08.11.2022, 00:44
Lesson activity: Hypothesis testing about a population proportion and difference between two population proportions.: Attem…
Started on Tuesday, 1 November 2022, 1:07 PM
State Finished
Completed on Tuesday, 8 November 2022, 12:44 AM
Time taken 6 days 11 hours
Marks 343.11/346.00
Grade 9.92 out of 10.00 (99.16%)
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Lesson activity: Hypothesis testing about a population proportion and difference between two population proportions.: Attem…
Question 1
Complete
Not graded
Chapter 10. Asking and Answering Questions About a Population Proportion
This lesson activity introduces you to the first inference test of the course.
One of the major struggles with this chapter will come from beginning and ending the
inference test. Students rarely struggle with the mechanics of the test, however, they do
struggle with translating contextual questions into formal hypotheses and with providing
contextual interpretations to the outcome of the test.
It is important to remember the four key questions that should be asked and answered prior to
any quantitative research analysis to ensure that you consider the context of the problem prior to
applying any procedures.
Also remember that conclusions to inference problems should not end with “reject” or “fail to reject” only
but rather that they should explain what the conclusion means in the context of the problem you solve.
This chapter introduces/reminds you of the p-value. The meaning of the p-value will be explained.
This will allow you to explore how this meaning lends itself to the conclusion to be drawn.
The lesson activity also presents the errors that can be made when performing a hypothesis test. Video and
exercises will guide you through a thorough discussion of the errors that can occur and why they can
occur. This will help you understand the relationship between these errors and the hypothesis test.
After completing this chapter, students should know the following information:
1. How to set up formal statistical hypotheses for your research problem
2. Understand how to compute a test statistic for a test for the population proportion
3. Understand how to find a p-value for a test for the population proportion
4. Understand how to draw formal and contextual conclusions from an inference test for the population
proportion
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Lesson activity: Hypothesis testing about a population proportion and difference between two population proportions.: Attem…
5. Understand why we reject for certain p-values and not for others
6. Understand the different types of error and why they can occur
Does Introduction to the lesson activity make sense? There is no wrong answer here.
No
Yes
Not quite yet
Your answer is partially correct.
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Lesson activity: Hypothesis testing about a population proportion and difference between two population proportions.: Attem…
Question 2
Correct
Mark 6.00 out of 6.00
A hypothesis test is:
I don't know
In a hypothesis test, we are using data from a sample to make a decision between two competing
claims about a population characteristic. Correct!
Mark 1.00 out of 1.00
A null hypothesis is
the claim about a population characteristic that is initially assumed to be true.
Correct!
the competing claim about the population characteristic.
Mark 1.00 out of 1.00
and alternative hypothesis is
the competing claim about the population characteristic.
Correct!
the claim about a population characteristic that is initially assumed to be true.
Mark 1.00 out of 1.00
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Lesson activity: Hypothesis testing about a population proportion and difference between two population proportions.: Attem…
If the null hypothesis is not rejected, there is strong statistical evidence that the null is true:
False Correct! A statistical hypothesis test is only capable of demonstrating strong support for the
alternative hypothesis (by rejecting the null hypothesis). When the null hypothesis is not rejected, it does
not mean strong support for H0—only lack of strong evidence against it.
True
Mark 1.00 out of 1.00
You are interested in determining whether there is strong evidence in support of the claim that less than
40% of retired adults have a part-time job. To answer this question, what null and alternative hypotheses
should you test?
H0:p=0.4 versus Ha:p≠0.4
H0:p<0.4 versus Ha:p>0.4
H0:p=0.4 versus Ha:p<0.4
Correct!
H0:p<0.4 versus Ha:p=0.4
H0:p=0.4 versus Ha:p>0.4
Mark 1.00 out of 1.00
Which of the following specify legitimate pairs of null and alternative hypotheses?
H0:p=0.2 versus Ha:p>0.2 Correct!
H0:p≠0.5 versus Ha:p=0.5
H0:p≠0.35 versus Ha:p>0.35
H0:p=0.4 versus Ha:p≠0.35
H0:p<0.40 versus Ha:p>0.65
Mark 1.00 out of 1.00
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Lesson activity: Hypothesis testing about a population proportion and difference between two population proportions.: Attem…
Question 3
Correct
Mark 4.00 out of 4.00
On Potential Errors in Hypothesis Testing
All other things being equal, choosing a smaller value of α will increase the probability of making a type II
error, β.
False
True
Correct!
Mark 1.00 out of 1.00
Assuming the same sample size is used, how does changing the significance level, α, from 0.2 to 0.05
affect the power of a test =1-β?
The power does not change.
The power decreases.
Correct!
It is not possible to predict whether the power will increase, decrease, or remain the unchanged.
The power increases.
Mark 1.00 out of 1.00
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Lesson activity: Hypothesis testing about a population proportion and difference between two population proportions.: Attem…
To decide on the significance level, α, the consequences of Type I and II errors should be considered. If a
Type I error is worse, a
large significance level (α>0.05)
small significance level (α<0.05)
Correct!
Mark 1.00 out of 1.00
should be chosen
AND
if a Type II error is worse, a
small significance level (α<0.05)
large significance level (α>0.05)
Correct!
Mark 1.00 out of 1.00
should be chosen.
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Lesson activity: Hypothesis testing about a population proportion and difference between two population proportions.: Attem…
Question 4
Correct
Mark 8.00 out of 8.00
In the video below I talk about Type I and Type II errors
in hypothesis testing in the context of particular research questions:
Children as young as 2 years of age, upon seeing an object placed under a pillow in a familiar
setting at home, will understand to look for it after an interval of time and be able to find it. Investigators
believe this capability will be less pronounced in a laboratory situation, where the child is away from the
familiar setting of home.
Let p denote the proportion of 2-year-olds than have this understanding in the home situation, and suppose
that p = 0.35. The investigators wish to determine whether the proportion that remembers is less when the
child is away from home.
What is the appropriate null hypothesis? H0: p
=0.35

