Transcribed Text
1.
According to a recent poll, 26% of adults in a certain area have high levels of cholesterol.
They report that such elevated levels "could be financially devastating to the regions
healthcare system" and are a major concern to health insurance providers. According to recent
studies, cholesterol levels in healthy adults from the area average about 210 mg/dL, with a
standard deviation of about 35 mg/dL. and are roughly Normally distributed. If the cholesterol
levels of a sample of 42 healthy adults from the region is taken, answer parts (a) through (d).
(a) What shape will the sampling distribution of the mean have?
A. The sampling distribution of the mean is normal.
8. The sampling distribution of the mean is skewed right.
C. The sampling distribution of the mean is skewed left.
D. There is not enough information to determine the shape of the sampling distribution
(b) What is the mcan of the sampling distribution?
um
mg/dL
(c) What is the standard deviation?
SD(y) 
mg/dL (Round to two decimal places as needed.)
(d) If the sample size were increased to 120. how would your answers to parts (a)(c) change?
also be normally distributed.
For part (a). the shape of the distribution would
become skewed left.
become skewed right.
For part (b), the mean of the sampling distribution would
change to
n
mg/dL.
remain as
1.
(cont.)
For part (c). the standard deviation of the sampling distribution would
change to
SD(y)=
remain as
mg/dL. (Round to two decimal places as needed.)
2.
A marketing researcher for a phone company surveys 150 people and finds that the proportion
of clients who are likely to switch providers when their contract expires is 0.14. Use this
information to complete parts (a) and (b).
(a) What is the standard deviation of the sampling distribution of the proportion?
SE(p) 
(Round to four decimal places as needed.)
(b) If the rescarcher wants to reduce the standard deviation by half. how large a sample would
the researcher need?
(Type a whole number.)
3.
A survey of 25 randomly selected customers found the ages shown
24
49
29
39
(in years). The mean is 32.84 years and the standard deviation is
39
20
19
43
19
8.88 years.
a) What is the standard error of the mean?
35 40 36 33 23
b) How would the standard crror change if the sample size had been
23 38 33 34 42
400 instead of 25? (Assume that the sample standard deviation didn't
40 29 46 38 30
change.)
a) The standard error of the mean is
.
(Round to two decimal places as needed.)
b) How would the standard crror change if the sample size was 400 instead of 25 with the
same sample standard deviation? Select the correct choice below and fill in any answer boxes
within your choice.
A. The standard crror would increase. The new standard crror would be
times the old.
B. The standard crror would decrease. The new standard error would be the old standard
error divided by
.
C. The standard error would not change.
4.
For parts a and b, use the t tables, software, or a calculator to estimate.
a) the critical value of t for a 90% confidence interval with df= 22.
b) the critical value oft for a 95% confidence interval with df=78.
a) What is the critical value of t for a 90% confidence interval with df= 22?
(Round to two decimal places as needed.)
b) What is the critical value of t for a 95% confidence interval with df= 78?
(Round to two decimal places as needed.)
5.
The histogram shows the ages (in years) of 25 customers that
were in the freezer aisle at a large grocery store. Check the
10
assumptions and conditions for an inference using Student's
8
tmodel
6
4
2
0
15
30
45
Which assumptions and conditions are satisfied by the sample?
The Independence Assumption
is
satisfied.
is not
The Randomization Condition
is not satisfied.
is
The 10% Condition is not satisfied
is
The Nearly Normal Condition
is not
satisfied.
is
6.
An analyst collected data from 25 randomly selected
21.34 46.91 18.22 18.37 19.34
transactions and found the purchase amounts (in $). The
5.78 27.24 30.78 19.02 18.18
mean is $22.03 and the standard deviation is $13.07. The
7.39 21.61 32.36 23.86 46.78
analyst wants to know if the mean purchase amount of all
transactions is at least $15. Use the given information to
3.24 10.57 43.42 18.98 4.49
complete parts a through c.
31.42
3.33 41.44 15.58 21.21
a) What is the null hypothesis?
Ho:
$

y
"
(Type an integer or a decimal.)
b) Is the alternative one or twosided?
The alternative hypothesis is
onesided.
twosided.
c) What is the value of the test statistic?
The test statistic is
.
(Round to two decimal places as needed.)
d) What is the Pvalue of the test statistic?
Pvalue 
(Round to four decimal places as needed.)
e) What do you conclude at a  0.005?
The Pvalue is
less than a,so fail to reject the null hypothesis. There is sufficient
greater than
reject
insufficient
evidence to conclude that the mean purchase amount of all transactions is greater than $15.
