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Stat 250.3 – Practice Quiz 4
Chapter 11
1.
If the results of a hypothesis test give a p-value of 0.03, then with a significance level (alpha) = .05 the
results are said to be
A. Not statistically significant because p  alpha
B. Statistically significant because p  alpha
C. Practically significant because p  alpha
D. Not practically significant because p  alpha
KEY: B
2.
The probability 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
3.
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
4.
The designated level (typically set at .05) for which the p-value is compared 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
Questions 5 to 6: A hypothesis test for a population proportion p is given below:
H0: p = 0.10
Ha: p  010
For each sample size n and sample proportion p̂ compute the value of the z-statistic:
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
5.
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
6.
7.
A drug shows promise as a new treatment for hay fever. The hypotheses below were tested for p =
proportion of patients taking this drug who are cured of hay fever.
H0: p  0.80
Ha: p >0.80
What are the Type 1 and Type 2 errors for this study?
KEY: A Type 1 error is to wrongly conclude that the proportion cured is greater than 80% when in fact the
proportion cured is truly no more than 80%. A Type 2 error is not to conclude that the proportion cured is
greater than 80% when in fact the proportion cured is truly greater than 80%.
Questions 8 to 12: A random sample of 800 high school boys showed that 420 had done strenuous exercise
during the past year. The researcher wants to test the hypotheses below for p = the proportion of all high school
boys who do strenuous exercise.
H0: p  0.50
Ha: p > 0.50
8. State the null hypothesis (H0) and the alternative hypothesis (Ha). in words.
KEY: Null hypothesis: At most 50% of high school boys do strenuous exercise. Alternative hypothesis: More
than 50% of high school boys do strenuous exercise.
9. Verify necessary data conditions for a z-statistic and calculate the z-statistic.
KEY: np0 = n(1-p0) = 400 > 10, so the sample size is large enough. The sample is also a random sample. z =
1.41
10. Assuming the null hypothesis is true, find the p-value.
KEY: p-value = 0.079
11. Are the results statistically significant at a significance level of
KEY: No, the results are not statistically significant.
 = 0.05?
12. Report the conclusion in the context of the situation.
KEY: The evidence is not strong enough to discount the possibility that the proportion of boys who do
strenuous exercise is 50% or less.
Chapter 12
Note: For approximate 95% confidence intervals of any type, use a multiplier (z* or t*) equal to 2!
1.
In a randomly selected sample of 500 college students, 80% do not own laptop computers. Calculate an
approximate 95% confidence interval for the proportion of all students who do not own a laptop computer.
A. (.746, .854)
B. (.782, .818)
C. (.764, .836)
D. None of the above
KEY: C
2.
A random sample of 60 mathematics majors spent an average of $200.00 for textbooks for a term, while a
random sample of 40 literature majors spent an average of $180.00 for textbooks that term. The standard
deviation for each sample was $20.00. Calculate an approximate 95% confidence interval for the difference
in average amounts spent on textbooks (average for math majors minus the average for literature majors).
A. ( 7.8, 32.2)
B. (11.8, 28.2)
C.
(15.9, 24.1)
D. None of the above
KEY: B (Hint: The standard error for the difference between the two sample means is 4.082.)
3.
In a random sample of 250 full time students, 80% preferred to have three one-hour lectures per week
rather than two 90-minute lectures. In a random sample of 500 part-time students, 50% said they preferred
the three one-hour lectures schedule. Calculate an approximate 95% confidence interval for the difference
in population proportions preferring the three one-hour lectures schedule (proportion for full-time students
minus proportion for part-time students).
A. (.199, .401)
B. (.232, .368)
C. (.266, .334)
D. None of the above
KEY: B (Hint: The standard error for the difference between the two sample proportions is 0.0338.)
4.
A randomly selected sample of n =51 men in Brazil had an average lifespan of 59 years. The standard
deviation was 10 years and the standard error was 1.400. Calculate a 98% confidence interval for the
average lifespan for all men in Brazil.
A. (55.6, 62.4)
B. (56.2, 61.8)
C. (35.0, 83.0)
D. None of the above
KEY: A
5.
Random samples from two airlines during normal weather (100 flights from Airline A and 200 flights from
Airline B) showed that .50 of Airline A’s flights were delayed, and .30 of Airline B’s flights were delayed.
The difference between the two proportions is .20, with a standard error of .0596. . What is a 99%
confidence interval for the difference between the population proportions of flights delayed (Airline A 
Airline B)?
