Download Ch 8 – One Population Confidence Intervals σz

Ch 8 – One Population Confidence Intervals
Section A: Multiple Choice
__C___1. A single number used to estimate a population parameter is
a. the confidence interval
b. the population parameter
c. a point estimate
d. the mean of the population
___a _ 2. A range of values constructed from sample data so that the parameter occurs within that range at a specified
probability is
a. a confidence interval
b. the population parameter
c. a point estimate
d. the mean of the population
__b___3. Suppose we select 1000 samples from a population. For each sample we construct a 95 percent confidence
interval. We would expect about 95 percent of these confidence intervals to contain
a. a point estimate
b. the population parameter
c. a sample mean
d. the size of the population
___C__4. If the level of confidence is decreased from 95 to 90 percent, the width of the corresponding interval will
a. be increased
b. stay the same
c. be decreased
d. not have an effect on the level of confidence
___a____5. The maximum (margin of) error of the estimate for  is

a. z  x or z s x

b. 

n or s
n

c.  x or s x 
d. All the above
___b____6. The parameter(s) of the t distribution is (are)
a. the sample size n
b. degrees of freedom c.  and degrees of freedom
d. the mean
___b___7. The t distribution is used when the population is normal, the sample is small, and
a. the population standard deviation is known
b. the population standard deviation is unknown
c. the point estimate is known
d. the mean of the population is unknown
___b___8. The 95% confidence interval for the population mean is
a. The range within which 95% of individuals lie
b. The range within which we can be 95% sure the true population mean lies
c. Dependent on whether the mean is biased or not
d. Narrower if the standard deviation is larger.
___d___9. We wish to develop a confidence interval for the population mean. The shape of the population distribution
is not known, but we have a sample of 95 observations. We decide to use the 98 percent level of confidence.
The appropriate value to represent the confidence level is
a. t = 1.69
b. z = 1.96
c. t = 2.44
d. z = 2.33
___C___10. The Confidence level is denoted by
a. 1   
b. 1    %
c. 1    100%
d. None of the above
___d___11. A random sample of married people were asked “Would you remarry your spouse if you were given the
opportunity for a second time?”; Of the 150 people surveyed, 127 of them said that they would do so.
Find a 95% confidence interval for the proportion of married people who would remarry their spouse.
a. 0.847± 0.002
b. 0.847± 0.048
c. 0.847± 0.029
d. 0.847± 0.058
___C__12. Suppose we wish to estimate the proportion of students who actually understand the contents of the course
“Statistics 220”.What is the approximate number of students need to be sampled so that the 95%
confidence interval has a width of 2 percentage points?
a. about 500
b. about 5000
c. about 9604
d. about 2500
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___C___ 13. Which of the following assumptions is not required to use the t distribution to make a confidence interval
for  ?
a. The population is (approximately) normally distributed
b. The sample size is small
c. The population standard deviation is known
d. All of the above
____C___14. A 95% confidence interval for the mean time taken to process new insurance policies is (11, 12) days.
This interval can be interpreted to mean that:
a. only 5 percent of all policies take less than 11 or more than 12 days to process
b. only 5 percent of all policies take between 11 and 12 days to process
c. about 95 percent of all such intervals constructed from random samples of the same size will contain the
population mean processing time
d. the probability is .95 that all policies take between 11 and 12 days to process
___d___15 For a t distribution with 16 degrees of freedom, the area in between -1.746 and 1.746 is
a. 0.10
b. 0.05
c. 0.95
d. 0.90
__C____16. As degrees of freedom increase, the t-distribution approaches the
a. Binomial distribution
b. Exponential distribution
c. Standard normal distribution
d. None of the above
___b___17. An electric firm which manufactures a certain type of bulb wants to estimate its mean life. Assuming that
the life of the light bulb is normally distributed and that the standard deviation is known to be 40 hours,
how many bulbs should be tested so that we can be 90 percent confident that the estimate of the mean
will not differ from the true mean life by more than 10 hours?
a. 7
b. 44
c. 8
d. 62
___d___18. A researcher wishes to estimate the mean growth of seedlings in a large plot. A random sample of n = 100
seedlings is selected and the growth for each is measured. The sample mean and standard deviation are, respectively,
5.62 cm and 2.5 cm. The 95% confidence interval for the mean growth is:
a. (3.12, 8.12)
b. (4.98, 6.26) c. (5.37, 5.87)
d. (5.13, 6.11)
___C___19. Which of the following statements is correct?
a. A point estimate is an estimate of the range of a population parameter.
b. An interval estimate is an estimate of the range for a sample statistic
c. A point estimate is a single value estimate of the value of a population parameter.
d. An interval estimate is an estimate of the value of a population parameter.
