis about a population.
KEY: D
52. The level of significance associated with a significance test is the probability
A. of rejecting a true null hypothesis.
B. of not rejecting a true null hypothesis.
C. that the null hypothesis is true.
D. that the alternative hypothesis is true.
KEY: A
53. A result is called statistically significant whenever
A. the null hypothesis is true.
B. the alternative hypothesis is true.
C. the p-value is less or equal to the significance level.
D. the p-value is larger than the significance level.
KEY: C
54. Which of the following is not one of the steps for hypothesis testing?
A. Determine the null and alternative hypotheses.
B. Verify data conditions and calculate a test statistic.
C. Assuming the null hypothesis is true, find the p-value.
D. Assuming the alternative hypothesis is true, find the p-value.
KEY: D
55. Which of the following is not a correct way to state a null hypothesis?
A. H0:
B. H0: d = 10
C. H0:  = 0
D. H0:  = 0.5
KEY: A
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56. In hypothesis testing for one mean, the "null value" is not used in which of the following?
A. The null hypothesis.
B. The alternative hypothesis.
C. The computation of the test statistic.
D. The (null) standard error.
KEY: D
57. A null hypothesis is that the average pulse rate of adults is 70. For a sample of 64 adults, the
average pulse rate is 71.8. A significance test is done and the p-value is 0.02. What is the most
appropriate conclusion?
A. Conclude that the population average is 70.
B. Conclude that the population average is 71.8.
C. Reject the hypothesis that the population average is 70.
D. Reject the hypothesis that the sample average is 70.
KEY: C
58. The p-value for a one-sided test for a mean was 0.04. The p-value for the corresponding two-sided
test would be:
A. 0.02
B. 0.04
C. 0.06
D. 0.08
KEY: D
59. A null hypothesis is that the mean cholesterol level is 200 in a certain age group. The alternative is
that the mean is not 200. Which of the following is the most significant evidence against the null
and in favor of the alternative?
A. For a sample of n = 25, the sample mean is 220.
B. For a sample of n = 10, the sample mean is 220.
C. For a sample of n = 50, the sample mean is 180.
D. For a sample of n = 20, the sample mean is 180.
KEY: C
60. A test of H0:  = 0 versus Ha:  > 0 is conducted on the same population independently by two
different researchers. They both use the same sample size and the same value of  = 0.05. Which of
the following will be the same for both researchers?
A. The p-value of the test.
B. The power of the test if the true  = 6.
C. The value of the test statistic.
D. The decision about whether or not to reject the null hypothesis.
KEY: B
61. Which of the following is not true about hypothesis testing?
A. The null hypothesis defines a specific value of a population parameter, called the null value.
B. A relevant statistic is calculated from population information and summarized into a “test
statistic.”
C. A p-value is computed on the basis of the standardized “test statistic.”
D. On the basis of the p-value, we either reject or fail to reject the null hypothesis.
KEY: B
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Questions 62 to 66: An investigator wants to assess whether the mean  = the average weight of
passengers flying on small planes exceeds the FAA guideline of average total weight of 185 pounds
(passenger weight including shoes, clothes, and carry-on). Suppose that a random sample of 51
passengers showed an average total weight of 200 pounds with a sample standard deviation of 59.5
pounds. Assume that passenger total weights are normally distributed.
62. What are the appropriate null and alternative hypotheses?
A. H0:  =185 and Ha:  < 185
B. H0:  = 185 and Ha:  >185.
C. H0:  = 185 and Ha:   185.
D. H0:   185 and Ha:  = 185.
KEY: B
63. What is the value of the test statistic?
A. t = 1.50
B. t = 1.65
C. t = 1.80
D. None of the above
KEY: C
64. What is the p-value?
A. p-value = 0.039
B. p-value = 0.053
C. p-value = 0.070
D. None of the above
KEY: A
65. For a significance level of  = 0.05, are the results statistically significant?
A. No, the results are not statistically significant because the p-value < 0.05.
B. Yes, the results are statistically significant because the p-value < 0.05.
C. No, the results are not statistically significant because the p-value > 0.05
D. Yes, the results are statistically significant because the p-value > 0.05.
KEY: B
66. Which of the following is an appropriate conclusion?
A. The results are statistically significant so the average total weight of all passengers appears to be
greater than 185 pounds.
B. The results are statistically significant so the average total weight of all passengers appears to be
less than 185 pounds.
C. The results are not statistically significant so there is not enough evidence to conclude the
average total weight of all passengers is greater than 185 pounds.
D. None of the above.
KEY: A
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67. Spatial perception is measured on a scale from 0 to 10. A group of 9 th grade students are tested for
spatial perception. SPSS was used to obtain descriptive statistics of the spatial perception scores in
the sample.
Using α = 0.01, is the mean score significantly different from 5?
A. Yes, because t = 2.34 and this is greater than t* = 2.11.
B. No, because t = 2.34 and this is smaller than t* = 3.97.
C. No, because t = 0.55 and this is smaller than t* = 2.11.
D. No, because t = 2.34 and this is smaller than t* = 2.90.
KEY: D
68. The amount of time the husband and the wife spend on house work is measured for 15 women and
their 15 husbands. For the wives the mean was 7 hours/week and for the husbands the mean was
4.5 hours/week. The standard deviation of the differences in time spent on house work was 2.85.
What is the value of the test statistic for testing the difference in mean time spent on housework
between husbands and wives?
A. 0.88
B. 2.40
C. 3.40
D. 4.80
KEY: C
69. The head circumference is measured for 25 girls and their younger twin sisters. The mean of the
older twin girls was 50.23 cm and the mean of the younger twins was 49.96 cm. The standard
deviation of the differences was 1 cm. Is this difference significant at a significance level of 5%?
A. Yes, because t = 1.35 and the p-value < 0.10.
B. Yes, because t = 6.25 and the p-value < 0.10.
C. No, because t = 1.35 and the p-value > 0.10.
D. No, because t = 0.27 and the p-value > 0.10.
KEY: C
70. An experiment is conducted with 15 seniors who are taking Spanish at Oak View High School. A
randomly selected group of eight students is first tested with a written test and a day later with an
oral exam. To avoid order effects, the other seven students are tested in reverse order. The
instructor is interested in the difference in grades between the two testing methods. SPSS is used to
obtain descriptive statistics for the grades of the two tests.
N
Talk
Write
Talk - Write
Mean
15
15
15
1.523
4.166
-2.643
Std. Deviation
1.530
2.047
2.182
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Chapter 12
Is there a significant difference between the mean grades using the two different testing methods?
Use α = 0.05.
A. Yes, because t = −4.69 and the p-value < 0.05.
B. No, because t = −1.21 and the p-value > 0.05.
C. Yes, because t = −6.63 and the p-value < 0.05.
D. No, because t = −1.48 and the p-value > 0.05.
KEY: A
Questions 71 to 74: It is known that for right-handed people, the dominant (right) hand tends to be
stronger. For left-handed people who live in a world designed for right-handed people, the same may not
be true. To test this, muscle strength was measured on the right and left hands of a random sample of 15
left-handed men and the difference (left - right) was found. The alternative hypothesis is one-sided (left
hand stronger). The resulting
t-statistic was 1.80.