What is the appropriate alternative hypothesis? Ha: p
<0.35

A type I error is made by
faling to reject a false null hypothesis.
rejecting the null hypothesis when the null is true
Correct!
Mark 1.00 out of 1.00
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Lesson activity: Hypothesis testing about a population proportion and difference between two population proportions.: Attem…
A type II error is made by
rejecting the null hypothesis when the null is true
faling to reject a false null hypothesis.
Correct!
Mark 1.00 out of 1.00
A type I error in this context is
to decide that children remember less when outside of the home environment, when in fact, they
remember equally well. Correct!
to decide that children remember as well in the laboratory setting as in the home setting, when in fact
they remember less away from home.
Mark 1.00 out of 1.00
A type II error in this context is
to decide that children remember less when outside of the home environment, when in fact, they
remember equally well.
to decide that children remember as well in the laboratory setting as in the home setting, when in fact
they remember less away from home. Correct!
Mark 1.00 out of 1.00
The level of significance of a test is the probability of making a type I error, given that the null hypothesis
is true.
False
True
Correct!
Mark 1.00 out of 1.00
One of the early decisions that must be made when performing a hypothesis test is the choice of
significance level. What are the considerations that go into making this decision?
I don't know.
To decide on the significance level, the consequences of Type I and II errors should be considered. If a
Type I error is worse, a small significance level should be chosen (α<0.05) and if a Type II error is worse, a
large significance level should be chosen (α>0.05). Correct!
Mark 1.00 out of 1.00
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Lesson activity: Hypothesis testing about a population proportion and difference between two population proportions.: Attem…
Question 5
Partially correct
Mark 44.00 out of 45.00
You know three things about the sampling distribution of the sample proportion p̂ when a random sample
of size n is selected:
1. The mean of the sampling distribution of possible sample proportions p̂ from all possible random
samples of size n is true population proportion, p
 . Hence, μp̂ =p where p denotes the proportion
in the population that possess some characteristic of interest.
−
−
−
−
−
2. The standard error (=standard deviation of any statistic) of p̂ is σp̂ =√
p(1−p)
n
3. If sample size n is large, then the sampling distribution of is approximately
.
normal

