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M10 L1
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1) Suppose you wanted to create a confidence interval at the 98% confidence level.
Explain the meaning of “98% confident.” (HINT: look on pages 539-541 in TPOS 2E
or pages 618-624 in TPOS 3E)
1) Suppose you wanted to create a confidence interval
at the 98% confidence level. Explain the meaning of
“98% confident.” (HINT: look on pages 539-541 in
TPOS 2E or pages 618-624 in TPOS 3E)
To be 98% confident means that if we computed a
98% confidence interval for a sample and then
repeated this many times over, 98% of the confidence
interval will contain the true population mean, µ.
Use the following information to answer #2
Suppose your class is investigating the weights of Snickers 1-ounce fun-size candy bars
to see if customers are getting full value for their money. Assume that the weights are
normally distributed with standard deviation of .005 ounces. Several candy bars are
randomly selected and weighed with sensitive balances borrowed from the physics lab.
The weights are:
.95 1.02 .98 .97 1.05 1.01 .98 1.00
ounces. We want to determine some possible confidence intervals for the true mean, µ.
2). Calculate a 90%, 95%, and a 99% Confidence Interval for the mean, µ, weight of
Snickers 1 ounce candy bars. Compare and contrast the three intervals. What happens
to the intervals from a 90% confidence level to a 99% confidence level? Justify your
answer
3)You have measured the systolic blood pressure of a random sample of 25 employees
of a company located near you. A 95% confidence interval for the mean systolic blood
pressure for the employees of this company is (122, 138). Which of the following
statements gives a valid interpretation of this interval? You must also explain why your
answer is correct
(a)Ninety-five percent of the sample of employees has a systolic blood pressure
between 122 and 138.
(b)Ninety-five percent of the population of employees has a systolic blood pressure
between 122 and 138.
Suppose your class is investigating the weights of
Snickers 1-ounce fun-size candy bars to see if customers
are getting full value for their money. Assume that the
weights are normally distributed with standard deviation
of .005 ounces. Several candy bars are randomly
selected and weighed with sensitive balances borrowed
from the physics lab. The weights are:
.95 1.02 .98 .97 1.05 1.01 .98
1.00
ounces. We want to determine some possible confidence
intervals for the true mean, µ.
2). Calculate a 90%, 95%, and a 99% Confidence
Interval for the mean, µ, weight of Snickers 1 ounce
candy bars. Compare and contrast the three intervals.
What happens to the intervals from a 90% confidence
level to a 99% confidence level? Justify your answer
90% = (.992, .998), 95% = (.991, .998), 99% =
(.990, .999)
The intervals become wider. This is because the
confidence levels increase
3)You have measured the systolic blood pressure of a
(c)If the procedure were repeated many times, only 5% of the resulting confidence
intervals would contain the population mean systolic blood pressure.
(d)The probability that the population mean blood pressure is between 122 and 138 is
.95.
(e)If the procedure were repeated many times, 95% of the sample means would be
between 122 and 138.
(f)None of the above. The answer is
.
4)An analyst, using a random sample of n = 500 families, obtained a 99% confidence
interval for mean monthly family income for a large population: ($600, $800). If the
analyst had used a 95% confidence coefficient instead, the confidence interval would
be: You must also explain why your answer is correct
(a)Narrower and would involve a larger risk of being incorrect
(b)Wider and would involve a smaller risk of being incorrect
(c)Narrower and would involve a smaller risk of being incorrect
(d)Wider and would involve a larger risk of being incorrect
(e)Wider but it cannot be determined whether the risk of being incorrect would be
larger or smaller
5)I collect a random sample of size n from a population and from the data collected
compute a 95% confidence interval for the mean of the population. Which of the
following would produce a new confidence interval with larger width (larger margin of
error) based on these same data? You must also explain why your answer is correct
(a)Use a larger confidence level.
(b)Use a smaller confidence level.
(c)Use the same confidence level, but compute the interval n times. Approximately 5%
of these intervals will be larger.
(d)Increase the sample size.
random sample of 25 employees of a company located
near you. A 95% confidence interval for the mean
systolic blood pressure for the employees of this
company is (122, 138). Which of the following
statements gives a valid interpretation of this interval?
You must also explain why your answer is correct
(a)Ninety-five percent of the sample of employees has
a systolic blood pressure between 122 and 138.
(b)Ninety-five percent of the population of employees
has a systolic blood pressure between 122 and 138.
(c)If the procedure were repeated many times, only 5%
of the resulting confidence intervals would contain the
population mean systolic blood pressure.