71. This is an example of
A. a two-sample t-test.
B. a paired t-test.
C. a pooled t-test.
D. an unpooled t-test.
KEY: B
72. Which of the following is true about the conditions necessary to carry out the t-test in this
situation?
A. Because the sample size is small (15), the population of differences must be assumed to be
approximately normal, but no assumption about the variances is required.
B. Because the sample size is not small (15 + 15 = 30), the population of differences need not be
assumed to be approximately normal, and no assumption about the variances is required.
C. Because the sample size is small (15), the population of differences must be assumed to be
approximately normal, and the variances of the right and left hand strengths must be assumed to
be equal.
D. Because the sample size is not small (15 + 15 = 30), the population of differences need not be
assumed to be approximately normal, but the variances of the right and left hand strengths must
be assumed to be equal.
KEY: A
73. Assuming the conditions are met, based on the t-statistic of 1.80 the appropriate decision for this
test using
 = 0.05 is:
A. df = 14, so p-value < 0.05 and the null hypothesis can be rejected.
B. df = 14, so p-value > 0.05 and the null hypothesis cannot be rejected.
C. df = 28, so p-value < 0.05 and the null hypothesis can be rejected.
D. df = 28, so p-value > 0.05 and the null hypothesis cannot be rejected.
KEY: A
74. Which of the following is an appropriate conclusion?
A. The results are statistically significant so the left hand appears to be stronger.
B. The results are statistically significant so the left hand does not appear to be stronger.
C. The results are not statistically significant so there is not enough evidence to conclude the left
hand appears to be stronger.
D. None of the above.
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Chapter 12
KEY: A
75. Suppose that a difference between two groups is examined. In the language of statistics, the
alternative hypothesis is a statement that there is __________
A. no difference between the sample means.
B. a difference between the sample means.
C. no difference between the population means.
D. a difference between the population means.
KEY: D
76. The maximum distance at which a highway sign can be read is determined for a sample of young
people and a sample of older people. The mean distance is computed for each age group. What's
the most appropriate null hypothesis about the means of the two groups?
A. The population means are different.
B. The sample means are different.
C. The population means are the same.
D. The sample means are the same.
KEY: C
77. Researchers want to see if men have a higher blood pressure than women do. A study is planned in
which the blood pressures of 50 men and 50 women will be measured. What's the most appropriate
alternative hypothesis about the means of the men and women?
A. The sample means are the same.
B. The sample mean will be higher for men.
C. The population means are the same.
D. The population mean is higher for men than for women.
KEY: D
78. When comparing two means, the situation most likely to lead to a result that is statistically
significant but of little practical importance is
A. when the actual difference is large and the sample sizes are large.
B. when the actual difference is large and the sample sizes are small.
C. when the actual difference is small and the sample sizes are large.
D. when the actual difference is small and the sample sizes are small.
KEY: C
79. When comparing two means, which situation is most likely to lead to a result that is statistically
significant? (Consider all other factors equal, such as significance level and standard deviations.)
A. x1  10 , x2  20 and the sample sizes are n1  25 and n2  25
B. x1  10 , x2  25 and the sample sizes are n1  25 and n2  25
C.
D.
KEY: D
x1  10 , x2  20 and the sample sizes are n1  15 and n2  20
x1  10 , x2  25 and the sample sizes are n1  25 and n2  35
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Questions 80 and 81: A null hypothesis is that the mean nose lengths of men and women are the same.
The alternative hypothesis is that men have a longer mean nose length than women.
80. Which of the following is the correct way to state the null hypothesis?
A. H 0 : p  0.50
B.
H 0 : x1  x2  0
H 0 : p1  p2  0
D. H0 : 1  2  0
KEY: D
C.
81. A statistical test is performed for assessing if men have a longer mean nose length than women.
The
p-value is 0.225. Which of the following is the most appropriate way to state the conclusion?
A. The mean nose lengths of the populations of men and women are identical.
B. There is not enough evidence to say that that the populations of men and women have different
mean nose lengths.
C. Men have a greater mean nose length than women.
D. The probability is 0.225 that men and women have the same mean nose length.
KEY: B
Questions 82 to 86: An airport official wants to assess if the flights from one airline (Airline 1) are less
delayed than flights from another airline (Airline 2). Let 1 = average delay for Airline 1 and  2 =
average delay for Airline 2. A random sample of 10 flights for Airline 1 shows an average of 9.5 minutes
delay with a standard deviation of 3 minutes. A random sample of 10 flights for Airline 2 shows an
average of 12.63 minutes delay with a standard deviation of 3 minutes. Assume delay times are normally
distributed, but do not assume the population variances are equal. Use the conservative “by hand”
estimate for the degrees of freedom.
82. What are the appropriate null and alternative hypotheses?
A. H0: 1   2 = 0 and Ha: 1   2  0
B. H0: 1   2  0 and Ha: 1   2 = 0
C. H0: 1   2 = 0 and Ha: 1   2 < 0
D. H0: 1   2 = 0 and Ha: 1   2 > 0
KEY: C
83. What is the value of the test statistic?
A. t =1.80
B. t =2.00
C. t =2.33
D. None of the above
KEY: C
84. What is the p-value?
A. p-value = 0.021
B. p-value = 0.022
C. p-value = 0.038
D. None of the above
KEY: B
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Chapter 12
85. For a significance level of  = 0.05, are the results statistically significant?
A. No, results are not statistically significant because the p-value < 0.05.
B. Yes, results are statistically significant because the p-value < 0.05.
C. No, results are not statistically significant because the p-value > 0.05
D. Yes, results are statistically significant because the p-value > 0.05.
KEY: B
86. Which of the following is an appropriate conclusion?
A. The average delay for Airline 1 does appear to be less than the average delay for Airline 2.
B. The results are not statistically significant so there is not enough evidence to conclude the
average delay for Airline 1 is less than the average delay for Airline 2.
C. The average delay for Airline 1 is at least 3 minutes less than Airline 2.
D. None of the above.
KEY: A
Questions 87 to 91: A bank official wants to assess if the average delays for two airlines are different.
Let 1 = average delay for Airline 1 and  2 = average delay for Airline 2. A random sample of 10 flights
for Airline 1 shows an average of 6 minutes delay with a standard deviation of 5 minutes. A random
sample of 10 flights for Airline 2 shows an average of 12 minutes delay with a standard deviation of 5
minutes. Assume delay times are normally distributed, but do not assume the population variances are
equal. Use the conservative “by hand” estimate for the degrees of freedom.