.
The following example illustrates the logic of testing hypotheses about a population proportion (when the
sample size is large):
Impact of Food Labels
An Associated Press survey was conducted to investigate how people use the nutritional information
provided on food package labels. Interviews were conducted with 1003 randomly selected adult Americans,
and each participant was asked a series of questions, including the following two:
Question 1: When purchasing packaged food, how often do you check the nutrition labeling on the
package?
Question 2: How often do you purchase foods that are bad for you, even after you’ve checked the nutrition
labels?
It was reported that 582 responded “frequently” to Question 1 and 441 responded “very often or somewhat
often” to Question 2.
Start by considering the responses to the first question. Based on these data, is it reasonable to conclude
that a majority of adult Americans (>50%) frequently check nutrition labels?
With p denoting the unknown
population

proportion of all American adults who frequently check
nutrition labels, this question can be answered by testing the following hypotheses:
H0: p
=

0.50
Ha: p
>

0.50
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Lesson activity: Hypothesis testing about a population proportion and difference between two population proportions.: Attem…
This alternative hypothesis states that the population proportion of adult Americans who frequently check
nutrition labels is greater than
 0.50.
This would mean that
more than

half of adult Americans frequently check nutrition labels. The null
hypothesis will be rejected only if there is convincing evidence against p =0.50.
For this sample, the sample proportion
p̂
who frequently check nutrition labels equals to
582

/
1003

=
0.58

. The observed sample proportion is certainly
greater than

0.50, but could this just be
due to sampling variability?
Assuming H0 is true, that is, when p=0.50, the sample proportion
usually differs somewhat
from 0.50 simply because of chance variability from one sample to another.
p̂
Is it plausible that the sample proportion of p̂ =
0.58
occurred only as a result of this sampling variability, or is it unusual to observe a sample proportion
this large when population proportion is assumed to be p = 0.50?

If the null hypothesis, H0: p= 0.50, is true, you know the following:
1. the values of p̂ from different random samples will center around
I don't know
zero
p, the actual population proportion of all adult Americans who frequently check nutrition labels
Correct!
Mark 1.00 out of 1.00
That is, μp̂ , the mean of the sampling distribution formed by all possible p̂ values from different random
samples of all possible sample sizes
 , is 0.50
 Correct!
Mark 1.00 out of 1.00
2. the standard deviation of statistic
samples of size n=
p̂
, which describes how much the p̂ values from different
1003
tend to vary from sample to sample, is the standard deviation(p̂ ) =σp̂ = [p*(1-p)/n]0.5 =
[0.50*(1
0.5

)/
1003
 ]0.5
=
0.01578

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Lesson activity: Hypothesis testing about a population proportion and difference between two population proportions.: Attem…
3. Because the expected number of "successes", in a sample of this size and assuming H0
true, is n*p=1003*
0.5
which is

more than

10 and the expected number of "failures" is n*(1-p)=1003*[1-
0.5

] which is
more than

10 THEN the sampling distribution of sample proportion p̂ of
those who frequently check nutrition labels
may be

approximated by normal
distribution.
You can now use what you know about normal distributions to judge whether the observed sample
proportion p̂ =
0.58
would be surprising if the null hypothesis H0: p=0.50 were true. Converting the observed sample
statistic p̂ into corresponding statistic gives:

test statistic z=[p̂ -μp̂ ]/σp̂ =[
0.58

-
0.5

]/
0.01578

=
5.07

A test statistic is
a measure of how much the sample estimate differs from the null hypothesis. 
a measure of how much the sample estimate differs from the null hypothesis.
Mark -1.00 out of 1.00
This means that p̂ =
0.58

is
5.07

standard errors/deviation(s)
above
p=0.50 were true.
For a normal distribution this would be

than what you would expect it to be if the null hypothesis H0:
very unlikely

and, hence, can be regarded as
inconsistent

with the null hypothesis H0: p=0.50.
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Lesson activity: Hypothesis testing about a population proportion and difference between two population proportions.: Attem…
Z
-3
-2
-1
0
1
2
3
You can evaluate the likelihood of this happening by calculating the following probability:
p-value=P[observing a value of p̂ ≥ 0.58 when H0 is true]=
= P[z≥
5.07
when H0 is true]