(d)The probability that the population mean blood
pressure is between 122 and 138 is .95.
(e)If the procedure were repeated many times, 95% of
the sample means would be between 122 and 138.
(f)None of the above. The answer is
answer choice “f” – none of the above
The best interpretation of the confidence
interval is “If the procedure were repeated many
times, 95% of the resulting confidence intervals
would contain the population mean systolic blood
pressure.
.
4)An analyst, using a random sample of n = 500
families, obtained a 99% confidence interval for mean
monthly family income for a large population: ($600,
$800). If the analyst had used a 95% confidence
coefficient instead, the confidence interval would be:
You must also explain why your answer is correct
(a)Narrower and would involve a larger risk of being
incorrect
(b)Wider and would involve a smaller risk of being
incorrect
(c)Narrower and would involve a smaller risk of being
incorrect
(d)Wider and would involve a larger risk of being
incorrect
(e)Wider but it cannot be determined whether the risk of
being incorrect would be larger or smaller
answer choice “a” - Narrower and would involve a
larger risk of being incorrect
5)I collect a random sample of size n from a population
and from the data collected compute a 95% confidence
interval for the mean of the population. Which of the
following would produce a new confidence interval with
larger width (larger margin of error) based on these same
data? You must also explain why your answer is correct
(a)Use a larger confidence level.
(b)Use a smaller confidence level.
(c)Use the same confidence level, but compute the
interval n times. Approximately 5% of these intervals
will be larger.
(d)Increase the sample size.
answer choice “a” - Use a larger confidence level.
M10 L2
Answer the questions given, then explain how you obtained your answers. Post
your answers to the Forum. After you make your post you'll be able to see others'
answers.
You can post your answers by clicking on "Reply" below. After you make your
post you'll be able to see others' answers. Check back periodically to see who else
has posted. Make comments on the correctness or errors to answers given by two
other classmates.
Use the following information to answer #1a - #1d
At the bakery where you work, loaves of bread are supposed to weigh 1 pound, with
standard deviation s = 0.13 pounds. You believe that new personnel are producing
loaves that are heavier than 1 pound. As supervisor of Quality Control, you want to test
your hypotheses at the 5% significance level. You weigh 20 loaves and obtain a mean
1a) The population of interest is all loaves of bread in
the bakery, and the parameter is the mean weight of
these loaves.
Ho: µ = 1.0 lb The mean weight of all the
loaves of bread is 1 lb.
Ha: µ > 1.0 lb The mean weight of all
loaves of bread is more than 1 lb.
1b) Use a one sample z-test
Conditions:
 Not clear if sample was randomly selected so
results may not be generalizable

weight of 1.05 pounds.
1a)
Identify the population and parameter of interest. State your null and alternative
hypotheses in both words and symbols.
1b)
Identify the statistical procedure you should use. Then state and verify all
conditions required for using this procedure.
1c) Calculate the test statistic and the P-value. Justify your answer by showing and or
explaining your work
1d) State your conclusions clearly in complete sentences.
2) A significance test gives a P-value of 0.02. From this we can
(a) Reject H0 at the 1% significance level
(b) Reject H0 at the 5% significance level
(c) Say that the probability that H0 is false is 0.02
(d) Say that the probability that H0 is true is 0.02
(e) None of the above. The answer is
Not clear if distribution of sample mean is
normally distributed – sample size is not
large enough for CLT to work and we do not
have actual sample data. Proceed with
caution
 σ is known so z-procedures are acceptable
1c) z = 1.72 and p-value is 0.0427 (student needs to
show and explain how these were obtained)
1d) Since the p-value is less than 5% there is
sufficient evidence to reject null and conclude that
actual mean weight of all bread is more than 1 lb.
2) “b” Reject at 5% level
.
NOTE: You must explain why your answer is correct
M10 L3
Answer the questions given, then explain how you obtained your answers. Post
your answers to the Forum. After you make your post you'll be able to see others'
answers.
You can post your answers by clicking on "Reply" below. After you make your
post you'll be able to see others' answers. Check back periodically to see who else
has posted. Make comments on the correctness or errors to answers given by two
other classmates.
Use the following information to answer #1a - #1d
Many homeowners buy detectors to check for the invisible gas radon in their homes.
We want to determine the accuracy of these detectors. To answer this question,
university researchers placed 12 radon detectors in a chamber that exposed them to 105
picocuries per liter of radon. The detector readings were as follows:
91.9
97.8
111.4
122.3
105.4
95.0
1a) A Type I error would be to conclude that the
mean radon reading was different from 105 when, in
fact, it isn’t.