87. What are the appropriate null and alternative hypotheses?
A. H0: 1   2 = 0 and Ha: 1   2  0
B. H0: 1   2  0 and Ha: 1   2 = 0
C. H0: 1   2 = 0 and Ha: 1   2 < 0
D. H0: 1   2 = 0 and Ha: 1   2 > 0
KEY: A
88. What is the value of the test statistic?
A. t = 2.68
B. t = 1.50
C. t = 1.50
D. t = 2.68
KEY: A
89. For a significance level of 0.05, what is the critical value?
A. The critical value = 1.83.
B. The critical value = 2.10.
C. The critical value = 2.26.
D. None of the above.
KEY: C
90. Are the results statistically significant?
A. No, results are not statistically significant because | t | < 1.83.
B. Yes, results are statistically significant because | t | > 2.26.
C. No, results are not statistically significant because | t | > 2.26.
D. None of the above.
KEY: B
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Chapter 12
91. Which of the following is an appropriate conclusion?
A. The average delay for Airline 1 does appear to be different than the average delay for Airline 2.
B. The results are not statistically significant so there is not enough evidence to conclude that the
average delays are different.
C. The average delay for Airline 1 is at least 3 minutes less than for Airline 2.
D. None of the above.
KEY: A
Questions 92 to 96: A researcher wants to assess if the average age when women first marry has increased
from 1960 to 1990. Let 1 = average age of first marriage for women in 1990 and  2 = average age of
first marriage for women in 1960. A random sample of 10 women married in 1990 showed an average age
at marriage of 24.95 years, with a sample standard deviation of 2 years. A random sample of 20 women
married in 1960 showed an average age at marriage of 23.1 years, with a sample standard deviation of 1.5
years. Assume that age of first marriage for women is normally distributed, but do not assume the
population variances are equal. Use the conservative “by hand” estimate for the degrees of freedom.
92. What are the appropriate null and alternative hypotheses?
A. H0: 1   2 = 0 and Ha: 1   2  0
B. H0: 1   2  0 and Ha: 1   2 = 0
C. H0: 1   2 = 0 and Ha: 1   2 < 0
D. H0: 1   2 = 0 and Ha: 1   2 > 0
KEY: D
93. What is the value of the test statistic?
A. t = 1.80
B. t = 2.00
C. t = 2.33
D. t = 2.58
KEY: D
94. What is the p-value?
A. p-value = 0.015
B. p-value = 0.022
C. p-value = 0.038
D. None of the above
KEY: A
95. For a significance level of  = 0.05, are the results statistically significant?
A. No, results are not statistically significant because the p-value < 0.05.
B. Yes, results are statistically significant because the p-value < 0.05.
C. No, results are not statistically significant because the p-value > 0.05
D. Yes, results are statistically significant because the p-value > 0.05.
KEY: B
96. Which of the following is an appropriate conclusion?
A. The average age of first marriage for women in 1990 does appear to be greater than the average
age in 1960.
B. The results are not statistically significant so there is not enough evidence to conclude that the
average age in 1990 is greater than the average age in 1960.
C. The average age of marriage is at least 23 years old for both 1960 and 1990.
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D. None of the above.
KEY: A
97. In each of the following cases, we wish to test the null hypothesis H0: μ =10 vs. Ha: μ ≠ 10. In which
case can this null hypothesis not be rejected at a significance level α = 0.05?
A. 90% confidence interval for μ is (9 to 12).
B. 95% confidence interval for μ is (11 to 21).
C. 98% confidence interval for μ is (4.4 to 9.4).
D. 99% confidence interval for μ is (5 to 9).
KEY: A
98. In each of the following cases, we wish to test the null hypothesis H0: μ =18. In which case can this
null hypothesis be rejected at a significance level α = 0.05?
A. 95% confidence interval for μ is (16 to 21), Ha: μ ≠ 18.
B. 90% confidence interval for μ is (15 to 20), Ha: μ < 18.
C. 90% confidence interval for μ is (19 to 26), Ha: μ > 18.
D. 90% confidence interval for μ is (15 to 22), Ha: μ ≠ 18.
KEY: C
99. Each of the following presents a two-sided confidence interval and the alternative hypothesis of a
corresponding hypothesis test. In which case can we not use the given confidence interval to make
a decision at a significance level α = .05?
A. 95% confidence interval for p1 − p2 is (−0.15 to 0.07), Ha: p1 − p2 ≠ 0.
B. 95% confidence interval for p is (.12 to .28), Ha: p > 0.10.
C. 90% confidence interval for μ is (101 to 105), Ha: μ ≠ 100.
D. 90% confidence interval for μ1 − μ2 is (3 to 15), Ha: μ1 − μ2 > 0.
KEY: C
100. A random sample of 25 college males was obtained and each was asked to report their actual
height and what they wished as their ideal height. A 95% confidence interval for d = average
difference between their ideal and actual heights was 0.8" to 2.2". Based on this interval, which one
of the null hypotheses below (versus a two-sided alternative) can be rejected?
A. H0: d = 0.5
B. H0: d = 1.0
C. H0: d = 1.5
D. H0: d = 2.0
KEY: A
101. As Internet usage flourishes, so do questions of security and confidentiality of personal
information. A survey of U.S. adults resulted in a 95% confidence interval for the proportion of all
U.S. adults who would never give personal information to a company of (0.22, 0.30). Based on this
interval, which one of the null hypotheses below (versus a two-sided alternative) can be rejected?
A. H0: p = 2/7
B. H0: p = 1/3
C. H0: p = 1/4
D. H0: p = 0.26
KEY: B
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Chapter 12
Questions 102 to 107: A small bakery is trying to predict how many loaves of bread to bake daily. They
randomly sample daily sales records from the past year. In these days, the bakery sold between 40 and 80
loaves per day with a 95% confidence interval for the population mean daily loaf demand given by (51,
54).
102. What was the mean number of loaves sold daily for the sample of 50 days?
A. 50
B. 60
C. 52.5
D. Cannot be determined.
KEY: C
103. How would a 99% confidence interval compare to the 95% confidence interval?
A. It would be wider.
B. It would be narrower.
C. It would be the same width.
D. Cannot be determined.
KEY: A
104. Based on this interval, what is the decision for testing H0:  = 50 versus Ha:   50 at the 5%
significance level?
A. Reject the null hypothesis.
B. Fail to reject the null hypothesis.
C. Cannot be determined.
KEY: A
105. Based on this interval, what is the decision for testing H0:  = 50 versus Ha:   50 at the 10%
significance level?
A. Reject the null hypothesis.
B. Fail to reject the null hypothesis.
C. Cannot be determined.
KEY: A
106. Based on this interval, what is the decision for testing H0:  = 52 versus Ha:   52 at the 5%
significance level?
A. Reject the null hypothesis.
B. Fail to reject the null hypothesis.
C. Cannot be determined.
KEY: B
107. Based on this interval, what is the decision for testing H0:  = 52 versus Ha:   52 at the 10%
significance level?