A. (.083, .317)
B. (.140, .260)
C. (.199,.204)
D. None of the above
KEY: A – that is not right…
It seems like the right interval is not in the possible answers (A-C). The interval in A is the 95%,
the 99% interval is .2 -/+(2.58)*.0596 = (0.046232, 0.353768).
6.
For a randomly selected sample of 20 new mothers in the year 2000, the mean age was 24.6 years. For a
randomly selected sample of 10 new mothers in 197, the mean age was 21.4 years. The difference between
the mean ages is 3.2 years, and the standard error of the difference is 1.366. Calculate a 90% confidence
interval for the difference in population mean ages of new mothers in the two years (mean age of new
mothers in year 2000 minus mean age of new mothers in year 1970). Assume that the ages of new mothers
are normally distributed, and use the conservative “by hand” (approximate) estimate for the degrees of
freedom.
A. (0.84, 5.56)
B. (0.78, 5.67)
C. (0.70, 5.70)
D. None of the above
KEY: C
7.
Based on a random sample of 41 flights, the average minutes of airborne time for an airline was 106
minutes, with a standard deviation of 5 minutes. What is a 90% confidence interval for the average
airborne time for this airline?
KEY: (104.7, 107.3)
8.
The confidence level for a confidence interval for a mean is
A. the probability the procedure provides an interval that covers the sample mean.
B. the probability of making a Type 1 error if the interval is used to test a null hypothesis about the
population mean.
C. the probability that individuals in the population have values that fall into the interval.
D. the probability the procedure provides an interval that covers the population mean.
KEY: D
9.
In a random sample of 100 students (60 women and 40 men), the average hours of sleep during finals week
were compared. Are the samples of men and women independent samples or paired samples?
KEY: They are independent samples.
10. In a random sample of 100 students, the change in hours of sleep for each student during and after finals
week were compared. Are the samples of students’ sleep during finals and after finals independent samples
or paired samples?
KEY: They are paired samples.
11. A researcher asked random samples of 50 kindergarten teachers and 50 12 th grade teachers how much
money they spent out-of-pocket on school supplies in the previous school year, to see if teachers at one
grade level spend more than the other. A 95% confidence interval for 12 is $30 to $50, where 1 and 2
are the population means for the amounts spent by kindergarten and 12 th grade teachers, respectively. Based
on this result, it is reasonable to conclude that
A. 95% of all kindergarten and 12th grade teachers spend between $30 and $50 on average.
B. 95% of all kindergarten teachers spend between $30 and $50 more then 95% of all 12 th grade
teachers.
C. Kindergarten teachers spend more on average than do 12th grade teachers.
D. 12th grade teachers spend more on average than do kindergarten teachers.
KEY: C
Chapter 13
Questions 1 to 5: An investigator wants to prove that  = 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.
1. 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
2.
What is the value of the test statistic?
A. 1.50
B. 1.65
C. 1.80
D. None of the above
KEY: C
3.
What is the p-value (computer or calculator for t-distribution required)?
A. p = 0.039
B. p = 0.053
C. p = 0.070
D. None of the above
KEY: A
4.
For a significance level of alpha = 0.05, are the results statistically significant?
A. No, results are not statistically significant because p < 0.05.
B. Yes, results are statistically significant because p < 0.05.
C. No, results are not statistically significant because p > 0.05
D. Yes, results are statistically significant because p > 0.05.
KEY: B
5.
Report your conclusion.
A. The results are statistically significant so that the average total weight of all passengers > 185 pounds.
B. The results are statistically significant so that the average total weight of all passengers < 185 pounds.
C. The results are not statistically significant: there is not enough evidence to conclude the mean > 185
pounds.
D. None of the above.
KEY: A
Questions 6 to 10: A counselor wants to show that for men who are married by the time they are 30,  =
average age when the men are married is not 21 years old. A random sample of 10 men who were married by
age 30 showed an average age at marriage of 22.2, with a sample standard deviation of 1.9 years. Assume that
the age at which this population of men get married for the first time is normally distributed.
6.
What are the appropriate null and alternative hypotheses?
A. H0:  = 21 and Ha:  < 21
B. H0:  = 21 and Ha:  >21.
C. H0:  = 21 and Ha:   21.
D. H0:   21 and Ha:  = 21.
KEY: C
7.
What is the value of the test statistic?
A. 1.80
B. 2.00
C.
2.33
D. None of the above
KEY: B
8.
What is the p-value (computer or calculator for t-distribution required)?
A. p = 0.022
B. p = 0.053
C. p = 0.076
D. None of the above
KEY: C
9.
For a significance level of alpha = 0.05, are the results statistically significant?