7.
Describe how the width of a 95% confidence interval for a mean changes as the standard
deviation (s) of a sample increases, assuming sample size remains the same.
Choose the correct answer below.
DA As the variability of a sample increases, the width of a 95% confidence interval
decreases, assuming that sample size remains the same.
B. As the variability of a sample increases, the width of a 95% confidence interval
increases, assuming that sample size remains the same.
OC. As the variability of a sample increases, the width of a 95% confidence interval remains
the same, assuming that sample size remains the same.
8.
Hoping to lure more shoppers downtown, a city builds a new public parking garage in the
central business district. The city plans to pay for the structure through parking fees. For a
random sample of 44 weekdays, daily fees collected averaged $126, with a standard deviation
of $15. Suppose that for budget planning purposes. the city needs a better estimate of the mean
daily income from parking fees. Complete parts (a) through (c).
(a) Somcone suggests that the city use its data to create a 95% confidence interval instead of
the 90% interval the city first created. Would this increased interval be better for the city?
(You need not actually create the interval.)
A. Yes. A 95% confidence level gives increased confidence that the mean parking revenue
is contained with the interval.
B.
Yes. A 95% confidence level means that more people were sampled so the interval is
more accurate.
C. No. There is no significant difference between using the 90% confidence level and the
95% confident level.
(b) Would the 95% confidence interval be worse for the planners?
A. Yes. A 95% confidence level creates a narrower interval and is, therefore. more precisc
and will cost the planners more moncy.
B. Yes. The increased confidence level creates a wider interval and is. therefore. less
precisc.
C. No. The increased confidence interval would not be worse for the planners.
(c) How could they achieve a confidence interval estimate that would better serve their
planning needs?
A. They could include the weekend parking fees in the sample.
3. They could collect a larger sample which would create a more precise interval without
sacrificing confidence.
C. The city officials and planner could compromise and use a 92.5% confidence interval.
9.
A certain region has been one of the fastest growing regions for a number of years.
Accompanying the rapid growth are massive new construction projects that create visible dust
pollution. As required by government regulation, researchers continually monitor pollution
levels. In the most recent test of pollution levels, 121 air samples were collected. The dust
particulate levels must be reported to the federal regulatory agencies. In the report sent to the
federal agency, it was noted that the mean particulate level is 57.6 micrograms/cubic liter of
air, and the 95% confidence interval estimate is (52.06 mg to 63.07 mg). A graph of the
distribution of the particulate amounts was also included and is shown below. Complete parts a
and b.
Click the icon to view the histogram of the particulate amounts.
a) Of the following choices, what are the assumptions and conditions we typically make when
preparing to use Student's t inference methods. Select all that apply.
A. The data arise from a random sample or suitably randomized experiment.
B. The data come from a nearly Normal distribution.
C. The sample size is at least 10.
D. The expected values for the numbers of successes and failures are at least 10.
b) Do you think the confidence interval noted in the report is valid? Briefly explain why or
why not. Choose the correct answer below.
A. Yes: although the data produce a histogram that is not incarly Normal, the sample size is
large enough.
B. No; the expected values for the numbers of successes and failures are less than 10.
DC. Yes: the sample is random, the histogram is nearly Normal, and the sample size is
greater than 10.
Do. No: the data do not appear to come from a nearly Normal distribution.
9.
20
(cont.)
15
10
5
o
20
40
60
80
100
Particulates (in mictograms per cubic liter of ait)
10.
A market researcher at a major clothing company that has traditionally relied on catalog
mailorder sales decides to investigate whether the amount of monthly online sales has
changed. She compares the mean monthly online sales of the past several months with a
historical figure for mean monthly sales for online purchases. She gets a Pvalue of 0.01.
Explain in this context what the 1% means.
What does the 1% Pvalue mean?
A. Online purchases make up 1% of all sales. Because the percentage of online sales is so
small, the resulting mean is insignificant.
B. Online purchases make up 1% of all sales. The resulting mean sales would not occur
assuming the historical mean for sales.
C.
If the mean monthly sales due to online purchases has not changed. there is a 1%
chance (or I out of every 100 samples) that the resulting mean sales would occur
assuming the historical mean for sales.
D. If the mean monthly sales due to online purchases has changed, there is a 1% chance
(or I out of every 100 samples) that the resulting mean sales would not occur assuming
the historical mcan for sales.
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1 a. Choice A
b. 210 mg/dL
c. 5.40 mg/dL
d. also be normally distributed
210 mg/dL
3.20...