A. Reject the null hypothesis.
B. Fail to reject the null hypothesis.
C. Cannot be determined.
KEY: C
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Chapter 12
108. For what problem is a one-sample t-test used?
A. To test a hypothesis about a proportion.
B. To test a hypothesis about a mean.
C. To test a hypothesis about the difference between two means for independent samples.
D. To test a hypothesis about the difference between two proportions for independent samples.
KEY: B
109. For what problem is a two-sample t-test used?
A. To test a hypothesis about a mean.
B. To test a hypothesis about the mean difference for paired data.
C. To test a hypothesis about the difference between two means for independent samples.
D. To test a hypothesis about the difference between two proportions for independent samples.
KEY: C
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MIXED MULTIPLE CHOICE PROBLEMS (Answers follow below)
1 – 5: In order to comply with the Environmental protection Agency (EPA) regulations of the Clean Water
Act, a large agricultural company wants to know the average nitrogen concentration in the soil of an
agricultural region it plans to purchase. The seller claims that the average nitrogen level does not exceed
0.49 units. To test this claim at 0.05 level of significance, nitrogen concentration of soil samples were
recorded at 51 sites in that agricultural region. The sample mean was found to be 0.505 and the sample
standard deviation 0.12.
1. State the null and the alternative hypotheses required to test the seller’s claim.
(a) H 0 :  .49, H a :  <.49
(b) H 0 :  <.49, H a :   .49
(c)
.49
(d)
H 0 :   .49, H a :  >.49
(e)
H 0 :  =.49, H a :  
H 0 :  <.49, H a :   .49
2. Which of the following produces a type II error?
(a) The mean nitrogen concentration  is .48 and the null hypothesis is retained
(b) The mean nitrogen concentration  is .52 and the null hypothesis is retained
(c) The mean nitrogen concentration  is .52 and the null hypothesis is rejected
(d) The mean nitrogen concentration  is .48 and the null hypothesis is rejected
(e) The mean nitrogen concentration  is .49 and the null hypothesis is retained
3. The value of the test statistic is
(a) -0.64 (b) -0. 475
(c) -0.89 (d) 0.89 (e) 0.12
4. The p-value is
(a) greater than .103
(d) between .012 and .025
(b) between .07 and .103
(e) less than 0.002
(c) between .039 and .05
5. What decision is reached at level  = .05?
(a) retain the null hypothesis
(b) reject the null hypothesis in favor of the alternative
6. Which of the following is an appropriate interpretation of the decision reached in the hypothesis
testing?
(a) These data show that the seller’s claim is wrong.
(b) These data show that the seller’s claim is correct.
(c) These data do not show sufficient evidence against seller’s claim.
7 – 9: To test the hypothesis
H 0 :  ≤20 versus H a :  > 20, a random sample of n = 30 units is drawn
from the population resulting in the sample mean 22.7 and sample standard deviation 5.4.
7. This is an example of a
(a) one-tailed test
(c) two-tailed test
8. Calculate the value of the test statistic.
(a) 5.7
(b) -2.7
(c) .5
(d) 2.96 (e) 2.74
9. What decision is reached at 0.05 level of significance?
(a) retain the null hypothesis
(b) reject the null hypothesis
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Chapter 12
10 – 13: A study was undertaken to investigate the claim of an insurance company that 15% of all claims
submitted are fraudulent. A sample of 100 claims revealed that 12 claims were fraudulent. Use this
sample to test H 0 : p = 0.15, H a : p 0.15 at 0.05 level of significance.
10. The test statistic is
(a) .84
(b) -.84
(c) .93
11. The p-value is approximately
(a) .20
(b) .40
(c) .18
12. The decision is
(a) reject the null hypothesis
(d) -.93
(e) .03
(d) .36
(e) .80
(b) retain the null hypothesis
13. If the 95% confidence interval for the population proportion was constructed based on this sample, it
would have contained 0.15.
(a) true
(b) false
14 – 16: A random sample of 45 packages of hot dogs produced by company 1 has 9 packages
failing to meet label weight requirements. A random sample of 50 packages of the same type
produces by company 2 has 12 packages failing to meet label weight requirements. Regard these
as independent random samples in answering the questions below. Let p1 and p2 denote the
respective population proportions.
13. The 98% CI estimate of the difference in population rates p1 - p2 is -.04 
(a) .08
(b) .25
(c) .20
(d) .12
(e) .42
14. For testing of H0: p1 - p2 = 0 v. Ha: p1 - p2 ≠ 0, the value of the test statistic is
(a) -2.03
(b) -1.96
(c) -2.16
(d) -0.89
(e) -0.47
15. The p-value for testing H0: p1 - p2 = 0 v. Ha: p1 - p2 ≠ 0, is
(a) .68
(b) .32
(c) .64
(d) .39
(e) .02
Answers:
1. d The expression in the null hypothesis should have “=”.
2. b The type II error is retaining false null hypothesis.
t
0.505  0.49
.
0.12 / 51
3.
d
4.
5.
6.
7.
a Using table A.3, the value of the test statistic is less than 1.28 (smallest value found in the
table). For 1.28, we find .103 in the 50 df row This is the amount in the right tail cut off by 1.28.
The amount in the right tail cut off by 0.89 is greater than .103. The amount in the tail that
gives the p-value would be then greater than 0.103.
a Since p-value is greater than 0.05, we retain the null hypothesis.
c We retain the null hypothesis, or fail to reject it, but do not prove the null hypothesis.
a The inequality in the alternative hypothesis corresponds to one tail, namely right tail.
8.
e
9.
b df=29, from the table A.3 the p-value is between .003 and .008. Since p-value is less than
0.05, reject the null hypothesis.
t
22.7  20
 2.74 .
5.4 / 30
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Chapter 12
10. b
z
0.12  0.15
0.15(1  0.15) / 100
11. b Look up the value of the test statistic on table A.1 to get p-value=2*0.2005=0.401.
12. b Since p-value is greater than 0.05, retain the null hypothesis.
13. a Since the null hypothesis was retained, it is possible that p=.15; the value .15 would be
inside the confidence interval.
14. c z*=2.33, margin of error= 2.33
(0.2)(0.8) (.24)(.76)

45
50
15. e The estimate of the common proportion (under the null hypothesis) is
pˆ  (9  12) /( 45  50)  0.22 .
The test statistic is t 
0.2  0.24
1 
 1
0.22(1  .22)  
 45 50 
.
16. c p-value=2*.3192
MORE MIXED MULTIPLE CHOICE PRACTICE EXERCISES (Answers follow below)
1. A researcher wants to show that a new teaching method increases the average number of words a child
can read per minute, namely that  is greater than 50. The appropriate alternative hypothesis H a is
(a)   50
(b)   50
(c)  >50
(d)  < 50
(e)   50
2 - 6. In order to comply with the new EPA regulations of the Clean Water Act, a large agricultural
company wants to know the nitrogen level in the soil of an agricultural region it plans to purchase. The
seller claims that the average nitrogen level does not exceed 0.49. To test this claim, nitrogen
concentration of soil samples were recorded at 51 sites in that agricultural region.
