Download Introduction to the Practice of Statistics

Introduction to the Practice of Statistics
Fifth Edition
Moore, McCabe
Section 6.1 Homework Answers
6.1
A study based on a sample of size 25 reported a mean of 76 with a margin error of 12 for 95%
confidence. Give the 95% confidence interval.
This problem is trying to remind you of the importance of recalling the structure of any confidence
interval.
point estimate ± margin of error
Notice, all the information is already given to you.
76 ± 12 which result in the confidence interval (64, 88)
6.3
Suppose that the sample mean is 50 and the standard deviation is assumed to be 5. Make a
diagram similar to Figure 6.5 (page 390) that illustrates the effect of sample size on the width
of a 95% interval. Use the following sample sizes: 10, 20, 40, and 100. Summarize what the
diagram shows.
Information given, x = 50, and σ = 5 (please note that in the exams I will either use the symbol σ to
make sure you realize the information I am giving you is the population standard deviation, or I will
write population standard deviation. If I write standard deviation in reference to a sample then I mean
s. More on this later).
The formula σ x =
n
10
20
40
100
σ
will provide me with a measure of the expected variation for my sample.
n
σx =
5
10
5
20
5
40
5
100
σ
n
1.96
σ
n
3.099
2.1913
0.98
0.74081
46
48
50
52
54
As the sample size increases the margin of error
decreases, but not at a linear rate. Notice that doubling
the sample size does not half the margin of error.
6.5 For many important processes that occur in the body, direct measurement of characteristics of the
process is not possible in live individuals. In many cases, however, we can measure a biomarker, a
biochemical substance that is relatively easy to measure and is associated with the process of interest.
Bone turnover is the net effect of two processes: the breaking down of old bone, called resorption, and
the building of new bone, called formation. One biochemical measure of bone resorption is tartrate
resistant acid phosphatanese (TRAP), which can be measured in blood. In a study of bone turnover
young women, serum TRAP was measured in 31 subjects.4 The units are units per liter (U/l). The
mean was 13.2 U/l. Assume that the standard deviation is known to be 6.5 U/l. Give the margin of error
and find a 95% confidence interval for the mean for young women represented by this sample.
Margin of error = 1.96
6.5
= 2.288 U/l
31
13.2 ± 2.288
(10.9, 15.5) All this interval suggests is that we are 95% confident that the mean TRAP level in young
women is in the indicated interval. The 95% represents the probability that our interval contains, µ, the
population parameter, which represents the mean TRAP level in young women.
6.7
Refer to Exercise 6.5. Repeat the calculations for a 99% confidence interval.
How do the results compare with those in Exercise 6.5?
Margin of error = 2.676
6.5
= 3.007 U/l
31
13.2 ± 3.007
(10.2, 16.2) You can see that the interval is now wider. The lesson to learn here is that the reason the
confidence level has gone up from 95% to 99% is you now are including more numbers as possibilities
for µ.
6.11
Refer to the previous exercise. Will the 95% confidence interval include approximately 95% of
the rents of all unfurnished one-bedroom apartments in this area? Explain why or why not.
No, the 95% does not have anything to do with the percentage of apartments between certain price
ranges. The only thing the confidence level represents is how often, in the long run, µ will be found in
the interval we create by following the outlined procedures. The purpose of the confidence interval is
to estimate µ and nothing else.
6.13 A questionnaire about study habits was given to a random sample of students taking a large
introductory statistics class. The sample of 25 students reported that they spent an average of 80
minutes per week studying statistics. Assume that the standard deviation is 35 minutes.
(a) Give a 95% confidence interval for the mean time spent studying statistics by students in this class.
80 ± 1.96
35
25
(66.28, 93.72)
(b) Is it true that 95% of the students in the class have weekly study times that lie in the interval you
found in part (a)? Explain your answer.
No, the result indicates that we are 95% certain the mean study time for students in this large
introductory statistics time lies between 66.28 minutes to 93.72 minutes per week. In other words
suppose from day one we logged the study time for every student in the class. At the end of the term,
we would calculate the number of minutes studied each week for each student. Then we would take the
average of all those numbers. This value is µ. So by taking a sample for 25 students we hope to
understand what value µ could possibly be. Is this method reasonable? Depends how the sample was
taken and the details are sketchy.
6.15 The mean number of calories consumed by women in the United States who are 19 to 30 years of
age is x = 1791 calories per day. The standard deviation is 31 calories.5 You will study a sample of 200
women in this age range, and one of the variables you will collect is calories consumed per day.
(a) What is the standard deviation of the sample mean x ?
31
σ
= 2.19 The purpose of this question is to make sure you understand what σ x =
200
n
really means. Study how this question was phrased.
(b) The 68-95-99.7 rule says that the probability is about 0.95 that x is within
__2_standard deviations__ of the population mean µ. Fill in the blank.
The purpose of this question is to tie in the visual that we had in section 1.3, when the normal
distribution was first introduced along with the 68-95-99.7 rule. We know now that the numbers we
associated with 68-95-99.7 are just approximations so it would be easier to remember.
6.17 Computers in some vehicles calculate various quantities related to performance. One of these is
the fuel efficiency, or gas mileage, usually expressed as miles per gallon (MPG). For one vehicle
equipped in this way, the MPG were recorded each time the gas tank was filled, and the computer was
then reset.6
Refer to the previous exercise. Here are the values of the average speed in miles per hour (MPH) for the
same sample:
21.0
19.0
18.7
39.2 45.8
31.6 49.0
16.0
34.6
36.3
19.8 48.4 21.0 29.1
19.0 43.3
37.5
35.7
16.5
34.5
Assume that the standard deviation, σ, is 10.3 MPH. Estimate the mean speed at which this vehicle
was driven with a margin of error for 95% confidence.
x = 30.8 mph
 10.3 
30.8 ± 1.96 