1b) The probability of a Type I error is 10% because
that is the significance level
1c) The power of the test is 1 - .3898 = 0.6102
1d) If sample size is increased then the power also
increases
2) “e” III only
103.8
99.6
96.6
119.3
104.8
101.7
Assume that the population standard deviation is 9 picocuries per liter of radon for the
population of all radon detectors. We want to determine if there is convincing evidence
at the 10% significance level that the mean reading of all detectors of this type differs
from the true value 105, so our hypotheses are:
H0: µ = 105
Ha: µ does not equal 105
A significance test to answer this question was carried out. The test statistic is
z = –0.3336, and the P-value is 0.74.
1a) Describe what a Type I error would be in this situation. Be sure to do this in the
context of the problem
1b) Calculate the probability of a Type I error for this problem. Justify your answer
1c) The researchers who carried out the study suspect that the large P-value is due to
low power. It was found that the Probability of a Type II error was 0.3898, when in
fact µ = 100. The power of the test against the alternative would be ______? Justify
your answer
1d) If the sample size is increased to n = 30, what would happen to the power against
the alternative, µ = 100? Justify your answer (HINT: look on page 601-602 TPOS 2E
or on page 730-731 TPOS 3E)
2) Which of the following are correct?
I. The power of a significance test doesn’t depend on the alternative value of the
parameter.
II. The probability of a Type II error is equal to the significance level of the test.
III. Type I and Type II errors only make sense when a significance level has been
chosen in advance.
(a) I and II only
(b) I and III only
(c) II and III only
(d) I, II, and III
(e) III only
(f) None of the above gives the complete set of true responses
NOTE: You must explain why your answer is correct
M11 L1
t-dist
part 1M
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make your post you'll be able to see others' answers.
1. Vitamin C Content - TPOS 2E: Exercise 11.9, p. 628, parts
(a) and (b) only or TPOS 3E: Exercise 10.30, pp. 649-650.
2. Healthy Bones: Here are estimates of the daily intakes of
calcium (in milligrams) for 38 women between the ages of 18
and 24 years who participated in a study of women's bone
health:
808
651
626
115
6
88
2
71
6
77
4
68
4
106
2
438
125
3
193
3
970
909
802
374
416
142
0
549
142
5
132
5
120
3
948
976
446
105
0
465
243
3
125
5
748
126
9
110
0
78
4
57
2
67
1
99
7
40
3
69
6
(a) Display the data using a histogram and boxplot and
make a normal probability plot. Describe the distribution of
calcium intakes for these women.
(b) Calculate the mean, the standard deviation, and the
standard error.
TPOS 2nd Edition: 11.9 or TPOS 3rd edition: 10.30
It is stated in the problem that we have a random sample.
This means SRS. To check to see if the normal
approximation is appropriate, we recognize that the
sample is too small to use the Central Limit Theorem.
Our best bet is to make a normal probability plot. recall
that this is the last type of plot on your calculator. The xaxis is Vitamin C content, and the Y-axis is the hard to
define quantity (Z-score of the quantile). If you label the
plot as a Normal Probably Plot, you do not have to label
the y-axis. That will be understood.
This plot is roughly linear,
consistent with this data fitting a normal distribution. It
is important that we have no outliers and a boxplot
shows that, also.
This boxplot shows no outliers and is consistent
with a reasonably symmetric distribution. We are
justified in our use of the normal approximation.
Confidence Interval:
(c) Using your calculator as described in the Technology
Toolbox (TPOS 2E pp. 558-559 or TPOS 3E pp. 641-642) Find
a 95% confidence interval for the mean. Interpret your
results in the context of the problem.
(d) Suppose that the recommended daily allowance (RDA)
of calcium for women in this age range is 1200 mg. Doctors
involved in the study suspected that participating subjects
had lower calcium intakes than the RDA. Test this claim at
the alpha = 0.05 significance level.
(e) Eliminate the two largest values and recompute the
95% confidence interval. What do you notice?
The are 95% confident that the true mean Vitamin C
content is between 16.49 and 28.51 mg/100g. In repeated
sampling this method produces confidence intervals that
capture the true mean an average of 95% of the time.
Test of Significance:
H0: The mean Vitamin C content in CSB is 40 mg/100g.
Ha: The mean Vitamin C content in CSB is not equal to
40 mg/100g.