2. State the null and the alternative hypotheses required to test the seller’s claim.
(a) H 0 :  .49, H a :  <.49
(b) H 0 :  <.49, H a :   .49
(c)
.49
(d)
H 0 :   .49, H a :  >.49
(e)
H 0 :  =.49, H a :  
H 0 :  <.49, H a :   .49
3. Which of the following results produces type II error?
(f) The mean nitrogen concentration  is .48 and the null hypothesis is retained
(g) The mean nitrogen concentration  is .52 and the null hypothesis is retained
(h) The mean nitrogen concentration  is .52 and the null hypothesis is rejected
(i) The mean nitrogen concentration  is .48 and the null hypothesis is rejected
(j) The mean nitrogen concentration  is .49 and the null hypothesis is retained
4. Suppose that from a sample of size 51 the sample mean was found to be 0.505 and the sample standard
deviation 0.12. The observed value of the test statistic is
(a) -.64
(b) -. 475
(c) -.89 (d) .89
(e) .12
5. For the sample considered in #19, the p-value is
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Chapter 12
(a) between .1 and .2
(b) between .1 and .05
(c) between .05 and .025
(d) between .025 and .01
(e) between .01 and .005
6. For the sample considered in #19, what decision is reached at level  = .05?
(a) retain the null hypothesis
(b) reject the null hypothesis in favor of the alternative
7 - 9. A researcher wants to test the hypothesis
H 0 :   20 versus H a :  > 20. Suppose that the
population standard deviation is known to be 5.4. A random sample of
population resulting in y = 22.7.
7. This is an example of a
(a) left-tailed test
(b) right-tailed test
(c) two-tailed test
8. Calculate the value of the test statistic.
(a) 5.7
(b) 2.7
(c) .5
n = 35 units is drawn from the
(d) 2.96 (e) –2.7
9. What decision is reached at 1% level of significance?
(a) retain the null hypothesis
(b) reject the null hypothesis
10 – 12: Consider a t-test with 15 degrees of freedom and the observed value of the test statistic 2.131.
10. If this is a right-tailed test, then the p-value is
(a) .025 (b) .05
(c) .95
(d) .975 (e) .1
11. If this is a left-tailed test, then the p-value is
(a) .025 (b) .05
(c) .95
(d) .975 (e) .1
12. If this is a two-tailed test, then the p-value is
(a) .025 (b) .05
(c) .95
(d) .975 (e) .1
13 - 15. A study was undertaken to investigate the claim of an insurance company that 15% of all claims
submitted are fraudulent. A sample of 100 claims revealed that 12 claims were fraudulent. Use this
sample to test H 0 :  = .15, H a :  .15 at 2% level of significance.
13. The test statistic is
(a) .84
(b) -.84
(c) .93
14. The p-value is approximately
(a) .20
(b) .40
(c) .18
15. The decision is
(a) reject the null hypothesis
(d) -.93
(e) .03
(d) .36
(e) .80
(b) retain the null hypothesis
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Chapter 12
ANSWERS
1.
2.
3.
4.
5.
c The expression in the alternative hypothesis should not contain “=”.
d The expression in the null hypothesis should have “=”.
b The type II error is retaining false null hypothesis.
d t_obs=(0.505-0.49)/(0.12/sqrt(51)) .
a The value of the test statistic falls between 0.849 and 1.299 (50 df on the t-table), the amount in
right tail is between 0.1 and 0.2.
6. a Since p-value is greater than 0.05, we retain the null hypothesis.
7. b The inequality in the alternative hypothesis corresponds to a right tail.
8. d Since sigma is known , use z_obs=(22.7-20)/(5.4/sqrt(35)) .
9. b P-value=1-0.9985=0.0015. Since p-value is less than 0.01, reject the null hypothesis.
10. a Look up 2.131 on the t0table with 15 df.
11. d 1-0.025 .
12. b 0.025*2 .
13. b
z obs 
0.12  0.15
0.15(1  0.15) / 100
14. b Look up the value of the test statistic on the z-table to get p-value=2*0.2005=0.401.
15. b Since p-value is greater than 0.02, retain the null hypothesis.
More! More! More! Example Multiple Choice Questions
1-2. For each of the following testing situations, give the p-value of the observed test
statistic. Pick the closest.
1. Two-sample t-test of H0: 1 - 2  0 versus Ha: 1 - 2 < 0, sample sizes n1 = 12,
n2 = 10, tobs = 1.38. Adopt the conservative approach to the degrees of freedom.
(a) .10
(b) .05
(c) .20
(d) .50
(e) .90
2. Two-sample (large samples) z-test of H0: 1 - 2  0 v. Ha: 1 - 2 > 0 if p1 = .47,
p2 = .33 and zobs = 1.15.
(a) .085
(b) .071
(c) .054
(d) .125
(e) .101
3-8. A veterinarian believes that the rate of hip dysplasia in Boxers (breed of dog) is less
than the rate hip dysplasia in American Bulldogs (breed of dog) and decides to test the
hypothesis H0: 1-2  0 v. Ha: 1-2 < 0. A random sample of 80 Boxers has 8 dogs
with hip dysplasia. A random sample of 140 American Bulldogs has 21 dogs with hip
dysplasia.
3. What is the standard error of the estimate p1 – p2 used in calculating the CI for 1
- 2?
(a) .085
(b) .071
(c) .045
(d) .037
(e) .101
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4. What is the 99% CI estimate of the difference in population rates 1 - 2?
(a) -.05  .12 (b) -.05  .07 (c) -.05  .09 (d) -.05  .20 (e) -.05  .17
5. What is the standard error of the estimate p1 – p2 used in calculating the z-test
statistic? (Hint: This is the denominator of z-test statistic.)
(a) .085
(b) .071
(c) .064
(d) .047
(e) .101
6. Compute the value zobs of the z-test statistic.
(a) 0.52
(b) -0.90
(c) 1.96
(d) –1.06
(e) -1.96
7. What is the p-value of the test statistic zobs that is appropriate for this test? Pick
the closest.
(a) 0.30
(b) 0.18
(c) 0.14
(d) 0.05
(e) 0.025
8. What decision is reached at 2% level of significance?
(a) reject the null hypothesis
(b) retain the null hypothesis
9-13. Fifty-two healthy women subjects ages 55 through 60 take part in a study to
compare the mean gain in bone density from two different regimens involving calcium
supplements and estrogen. Regimen A was administered to 29 subjects chosen at
random and the remaining 23 received regimen B. Below is the Minitab output for the
basic statistics on the measurements of bone density gain for the two independent
samples. Adopt the conservative approach to the degrees of freedom.