A. No, results are not statistically significant because p < 0.05.
B. Yes, results are statistically significant because p < 0.05.
C. No, results are not statistically significant because p > 0.05
D. Yes, results are statistically significant because p > 0.05.
KEY: C
10. Report your conclusion.
A. The results are statistically significant so that the average age > 21.
B. The results are statistically significant so that the average age < 21.
C. The results are not statistically significant: there is not enough evidence to conclude average age is
different from 21.
D. None of the above.
KEY: C
Questions 11 to 15: 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 (.50) of Airline A’s flights were delayed, and 70 out of 200
(.35) of Airline B’s flights were delayed.
11. What are the appropriate null and alternative hypotheses?
A.
B.
C.
D.
KEY: A
H0: p1 – p2 = 0 and Ha: p1 – p2  0
H0: p1 – p2  0 and Ha: p1 – p2 =0
H0: p1 – p2 = 0 and Ha:p1 – p2 < 0
H0: p1 – p2= 0 and Ha:p1 – p2 > 0
12. What is the value of the test statistic?
A. 0.79
B. 2.00
C.
2.50
D. None of the above
KEY: C
13. What is the p-value (computer or calculator for z-distribution required)?
A. p = 0.2148
B. p = 0.0124
C. p = 0.0456
D. None of the above
KEY: B
14. For a significance level of alpha = 0.05, are the results statistically significant?
A. No, results are not statistically significant because p < 0.05.
B. Yes, results are statistically significant because p < 0.05.
C. No, results are not statistically significant because p > 0.05
D. Yes, results are statistically significant because p > 0.05.
KEY: B
15. Report your conclusion.
A. The proportion of delayed flights after a storm for Airline 1 is 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. None of the above.
KEY: A
Questions 16 and 17: 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.
16. Which of the following is the correct way to state the null hypothesis?
A. p = 0.5
B. x1  x 2  0
C. p1 - p2 = 0
D. 1 - 2 = 0
KEY: D
17. A statistical test is done and 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.
D. The probability is 0.225 that men and women have the same mean nose length.
KEY: B
Questions 18 and 19: A newspaper article reported on a study comparing the proportions experiencing pain
relief using a new medication compared to using a placebo.
18. The researcher is quoted as saying that there was less than a 1 in 100 probability that the observed
difference in proportions was due to chance. In statistical terms, the researcher is saying the results are
____.
A. probably due to non-sampling errors.
B. probably due to chance.
C. statistically significant using a .01 level of significance.
D. not statistically significant using a .01 level of significance.
KEY: C
19. The null hypothesis for this study is:
A.
p1  p2  0
B.
C.
D.
KEY: A
p1  p2  0
pˆ 1  pˆ 2  0
pˆ 1  pˆ 2  0
20. 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
21. An airport official wants to prove that p1 = proportion of delayed flights after a storm for Airline 1 was
different from p2 = proportion of delayed flights for Airline 2. Random samples from two airlines after a
storm showed that 50 out of 100 (.50) of Airline A’s flights were delayed, and 80 out of 200 (.40) of
Airline B’s flights were delayed. What is the test statistic and p-value for testing whether the proportion of
delayed flights are the same or different?
KEY: The test statistic z = 1.65 p-value = 0.099
22. The average time in years to get an undergraduate degree in computer science was compared for men and
women. Random samples of 100 male computer science majors and 100 female computer science majors
were taken. Choose the appropriate parameter(s) for this situation.
A. One population proportion p.
B. Difference between two population proportions p 1  p2.
C. One population mean  1
D. Difference between two population means
KEY: D
Inference
Parameter Statistic Standard Error
1   2
Multiplier
Test Statistic
One Mean
(1-sample t)
Difference of
two means
(2-sample t)
One proportion
s
or
sd
µ or µd
x or d
µ1-µ2
x1  x 2
s12 s 2 2

n1 n2
p̂
pˆ (1  pˆ )
n
P
Difference of
two proportions p1-p2
pˆ 1  pˆ 2
n
n
pˆ 1(1  pˆ 1) pˆ 2(1  pˆ 2)

n1
n2
t*
t
df=n-1
t*
x  0
d 0
or t 
sd
s
n
n
t
df=min(n1-1, n2-1)
z
z*
z*
z
( x1  x 2 )  0
s1 2 s 2 2

n1 n 2
pˆ  p0
p0 (1  p0 )
n
pˆ 1  pˆ 2
pˆ (1  pˆ ) pˆ (1  pˆ )

n1
n2
,
Summary Table for Statistical Techniques: It will be provided in the quiz.
pˆ 
n1 pˆ1  n2 pˆ 2
n1  n2