= area under the z curve to the
right

of z=
5.07
and this area is equal to

0

A p-value is
the probability of observing a test statistic as or less extreme as the one observed in the sample
assuming that the null hypothesis is true.
I don't know
the probability of observing a test statistic as or more extreme as the one observed in the sample
assuming that the null hypothesis is true. OK!
Mark 1.00 out of 1.00
Therefore, the P-value you computed above indicates the
probability that you would observe this particular p-hat≥0.58 if the null hypothesis of H0:p=0.5 were
true
I have no idea
Mark 1.00 out of 1.00
is equal to
0
% (Hint: multiply the p-value above by 100%!)
Therefore, if H0 is true, there is virtually no


chance of seeing the sample proportion
=0.58 and corresponding z value this large and even larger as a result of sample-tosample variability alone. In this case,
the evidence for rejecting H0 in favor of Ha is very compelling.
p̂
there is no evidence from sample data for rejecting H0 in favor of Ha
Mark 1.00 out of 1.00
Conclude that, based on the sample data, there is a convincing evidence that the true
population proportion of all American adults who frequently check nutrition labels, p, is
greater than Correct!
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Lesson activity: Hypothesis testing about a population proportion and difference between two population proportions.: Attem…
likely equal to
Mark 1.00 out of 1.00
0.5.
Interestingly, in spite of the fact that there is strong evidence that a majority of American adults frequently
check nutrition labels, the responses to the second question suggest that the percentage of people who
then ignore the information on the label and purchase “bad” foods anyway is not small—the sample
proportion of adults who gave a response of very often or somewhat often was 0.440
You may use the applet by dragging the values of lower and upper boundaries of corresponding intervals
along the lines:
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Lesson activity: Hypothesis testing about a population proportion and difference between two population proportions.: Attem…
Question 6
Correct
Mark 30.00 out of 30.00
Breastfeeding
In the report “Healthy people 2020 objectives for the nation,” The Centers for Disease Control
and Prevention (CDC) set a goal of 0.341 for the proportion of mothers who will still be
breastfeeding their babies one year after birth (www.cdc.gov/breastfeeding/policy/hp2020.htm,
April 11, 2016, retrieved November 28, 2016). The CDC also estimated the proportion who
were still being breastfed one year after birth to be 0.307 for babies born in 2013
(www.cdc.gov/breastfeeding /pdf/2016breastfeedingreportcard.pdf, retrieved November 28,
2016). This estimate was based on a survey of women who had given birth in 2013. Suppose
that the survey used a random sample of 1000 mothers and that you want to use the survey
data to decide if there is evidence that the goal is not being met. Let p denote the population
proportion of all mothers of babies born in 2013 who were still breast-feeding at 12 months.
What is the shape, center, and variability of the sampling distribution of p̂ for random samples of size
1000 if the null hypothesis H0: p =0.341 is true?
The shape of the sampling distribution
is

approximately normal because
the expected number of "successes" in a sample of size 1000 and assuming H0 true is n*p
=1000*
0.341

which is
more than

10 "successes" AND/OR expected number of "failures" in a
sample of size 1000 and is n*(1-p)=1000*[10.341

] expected number of " failures" in a sample of 1000 mothers is
more than

10.
the center of the sampling distribution of the p̂ values from all possible random samples of
size 1000 is
zero
I don't know
p, the actual population proportion of all mothers of babies born in 2013 who were still
breast-feeding at 12 months Correct!
Mark 1.00 out of 1.00
and the expected value (or mean) of all possible sample proportions from this population, μ
 Correct!
p̂ , is equal to 0.341
Mark 1.00 out of 1.00
the variability of the sampling distribution of p̂ values, as measured by the standard deviation, is σp̂=
[p*(1-p)/n]0.5 =[0.341*(10.341
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
Lesson activity: Hypothesis testing about a population proportion and difference between two population proportions.: Attem…
)/
1000
 ]0.5
=
0.015

Would you be surprised to observe a sample proportion as small as p̂ =0.333 for a sample of size
1000 if the null hypothesis H0: p=0.341 were true?
I would
not be

Correct!
Mark 1.00 out of 1.00 surprised to observe a sample proportion p̂ =0.333 for a sample of
size n=1000 if the null hypothesis H0: p =0.341 were true because:
statistic z=[p̂ -μp̂]/σp̂ =[0.330.341