Descriptive Statistics
Variable
n Mean
StDev
Regimen A 29 16.3
8.0
Regimen B 23 12.1
5.0
9. What is the confidence factor for the 95% CI estimate of A - B ?
(a) 2.074
(b) 1.960
(c) 2.508
(d) 2.048
(e) 2.009
10. What is the 95% CI estimate of A - B?
(a) 4.2  2.7 (b) 4.2  4.1 (c) 4.2  5.6 (d) 4.2  3.8 (e) 4.2  1.7
11. What is the value of tobs for testing H0: A - B = 0 versus Ha: A - B  0?
(a) 2.77
(b) 2.31
(c) 3.13
(d) 1.96
(e) 4.20
12. The p-value of tobs for testing H0: A - B = 0 versus Ha: A - B  0 is
(a) between .01 and .025 (b) between .1 and .2
(c) between .02 and .05
(d) between .001 and .005 (e) between .002 and .01
13. What decision is reached at 1% level of significance?
(a) reject the null hypothesis
(b) retain the null hypothesis
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14. A clean standard requires that vehicle exhaust not exceed specified limits for various
pollutants. Many states require that cars be tested annually to be sure they meet these
standards. Suppose state regulators double check a random sample of cars that a
suspect repair shop has certified as okay. They will revoke the shop’s license if they find
significant evidence that the shop is certifying vehicle that do not meet standards.
The appropriate hypotheses are:
A. Null hypothesis: The regulators decide that the shop is meeting standards.
Alternative hypothesis: The regulators decide that the shop is not meeting
standards.
B. Null hypothesis: The shop’s emission standards are higher than for other shops.
Alternative hypothesis: The shop’s emission standards are not higher than for other
shops.
C. Null hypothesis: The shop is meeting the emissions standards.
Alternative hypothesis: The shop is not meeting the emission standards.
D. Null hypothesis: The repair shop’s license is revoked.
Alternative hypothesis: The repair shop’s license is not revoked.
QUESTIONS 15 - 17
The National center for Education Statistics monitors many aspects of elementary
and secondary education nationwide. Their 1996 numbers are often used as a
baseline to assess changes. In 1996 31% of students reported that their mothers had
graduated from college. In 2000, responses from 8368 students found that this figure
had grown to 32%. Is this evidence that there has been an increase in education level
among mothers?
15. The appropriate hypotheses are:
A. Ho: p = 0.31; HA: p > 0.31
B. Ho: p>0.31; HA: p<0.31
C. Ho: p = 0.32; HA: p  0.32
D. Ho: p<0.31; HA: p>0.32
16. Let all necessary conditions be satisfied. Find the test statistic (z), and the Pvalue.
A. z = 1.978, P-value = 0.0047
B. z = 1.978, P-value = 0.024
C. z = 0.0051, P-value = -1.978
D. z = 0.048, P-value = 1.978
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17. State your conclusion.
A. With a P-value of 1.978, we fail to reject the null hypothesis.
B. No conclusion can be drawn because the P-value is too big to be considered in this
context.
C. With a P-value of 0.024, we reject the null hypothesis. There is evidence to
suggest that the percentage of students whose mothers are college graduates has
changed since 1996. In fact, the evidence suggests that the percentage has
increased.
D. The confidence level was not given, so no conclusion can be arrived at.
18. National data in the 1960s showed that about 44% of the adult population had
never smoked cigarettes. In 1995 a national health survey interviewed a random
sample of 881 adults and found that 52% had never been smokers.
(a) Create a 95% confidence interval for the proportion of adults (in 1995) who had
never been smokers.
(b) Does this provide evidence of a change in behavior among Americans? Using
your confidence interval, test an appropriate hypothesis and state your
conclusion.
ANSWERS
(a) (0.487, 0.553) (b) Null Hypo: p = 0.44; Alter Hypo: p not equal to 0.44. Since 44%
is not in the 95% CI, we will reject the Null Hypo. There is strong evidence that, in
1995, the percentage of adults who have never smoked was not 44%.
19. A company hopes to improve customer satisfaction, setting as a goal no more than
5% negative comments. A random survey of 350 customers found only 10 with
complaints.
(a) Create a 95% confidence interval for the true level of dissatisfaction among
customers.
(b) Does this provide evidence that the company has reached its goal? Using your
confidence interval, test an appropriate hypothesis and state your conclusion.
ANSWERS
(a) (0.011, 0.046) (b) Null Hypo: p = 0.05; Alter. Hypo: p < 0.05. Since 5% is not in the
95% CI, we will reject the Null Hypo. There is strong evidence that less that 5% of
customers have complaints.
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20. A company with a fleet of 150 cars found that the emissions systems of 7 out of
22 they tested failed to meet pollution control guidelines. Is this strong evidence that
more than 20% of the fleet might be out of compliance? Test an appropriate
hypothesis and state your conclusion. Be sure the appropriate assumptions and
conditions are satisfied before you proceed.
ANSWER
Null Hypo: p = 0.20; Alter. Hypo: p > 0.20. Two conditions are not satisfied
(verify!!!!!!!) We cannot proceed with a hypothesis test.
21. In a rural area only about 30% of the wells that are drilled find adequate water at a
depth of 100 feet or less. A local man claims to be able to find water by “dowsing” –
using a forked stick to indicate where the well should be drilled. You check with 80 of
his customers and find that 27 have wells less than 100 feet deep. What do you conclude
about this claim? (We consider a P – value of around 5% to represent strong evidence)
(a) Write appropriate hypotheses.
(b) Check the necessary assumptions.
(c) Perform the mechanics of the test. What is the P – value?
(d) Explain carefully what the P – value means in this context.
(e) What’s your conclusion?
FOR SOLUTION TO PROBLEM 21, SEE CLASS NOTES
22. In the 1980s it was generally believed that congenital abnormalities affected about
5% of the nation’s children. Some people believe that the increase in the number of
chemicals in the environment has led to an increase in the incidence of abnormalities. A
recent study examined 384 children and found that 46 of them showed signs of an
abnormality. Is this strong evidence that the risk has increased? (We consider a P –
value of around 5% to represent strong evidence)
(a) Write appropriate hypotheses.
(b) Check the necessary assumptions.
(c) Perform the mechanics of the test. What is the P – value?
(d) Explain carefully what the P – value means in this context.
(e) What’s your conclusion?
(f) Do environmental chemicals cause congenital abnormalities?
FOR SOLUTION TO PROBLEM 22, SEE CLASS NOTES
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QUESTIONS 23 – 25
The National Center for Education Statistics monitors many aspects of elementary and
secondary education nationwide. Their 1996 numbers are often used as a baseline to
assess changes. In 1996, 34% of students had not been absent from school even once
during the previous month. In the 2000 survey, responses from 8302 students showed
that this figure had slipped to 33%. Officials would of course be concerned if student
attendance were declining. Do these figures give evidence of a decrease in student
attendance?
23. Write appropriate hypotheses.
A.
B.
C.
D.
E.
Ho : p = 0.33;
Ho : p > 0.34;
Ho : p = 0.34;
Ho : p < 0.33;
Ho : p = 0.34;
HA: p < 0.33
HA: p < 0.34
HA: p < 0.34
HA: p = 0.34
HA: p = 0.33
24. Perform the test by finding the Test statistics and the P – value.
A.