The assumptions are the same as above: we must have
SRS, normal population, and also since n=8 is less than
15, we must have no outliers .
df=7
The shaded regions are too
small to see, but should be on both sides.
Reject H0, a value this extreme will rarely occur by
chance alone. We have strong evidence that that the
mean Vitamin C content in CSB is different from 40
mg/100g.
Healthy Bones:
(a)
The distribution of calcium intake is skewed to the right
with two outliers. The median value is 845 mg.
(b) Mean = 926 mg; standard deviation = 427 mg;
standard error = 69.3.
(c) 95% confidence interval: 785.6 to 1066.5 mg
calcium. We are 95% confident that the true mean
calcium intake for women between the ages of 18 and 24
is between 785.6 mg and 1066.5 mg.
(d) A test of H0: µ = 1200 mg vs Ha: µ < 1200 mg yields
t = -3.953 and P = 0.000167. We reject H0 and conclude
that there is strong evidence that participating subjects
did have lower calcium intakes than recommended.
(e) 95% confidence interval: 752.41 to 959.98 mg. The
range of the interval decreased because the standard
error decreased. Removing the two outliers decreased
the sample standard deviation which decreased the
standard error.
M11 L1
t-dist
part 2
In your post, answer the following questions. After you make
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TPOS 2nd edition:
Problems 11.1, 11.2.
TPOS 3rd edition:
Problems 10.27, 10.28.
M11 L1 no post required
t-dist
part 3
M11 L2 In your post, answer the following questions. After you make
t-dist
your post you'll be able to see others' answers.
matched
pairs
TPOS 2nd Edition:
11.1 Answers in back of book
11.2 (a) 2.015
(b) 2.518
TPOS 3rd Edition:
10.27 answers are in back of book.
10.28 (a) df = 11, t* = 1.796.
(b) df = 29, t* = 2.045.
(c) df = 17, t* = 1.333.
TPOS 2nd Edition
Problem 11.13
a)
TPOS 2nd edition:
Work problem 11.13.
TPOS 3rd edition:
Work problem 12.21.
H0: The mean difference in MLA listening
score is zero.
Ha: The mean difference in MLA listening
score is greater than zero.
b)
I've entered the data in
lists 1 and 2. The difference is found in list
3.
This is the Normal Probability Plot of
the differences. It is roughly linear,
suggesting that the data are consistent
with a normal model.
This modified boxplot
shows that there are no outliers.
c) In addition to the info in parts a and b,
above, we are uncertain that the group of
Spanish teachers are an SRS.
df=19
Reject H0, at a 5% significance level. A
value this extreme will occur by chance
alone less than 3% of the time.
We have strong evidence that the mean
difference in MLA listening score is
positive. That is, attending the institute
seems to improve listening scores.
At a 1% significance level, however, we
would fail to reject the null hypothesis.
d) t-interval
Assumptions above.
df=19
We are 90% confident that the true mean
difference is between .2116 and 2.6883. In
repeated sampling, this method will
capture the true mean an average of 90%
of the time.
TPOS 3rd Edition #12.21 answer is in
back of book. Follow pattern from # 11.13
above, 2nd Edition
M11 L2
two
sample t
interval
s and
tests
In your post, answer the following questions. After you make your post you'll be
able to see others' answers.
TPOS 2nd edition:
Work problem 11.40.
TPOS 3rd edition:
Work problem 13.24
TPOS 2nd Edition:
11.40 A) if the loggers had known a study
might be done, then they might have cut
down fewer trees in order to reduce the
impact of logging.
B) We want to test Ho: μu = μL versus Ha:
μu > μL, where μu and μL are the mean
number of species in unlogged and logged
plots, respectively. The test statistic is
t = (17.5-13.67) / (sqrt(13.3sq/12 + 4.5sq
/9)) = 2.11
So, with df = 8 from table C: 0.025 < pvalue < 0.05. Logging does significantly
reduce the mean number of species in a
plot after eight years at the 5% level but
not at the 1% level.
Using the formula for confidence intervals
for μ1 - μ2 given in the textbook, the
confidence interval is (0.46, 7.21). I am 90
percent confidents that the difference in
the means for unlogged and logged plots
is between 0.46 and 7.21 species.
TPOS 3rd Edition:
13.24 a) answer is the same as #11.40
above, TPOS 2nd Edition.
b) Random assignment allows us to
make a cause-and-effect conclusion
c) answer is the same as #11.40b
above, TPOS 2nd Edition.
d) answer is the same as #11.40c
above, TPOS 2nd Edition.
Presentation using matched pairs and the t-distribution.