___d___20 Which of the following is incorrect in regards to confidence intervals?
a. A narrower interval can be achieved if we are willing to be less confident in our estimates.
b. A narrower interval can be achieved if we take a larger sample
c. A narrower interval can be achieved if the data are less variable
d. A narrower interval can be achieved if we use t instead of z
___b___21. A random sample of 4 animals gave a sample mean weight of 452 kg and a ample standard deviation of
12 kg. Assuming normality, a 95% confidence interval for the average weight of all animals is
a. (435.3, 468.7)
b. (432.9, 471.1)
c. (440.2, 463.8)
d. (4228.5, 475.5)
___a___22. Referring to the previous question, about how many animals should be sampled in order to be 95%
confident of determining the true mean weight within 2 kg?
a. 139
b. 170
c. 550
d. 100
___a___23. Suppose we wish to estimate the percentage of regular users of vitamins in a large population and we
would like our estimate to be accurate within 4 percentage points, with confidence level 95%.
Approximately how large should your sample size be?
a. 601
b. 2400
c. 400
d. 1500
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___C___24. In a random sample of 800 automobile owners, it was found that 480 would like to see the size of the cars
reduced. A 95% confidence interval for the proportion of all car owners who would like to see smaller
cars is:
a. (0.572, 0.628)
b. (0.555, 0.645)
c. (0.566, 0.634)
d. (0.572, 0.628)
Section B: Solve the Following Questions
1. A researcher wanted to know the percentage of judges who are in favor of the death penalty. He took a random
sample of 15 judges and asked them whether or not they favor the death penalty. The responses of these judges are
given below.
Yes No Yes Yes No No No Yes Yes No Yes Yes Yes No Yes
a. What is the point estimate of the population proportion?
b. Make a 95% confidence interval for the percentage of all judges who are in favor of the death penalty.
2. A simple random sample of 800 elements generates a sample proportion P = 0.70.
a. What is the point estimate of the population proportion?
b. What is your estimate of the standard error of the point estimate,  p ?
c. At 95% confidence, what is the margin of error?
d. Compute a 95% confidence interval for the population proportion.
3. The mean number of hours of flying time for pilots at continental Airlines is 49 hours per month. Assume that this
mean was based on actual flying times for a sample of 20 Continental pilots and that the sample standard deviation
was 8.5 hours.
a. At 95% confidence, what is the margin of error for estimating the population mean flying time?
b. What is the 95% confidence interval estimate of the population mean flying time?
4. A machine is set to fill 32-ounce milk cartons. When the machine is working properly, the mean net weight of these
cartons is 32 ounces. The standard deviation of the amount of milk in all such cartons is equal to 0.15 ounce. The
quality control department takes a sample of 35 such cartons every week, calculates the mean net weight of these
cartons, and makes a 99% confidence interval for the population mean. If either the upper limit of this confidence
interval is greater than 32.15 ounces or the lower limit of this confidence interval is less than 31.85 ounces, the
machine is stopped and adjusted. A recent sample of 35 such cartons produced a mean net weight of 31.94 ounces.
Based on this sample, will you conclude that the machine needs an adjustment? Justify your answer (Hint:
Construct 99% confidence interval and compare it with the acceptable limits of the machine (31.85, 32.15). Based
on the sample, the machine does not need any adjustment if the sample confidence interval is inside the acceptable
limits).
5. A survey of 20 randomly selected adult men showed that the mean time they spend per week watching sports on
television is 9.75 hours with a standard deviation of 2.2 hours. Assume that the time spent per week watching
sports on television by all adult men is (approximately) normally distributed.
a. Find a point estimate for the mean time spent per week by all adult men. Find the standard deviation (error) for
this point estimate.
b. Make a 90% confidence interval for the mean time spent per week by all adult men.