 20 
6.19 Here are the Degree of Reading Power (DRP) scores for a sample of 44 thirdgrade students:
40 26 39 14 42 18 25 43 46 27 19
47 19 26 35 34 15 44 40 38 31 46
52 25 35 35 33 29 34 41 49 28 52
47 35 48 22 33 41 51 27 14 54 45
Suppose that the standard deviation of the population of DRP scores is known to be σ = 11. Give a 95%
confidence interval for the population mean score.
x = 35.1 n = 44
 11 
35.1 ± 1.96 

 44 
(31.8, 38.4)
6.22 You are planning a survey of staring salaries for recent liberal arts major graduates form your
college. From a pilot study you estimate that the standard deviation is about $9000. What
sample size do you need to have a margin of error equal to $400 with 95% confidence?
2
 1.96(9000) 

 ≈ 1945. We would need a sample size of 1945.
400


6.27 An advertisement in the student newspaper asks you to consider working for a telemarketing
company. The ad states, “Earn between $500 and $1000 per week.” Do you think that the
ad is describing a confidence interval? Explain your answer.
I would venture to guess that the two values given are more likely to be the minimum and maximum
values (rounded) that people earn while working at this particular job. My reasoning is that
most of the population is unaware of the use of confidence intervals and if it was a
confidence interval this was displaying then this means that there are people earning a lot o
money and those earning very little. After all the margin of error is $500, a huge margin.
The $500 to $1000 is the average of many employees.
This is the type of problem for which your instructor has totally forgotten why he even
assigned it.
6.30
A newspaper invites readers to send email stating whether or not they are in favor of making
full-day kindergarten available to all students in the state. A total of 320 responses are
received and, of these, 80% are in favor of the new program. In an article describing the
results, the authors state that the margin of error is 4% for 95% confidence. Assume that they
have computed this number correctly.
(a) Use the sample proportion and the margin of error to compute the 95%
confidence interval.
0.8 ± 0.04
(0.76, 0.84)
Here is another problem trying to emphasize the structure of a
confidence interval
point estimate ± margin of error
(b) Do you think that these results are trustworthy? Discuss your answer.
The confidence interval created is worthless. All we can say is that of the 320 people who responded
80% were in favor of full-day kindergarten; no need for a margin of error since this result is exact for
our population of the 320 respondents.
We can not extrapolate to the population in general. So if the author is claiming that we are 95%
confident that between 76% to 84% of the population is in favor of full-day kindergarten this is not
true. Our sample was not random but rather a sample were the population self selected to participate.
We can not guarantee the sample is representative of the population, thus making the confidence
interval worthless.
6.31
A recent Gallup poll conducted telephone interviews with a random sample of adults aged 18
and older. Data were obtained for 1011 people. Of these, 37% said that football is their
favorite sport to watch on television.
(a) The poll announced a margin of error of ±3 percentage points for 95% confidence.
Explain to someone who knows no statistics why we can't just say that 37% of
Americans would say that football is their favorite sport to watch on television.
I will assume that the sample was conducted using a technique that produced a
representative sample. But even if the sample is random the problem with the statement
in (a) is the definitive guarantee for the true proportion, “37% of Americans would say
that football is their favorite sport to watch on television.”
If we sample again, then we would get a different result; hopefully, close to 37%. The
value is only an estimate.
“About 37% of Americans would say that football is their favorite sport to watch on
television.”
(b) Then explain clearly what "95% confidence" means. In the long run 95% of the
confidence intervals created using this technique will contain the parameter we are
looking for; in this case p, population proportion (sections 5.1, 8.1).
(c) Give the 95% confidence interval. 0.37 ± 0.03 which results in (0.34, 0.40).
(d) The poll was taken in December, an, exciting part of the football season. Do you think
that a similar poll conducted in June might produce different results? Explain why or
why not.
An interesting question. If the poll was taken in December, this is the time when the “drama” is
unfolding as to what team is doing great and will make the playoffs, issues with players, sportscasters
making predictions and so forth; as was mentioned an exciting part of the season for serious football
fans and casual fans of the sport.
In June baseball season is in its halfway point and for loyal baseball fans that also enjoy football this
may cause a change of heart to say football is your favorite sport during baseball season. Who knows,
but it is something to consider.
An interesting way to find out is to poll at different points throughout the year and see what occurs.
Certainly, the concern stated is very real and could bias the sample taken.