B.
C.
D.
E.
Zo = 0.0052; P – value = 0.027
Zo = - 1. 923; P – value = 0.027
Zo = 0.048; P – value = 0.0052
Zo = 0.01; P – value = 0.048
Zo = 1.923; P – value = 0.027
25. State your conclusion.
A. With a P – value of 0.027, we fail to reject the null hypothesis. There is no evidence
to suggest that the percentage of students with perfect attendance in the previous
month has decreased in 2000.
B. With a P – value of 0.0052, we reject the null hypothesis. There is evidence to
suggest that the percentage of students with perfect attendance in the previous
month has decreased in 2000.
C. With a P – value of 0.048, essentially 0.05, equal to the significance level, we retain
the null hypothesis. There is no evidence to suggest that the percentage of
students with perfect attendance in the previous month has decreased in 2000.
D. With a P – value of 0.027, we reject the null hypothesis. There is evidence to
suggest that the percentage of students with perfect attendance in the previous
month has decreased in 2000.
E. With a P – value of 0.027, we retain the null hypothesis. There is evidence to
suggest that the percentage of students with perfect attendance in the previous
month has decreased in 2000.
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26. Highway safety engineers test new road signs hoping that the increased reflectivity will make
them more visible to drivers. Volunteers drive through a test course with several of the new –
and old – style signs and rate which kind shows up the test.
(i). Write the appropriate hypotheses.
(ii) In this context, what would a Type I error be?
(iii) In this context, what would a Type II error be?
(iv) In this context, describe what is mean by the power of the test.
(v) Is this a one – tailed (left – tailed or right – tailed) or a two – tailed test? Why?
(vi) If the hypothesis is tested at the 1% level of significance instead of 5%, how will this affect
the power of the test?
(vii) The engineers hoped to base their decision on the reactions of 50 drivers, but time and
budget constraints may force them to cut back to 20. How would this affect the power of the
test? Explain.
27. A company with a large fleet of cars hopes to keep gasoline costs down and sets a goal of
attaining a fleet average of at least 26 miles per gallon. To see if the goal is being met, they
check the gasoline usage for 50 company trips chosen at random, finding a mean of 25.36 mpg
and a standard deviation of 4.21 mpg. Is this strong evidence they have failed to attain their fuel
economy goal?
(i) Write appropriate hypotheses.
A.
B.
C.
D.
E.
H0: µ = 26;
H0: µ > 26;
H0: µ < 26;
H0: µ = 26;
H0: µ = 26;
HA: µ > 26
HA: µ = 26
HA: µ = 26
HA: µ < 26
HA: µ ≠ 26
(ii) Are the necessary assumptions to perform inference satisfied?
A.
B.
C.
D.
E.
No, the 10% condition is not satisfied.
No, the randomization condition is not satisfied.
No, the nearly normal condition is not satisfied.
No, the success failure condition is not satisfied.
Yes, all the necessary assumptions and conditions are satisfied or can be assumed to
be satisfied.
(iii) Find the P – value (round to three decimal places) and explain what the P – value means in
this context.
A. P – value = 0.144; The P – value is the probability of obtaining a sample mean of 25.36
mpg or more if the mean mileage of the fleet is 26 mpg.
B. P – value = 1.07494; The P – value is the probability of obtaining a sample mean of
25.36 mpg or less if the mean mileage of the fleet is 26 mpg.
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C. P – value = - 1.07494; The P – value is the probability of obtaining a sample mean of
25.36 mpg or less if the mean mileage of the fleet is 26 mpg.
D. P – value = 0.59538; The P – value is the probability that the mean mileage of the fleet is
greater than or equal to the proposed mean of 26 mpg.
E. P – value = 0.144; The P – value is the probability of obtaining a sample mean of
25.36 mpg or less if the mean mileage of the fleet is 26 mpg.
(iv) State an appropriate conclusion.
A. Since the P – value = 0.144, we reject the null hypothesis. There is little evidence to
suggest that the mean mileage of cars in the fleet is less than 26 mpg.
B. Since the P – value = 0.144, we fail to reject the null hypothesis. There is little
evidence to suggest that the mean mileage of cars in the fleet is less than 26 mpg.
C. Since the P – value = - 1.07494, a negatively high value, we fail to reject the null
hypothesis. There is little evidence to suggest that the mean mileage of cars in the fleet is
less than 26 mpg.
D. Sine the P – value = 1.07494, a positively high value, we fail to reject the null hypothesis.
There is little evidence to suggest that the mean mileage of cars in the fleet is less than 26
mpg.
E. Since the P – value = 0.59538, a relatively small value, we reject the null hypothesis.
There is strong evidence to suggest that the mean mileage of cars in the fleet is more than
26 mpg.
28. Production managers on an assembly line must monitor the output to be sure that
the level of defective products remains small. They periodically inspect a random
sample of the items produced. If they find a significant increase in the proportion of
items that must be rejected, they will halt the assembly process until the problem can be
identified and repaired.
(a) In this context, what is a Type I error?
(b) In this context, what is a Type II error?
(c) Which type of error would the factory owner consider more serious?
(d) Which type of error might customers consider more serious?
29. Consider again the task of the quality control inspectors in Exercise 28.
(a) In this context, what is meant by the power of the test the inspectors conduct?
(b) They are currently testing 5 items each hour. Someone has proposed that they test 10
each hour instead. What are the advantages and disadvantages of such a change?
(c) Their test currently uses a 5% level of significance. What are the advantages and
disadvantages of changing to an alpha level of 1%?
(d) Suppose that, as a day passes, one of the machines on the assembly line produces
more and more items that are defective. How will this affect the power of the test?
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30. A governor is concerned about his “negatives” – the percentage of state residents
who express disapproval of his job performance. His political committee pays for a
series of TV ads, hoping that they can keep the negatives below 30%. They will use
follow – up polling to assess the ad’s effectiveness. After the political campaign, the
pollsters check the governor’s negatives. They test the hypothesis that the ads produced
no change against the alternative that the negatives are now below 30%, and find a P –
value of 0.22. What conclusion is appropriate?
A. There is a 22% chance that the ads worked.
B. There is a 78% chance that the ads worked.
C. There is a 22% chance that the poll they conducted is correct.
D. There is a 22% chance that natural sampling variation could produce poll results
like these if there is really no change in public opinion.
E. None of the above is appropriate.
31. The seller of a loaded die claims that it will favor the outcome 6. We don’t believe
that claim, and roll the die 200 times to test an appropriate hypothesis. Our P – value
turns out to be 0.03. Which conclusion is appropriate?
A. There’s a 3% chance that the die is fair.
B. There’s a 97% chance that the die is fair.
C. There’s a 3% chance that a loaded die could randomly produce the results we
observed, so it’s reasonable to conclude that the die is fair.
D. There’s a 3% chance that a fair die could randomly produce the results we
observed, so it’s reasonable to conclude that the die is loaded.