6. An insurance company selected a sample of 50 auto claims filed with it and investigated those claims carefully. The
company found that 12% of those claims were fraudulent (illegal).
a. What is the point estimate of the percentage of all auto claims filed with this company that are fraudulent? What
is the maximum (margin of) error associated with this estimate? Use 99% confidence level
b. Make a 99% confidence interval for the percentage of all auto claims filed with this company that are
fraudulent.
7. An economist wants to find a 90% confidence interval for the mean sale price of houses in a state. How large a
sample should he or she select so that the estimate is within $3500 of the population mean? Assume that the
standard deviation for the sale prices of all houses in this state is $31,500?
8. A company wants to estimate the mean net weight of its Top Taste cereal boxes. A sample of 16 such boxes
produced the mean net weight of 31.98 ounces with a standard deviation of 0.26 ounce. Assume that the net
weights of all such cereal boxes have a normal distribution.
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a. Find a point estimate for the mean net weight of all Top Taste cereal boxes. Find the standard deviation (error)
for this point estimate.
b. Make a 95% confidence interval for the mean net weight of all Top Taste cereal boxes.
9. According to the Retirement Confidence Survey, 17% of the workers included in the survey said that they have
saved $100,000 or more for their retirement. Suppose that this percentage is based on a random sample of 1200
workers.
a. What is the point estimate of the population proportion? What is the maximum error associated with this point
estimate?
b. Make a 95% confidence interval for the percentage of all workers who have saved $100,000 or more for their
retirement.
10. A college registrar has received complaints about the online registration procedure at her college. She wants to
estimate the proportion of all students at this college who are issatisfied with the online registration procedure.
a. What is the most conservative estimate(maximum value) of the sample size that would limit the maximum
(margin of) error to be within .05 of the population proportion for a 90% confidence interval?
b. Assume that a preliminary study has shown that 70% of the students surveyed at this college are dissatisfied with
the current online registration. How large a sample should be taken in this case so that the maximum error is
within .05 of the population proportion for a 90% confidence interval?
11. The average cost of a movie ticket in the United States was $5.7 in 2002. Suppose that a random sample of 25
theaters in the United States yielded a mean movie ticket price $5.7 with a standard deviation $1.05. Assuming
that movie ticket prices are normally distributed, find a 95% confidence interval for the mea price of movie tickets
for all theaters in the United States.
12. Determine the margin of error for a confidence interval estimate for the population mean of a normal distribution
given the following information: Confidence level = 0.98, n = 13, S = 15.68
13. A company operates retail pharmacies. The company's internal audit department selected a random sample of 300
prescriptions. The objective was to estimate the average dollar value of all prescriptions issued by the company.
The following data were collected: x  $14.23 , S = 3.0. Determine the 90% confidence interval estimate for the
true average sales value for prescriptions issued by the company. Interpret the interval estimate.
14. One of the affordable automobiles to receive additions is BMW's Mini Cooper. A sample of 179 recent Mini
purchasers yielded a sample mean of $5,000. Suppose that the cost of accessories purchased for all Mini Coopers
has a standard deviation of $1,500.
a. Calculate a 95% confidence interval for the average cost of accessories on Mini Coopers.
b. Determine the margin of error in estimating the average cost of accessories on Mini Coopers.
c. What sample size would be required to reduce the margin of error by 50%?
15. Sales personnel for Skillings Distributor submit weekly reports listing the customer contacts made during the
week. A sample of 65 weekly reports showed a sample mean of 19.5 customer contacts per week. The sample
standard deviation was 5.2.
a. At 95 percent confidence interval, what is the margin of error for estimation the population mean.
b. Provide 95 percent confidence interval for the population mean number of weekly customer contacts for the
sales personnel.
16. In s survey conducted in the UK, 40 percent of children who participated said they would get hold of a copy of the
new book in the first weekend after publication. If 1000 children participated in the survey,
a. what is the margin of error using 95% confidence.
b. What is the interval estimate of the population proportion of children who expect to get hold of a copy of
the new book in the first weekend after publication?