]/
0.015

=
-0.73

This z-statistic implies that p̂ =0.33 is
0.73

standard deviation(s)
below

Correct! negative z-statistic indicates for negative number of standard
deviations from the mean
Mark 1.00 out of 1.00 than what you would expect it to be if the null hypothesis H0:p=0.341 were true.
Observing the sample proportion p̂ as small as, or even smaller than, -0.33 would be likely
 for
a normal distribution.
The likelihood of the sample proportion p̂ as small as, or even smaller than, -0.33 happening is equal to:
P[observing a value of p̂≤ 0.33 when H0 is true]=
= P[z≤
-0.73

when H0 is true]
= area under the z curve to the
left

of z and this area is equal to
0.2327

Would you be surprised to observe a sample proportion as small as p̂=0.310 for a sample of
size n=1000 if the null hypothesis H0: p=0.341 were true?
I would
be

Correct!
Mark 1.00 out of 1.00 surprised to observe a sample proportion p̂ =0.31 for a sample of
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Lesson activity: Hypothesis testing about a population proportion and difference between two population proportions.: Attem…
size 1000 if the null hypothesis H0: p=0.341 were true because:
the corresponding z-statistic =[p̂ -μp̂]/σp̂
=[0.310.341

]/
0.015

=
-2.067

This implies that p̂=0.31 is about
2.067

standard deviation(s)
below

Correct! Because the z-score corresponding to p=0.31 is
negative!
Mark 1.00 out of 1.00 than what you would expect it to be if the null hypothesis H0: p=0.341
were true.
For (standard) normal distribution, this would be
unlikely

.
The likelihood of the above happening is equal to:
P[observing a value of p̂≤ 0.31 when H0 is true]=
= P[z≤
-2.067

when H0 is true]
= area under the z curve to the
left

of z and this area is equal to
0.0192

. This probability indicates that it is
unlikely

we would observe a p̂ at least as low as
what we observed in the sample ( p̂≤ 0.31) if H0 were true.
You may use the applet by dragging the values of lower and upper boundaries of corresponding intervals
along the lines:
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Lesson activity: Hypothesis testing about a population proportion and difference between two population proportions.: Attem…
Question 7
Correct
Mark 24.00 out of 24.00
As children grow they develop what is known as "representational insight," a connection between an
object and a symbol for that object. A random sample of youngsters 24-months old were
shown a video of someone putting a toy under 1 of 4 randomly placed boxes in a room
familiar to the child. They were then taken to the room, and asked to "find the toy." The
investigators reasoned that a child with representational insight should pick the correct box on
the first try. If the children were to overturn the boxes randomly, they would find the toy on
the first turn 25% of the time. Thirty children out of 57 subjects found the toy by turning over
the correct box on the first try.
Does this sample provide sufficient evidence that the proportion of 24-month old children who choose
the correct box on the first try is greater than 0.25?
Use a significance level of α=0.05 to test the appropriate hypothesis.
Follow the Five-Step Process For Hypothesis Test to carry out the hypothesis test.
Note: When entering your answers in this problem and others, round the values up to third decimal place, e.g. 0.521
instead of 0.52
H Hypothesis
1. p0 = the true population proportion of 24-month old children who pick the correct box on the first try.
2. H0: p0
=0.25

vs. Ha: p0
>0.25

M Method
1. test statistic, z=[\(\widehat{p}\)-μ
]/σ
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Lesson activity: Hypothesis testing about a population proportion and difference between two population proportions.: Attem…
2. significance level α=
0.05

C Check
1. Sample is random?
Yes

2. observed sample proportion,
=
0.526

3. Large sample size? The expected number of "successes", in a sample of this size if H0 is true, is n*p0 =
57
*

0.25
more or equal to


10 AND the expected number of "failures" is n*(1-p0)=
57
*

0.75
]

more or equal to

10
C Calculate
1. test statistic z corresponding to the observed sample proportion
]/[p*(1-p)/n]0.5
=[
is equal to [
-μ
0.52
-

0.25
]/

0.057
=

4.736

2. Hence, P-value=P[z>
4.736
]=

0

C Communicate results
1. Therefore, the P-value you computed above indicates the
I have no idea
probability that you would observe this particular p-hat≥0.5263 if the null hypothesis
of H0:p=0.25 were true
Mark 1.00 out of 1.00
is equal to
0