E. None of the above.
32. A clean air standard requires that vehicles exhaust emissions not exceed specified
limits for various pollutants. Many states require that cars be tested annually to be sure
they meet these standards. Suppose state regulators double check a random sample of
cars that a suspect repair shop has certified as okay. They will revoke the shop’s license
if they find significant evidence that the shop is certifying vehicles that do not meet
standards.
(i) The appropriate hypotheses are:
(ii) In this context, what is a Type I error?
(iii) In this context, what is a Type II error?
(iv) Which type of error would the shop’s owner consider more serious, and why?
(v) Which type of error might environmentalists consider more serious, and why?
(vi) In this context, what is meant by the power of the test the regulators are
conducting?
(vii) Will the power of the test be greater if they test 20 or 40 cars? Why?
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Chapter 12
(viii) Will the power of the test be greater if they use a 5% or a 10% level of significance?
Why?
(ix) Will the power of the test be greater if the repair shop’s inspectors are only a little
out of compliance or a lot? Why?
(SEE CLASS NOTES FOR SOLUTION)
33. Testing for Alzheimer’s disease can be a long and expensive process, consisting of
lengthy tests and medical diagnosis. Recently, a group of researchers (Solomon et al.,
1998) devised a 7 – minute test to serve as a quick screen for the disease for use in the
general population of senior citizens. A patient who tested positive would then go
through the more expensive battery of tests and medical diagnosis. The authors
reported a false positive rate of 4%f and a false negative rate of 8%.
1. Put this in the context of a hypothesis test. What are the null and alternative
hypotheses?
2. What would a Type I error mean?
3. What would a Type II error mean?
4. Which is worse here, a Type I or a Type II error? Explain.
5. What is the power of the test?
6. Compute the numerical value of the power of the test.
7. Will the power of the test be greater if the medical team tests 50 or 100 Alzheimer’s
patients?
8. Will the power be greater if they use a 5% or a 10% level of significance? Why?
TYPE I, TYPE II ERRORS
SOLUTION TO No. 28
Null Hypo: The assembly line process is working fine; Alter Hypo: The assemble line
process is producing defective items.
(a) Type I error is when the production managers decide that there has been an
increase in the number of defective items and stop the assembly line, when the
assembly process is working fine.
(b) Type II error is when the production managers decide that the assembly process is
working fine, but defective items are being produced.
(c) Type II
(d) Type II
(SEE CLASS NOTES FOR COMPLETE SOLUTION)
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SOLUTION TO No. 33
(a) Null Hypo: The person is healthy; Alter Hypo: The person has A’s disease.
(b) Type I error is a false positive. It has been decided that the person has the A
disease when he has not.
(c) Type II error is a false negative. It has been decided that the person is healthy,
when he actually has A disease.
(d) Type II error.
(SEE CLASS NOTES FOR COMPLETE SOLUTION)
QUESTIONS 34 – 37
In 1960, census results indicated that the age at which American men first married had a
mean of 23.3 years. It is widely suspected that young people today are waiting longer to
get married. We want to find out if the mean age of first marriage has increased during
the past 40 years.
34. Write appropriate hypotheses.
H A :   23.3
A. H o :   23.3;
H A :   23.3
B. H o :   23.3;
H A :   23.3
C. H o :   23.3;
H A :   23.3
D. H o :   23.3;
E. None of the above
35. Describe the approximate sampling distribution model for the mean age in such
samples.
 

A. N  23.3, 
40 


 

B. t 39  23.3,
40 


s 

C. N  23.3,
40 


s 

D. t 39  23.3,
40 

E. None of the above
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36. The men in our sample married at an average age of 24.2 years, with a standard
deviation of 5.3 years. What is the P – value for this result? Explain in context what this
P – value means.
A. P – value = 0.8544; If the mean age at first marriage is still 23.3 years, there is a 0.8544
chance of getting a sample mean of 24.2 years or older simply from natural sampling
variation.
B. P – value = 1.07; If the mean age at first marriage is still 23.3 years, then, the P –
value is what we observe from data due to natural sampling variation.
C. P – value = 0.1456; If the mean age at first marriage is still 23.3 years, there is a
0.1456 chance of getting a sample mean of 24.2 years or older simply from natural
sampling variation.
D. P – value = 0.1456; If the mean age at first marriage is still 23.3 years, there is a 0.1456
chance of getting a sample mean of 24.2 years or younger simply from natural
sampling variation.
E. P – value = 1.07; If the mean age at first marriage is still 23.3 years, then, the P –
value is what we observe in a sample of size n = 40, with mean 24.2 years or older.
37. What is your conclusion?
A. We fail to reject the null hypothesis. There is no evidence to suggest that the
mean age at first marriage has changed from 23.3 years, the mean in 1960, to
24.2 years.
B. We fail to reject the null hypothesis. There is strong evidence to suggest that the
mean age at first marriage has changed from 23.3 years, the mean in 1960, to 24.2
years.
C. We reject the null hypothesis. There is strong evidence to suggest that the mean
age at first marriage has changed from 23.3 years, the mean in 1960, to 24.2 years.
D. We reject the null hypothesis. There is no strong evidence to suggest that the
mean age at first marriage has changed from 23.3 years, the mean in 1960, to 24.2
years.
E. We cannot draw any conclusion in this case because the P – value = 1.07 is
exceedingly big.
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QUESTIONS 38 – 45 (Solution: To be discussed in class)
In 2005 the U.S. Census Bureau reported that 68.9% of American families owned their
homes. Census data reveal that the ownership rate in one small city is much lower. The
city council is debating a plan to offer tax breaks to first-time home buyers in order to
encourage people to become homeowners. They decide to adopt the plan on a 2-year
trial basis and use the data they collect to make a decision about continuing the tax
breaks. Since this plan costs the city tax revenues, they will continue to use it only if
there is strong evidence that the rate of home ownership is increasing.
38. In words, what will their hypotheses be?
39. In this context, what would a Type I error be?
40. In this context, what is a Type II error?
41. Which error will hurt the city council? Explain.
42. Which error will hurt the first-time home buyers? Explain.
43. What would the power of the test represent in this context?
44. Will the power of the test increase if 300 first-time home buyers are considered
instead of 200? Why?
45. Will the power of the test be greater if they use a 5% or a 10% level of significance?
Why?
PRACTICE EXERCISES FROM THE RECOMMENDED TEXTBOOK
CHAPTER 12
Nos. 12.1, 12.3, 12.6 – 12.9, 12.19, 12.21, 12.41, 12.43, 12.46, 12.47, 12.60 – 12.63, 12.74 –
12.75, 12.90,
CHAPTER 13
Nos. 13.1, 13.16 – 13.17, 13.19 – 13.21, 13.23, 13.24, 13.47 – 13.48, 13.51 – 13.53, 13.63 –
13.64, 13.67, 13.69, 13.77,
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