17. Fuel consumption tests are conducted for a particular model of car. If 98 percent confidence interval with a margin
of error 0.4 kilometre per litre is desired, how many cars should be used in the test? Assume that preliminary tests
indicate the standard deviation is 1.0 kilometres per litre.
18. In a survey, the planning value for the population proportion is   0.35 . How large a sample should be selected
to provide a 95% confidence interval with a margin of error of 0.05?
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19. Dr. Garcia wanted to estimate the mean stress score before a statistics test for all students. For a random sample of
25 students, she found the mean and standard deviation for this sample to be 7.1 and 1.2, respectively.
a. Find a point estimate for the corresponding population mean for all students. Find the standard deviation
(error) for this point estimate.
b. Make a 95% confidence interval for the corresponding population mean for all students.
20. Suppose a survey of 20 first-time home buyers finds that the mean of annual household income is K. D. 40,000
and the sample standard deviation is KD 15,300.
a. At 95% confidence interval, what is the margin of error for estimation the population mean.
b. Provide 95 percent confidence interval for the population mean
21. A simple random sample of 400 individuals provides 100 yes responses.
a. What is the point estimate of the proportion of the population that would provide Yes responses?
b. What is your estimate of the standard deviation (error) of the point estimate?
c. Compute a 95% confidence interval for the population proportion
22. A well-known bank credit card firm wishes to estimate the proportion of credit card holders who carry a non-zero
balance at the end of the month and incur at interest charge. Assume that the desired margin of error is 0.03 at 98
percent confidence. How large a sample should be selected if no planning value for the proportion could be
specified?
23. A medical center wants to estimate the mean time that a staff member spends with each patient. How large a
sample should be selected if the desired margin of error is 2 minutes at a 95% level of confidence? Use a value for
the population standard deviation of 8 minutes.
24. A consumer agency wanted to estimate the mean hourly rate for all lawyers in New York City. A sample of 70
lawyers taken from New York City showed that the mean hourly rate charged by them is $420 and the standard
deviation of hourly charges is $110. Make a 99% confidence interval for the mean hourly charges for all lawyers
in New York City.
25. A sample of 25 malpractice lawsuits filed against doctors showed that the mean compensation awarded to the
plaintiffs was $410425 with a standard deviation $74820. Assume that the compensation awarded to the plaintiffs
of all such lawsuits are normally distributed. Find a 95% confidence interval for the mean compensation awarded
to the
26. A department store manager wants to estimate at a 90% confidence level the mean amount spent by all customers
at this store. From an earlier study, the manager knows that the standard deviation of the amounts spent by
customers at this store is $31. What sample size should he choose so that the estimate will be within $3 of the
population mean%?
27. A marketing researcher wants to find a 95% confidence interval for the mean amount that visitors to a theme park
spend per person per day. She knows that the standard deviation of the amounts spent per perso
n per day by all visitors to this park is $11. How large a sample should be selected so that the estimate will be within
$2 of the population mean%?
28. If a 95% confidence interval based on a large sample to estimate a population mean is (129.21, 172.33) then what
is the value of the standard error of the point estimate for the population mean?
Answers for the Exercises of Quiz 2 (Chapter 8) Stat 220
1. 0.60, (0.3521, 0.8479)
3. 3.978, (45.02, 52.978)
6. 0.12, 0.1186, (0, 0.2386)
9. 0.17, 0.02125, (0.1487, 0.1913)
13. (13.944, 14.516)
15. 1.264, (18.236, 20.764)
19. 7.1, 0.24, (6.6, 7.595)
21. 0.02165, (0.20756, 0.2924)
25. (379540.8, 441309.1996)
2. 0.7, 0.0162, 0.03176, (0.668, 0.732)
4. (31.875, 32.005) 5. 9.75, 0.492, (8.899, 10.6)
7. 221
8. 31.98, 0.065, (31.84, 32.119)
10. 273, 229 11. (5.267, 6.133)
12. 11.659
14. (4780.25, 5219.75), E = 109.8735, n = 716
16. 0.03, (0.3696, 0.4304)
17. 34 18. 350
20. 7160.53, (32839.47, 47160.53)
22. 1509
23. 62 24. (386.08, 453.92)
26. 291
27. 117
28. 11
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