% (Hint: multiply the p-value above by 100%!)
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Lesson activity: Hypothesis testing about a population proportion and difference between two population proportions.: Attem…
2. Therefore, if H0 is true, there is
virtually no

chance of seeing the sample proportion
=0.5263 and corresponding z value this large and even larger as a result of sample-tosample variability alone. In this case,
there is no evidence from sample data for rejecting H0 in favor of Ha
the evidence for rejecting H0 in favor of Ha is very compelling.
Mark 1.00 out of 1.00
3. Since P-value
<

than α, you
reject
month old children who pick the correct box, p0, is

H0. Conclude that the true proportion of 24-
greater than

0.25.
4. That is, based on the sample data, it appears that 24-month old children
indeed

have
representational insight.
In statistical hypothesis testing, a result has statistical significance when it is very unlikely to have occurred
given the null hypothesis. More precisely, a study's defined significance level, denoted by α, is the
probability of the study rejecting the null hypothesis, given that the null hypothesis was assumed to be
true; and the p-value of your estimate is the probability of obtaining an estimate at least as extreme
(=very large or very small), given that the null hypothesis (=your assumption about the population) is true.
Your estimate is statistically significant, by the standards of the study, when P-value≤α. The significance
level for a study is chosen before data collection, and is typically set to 5%[or much lower—depending on
the field of study.
Conclude that based on the sample above, the estimated coefficient,
just appears to be greater than 0.25 only because of sampling/chance variation but true population
proportion for all babies actually may be not greater than 0.25
is "real" (that is, true population proportion of all babies may actually be greater than 0.25), 
Mark 1.00 out of 1.00
so based on the hypothesis testing you have performed above,
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is

statistically significant
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Lesson activity: Hypothesis testing about a population proportion and difference between two population proportions.: Attem…
Question 8
Correct
Mark 43.00 out of 43.00
Large-Sample Hypothesis Test for a Population Proportion
Example: Attention Span of Canadians
The article “You Now Have a Shorter Attention Span Than a Goldfish” describes a study of 2000 Canadians
over the age of 18 that was carried out by Microsoft. Study participants were asked whether the following
statement described them: “When nothing is occupying my attention, the first thing I do is reach for my
phone.”
Of the study participants in the age group 18 to 24 years old, 77% responded “yes” to this question.
Suppose that this group of 18- to 24-year-olds can be considered as a representative sample of Canadians
in this age group, and also suppose that 800 of the study participants were in this age group.
Does this sample support the claim that more than 75% of all Canadians in this age group would respond
“yes” to the given statement?
Note: When entering your answers in this problem and others, round the values up to third decimal place, e.g.
0.521 instead of 0.52
Begin by answering the four key questions for this problem:
Question type:
Hypothesis testing problem.
Correct! You are asked to test a claim about a population.
Estimation problem.
Mark 1.00 out of 1.00
Study type:
Sample data.
Correct! The data are from sample of Canadians age 18 to 24.
Experimental data.
Mark 1.00 out of 1.00
Type of data:
one

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categorical
Lesson activity: Hypothesis testing about a population proportion and difference between two population proportions.: Attem…

variable(s).
Number of Samples/Treatments:
one Correct!
two
more than two
Mark 1.00 out of 1.00
sample(s) or treatment(s).
Now use the five-step process for hypothesis testing:
Hypothesis
The claim is about the proportion of Canadians age 18 to 24 years who would say that the given statement
describes them, so the population characteristic of interest is a population proportion:
p =proportion of Canadians age 18 to 24 years who would say that the given statement describes them.
So, the relevant hypothesis will be
H0: p
=0.75

Ha: p
>0.75

Method
Select a potential method: Based on the correct answers to the 4 key questions, Table 7.1 directs you to the
following method: One-Sample z Hypothesis Test for a Population Proportion described in chapter 10

The test statistic is
z=[
-
0.75
]/[

0.75

*(1-
0.75

)/
800
 ]0.5
You also need to select a significance level for the test. In some cases, a significance level will be specified. If
not, you should choose a significance level based on a consideration of the consequences of Type I and
Type II errors.
In this example, a Type I error would be
deciding that more than 75% of Canadians age 18 to 24 years would say that the given statement
describes them, when in fact the actual percentage is 75% or less.  Correct! Type I error is the error of
rejecting the H0 when H0 is true!
thinking that 75% or fewer would respond “yes” when the actual percentage is greater than 75%.
Mark 1.00 out of 1.00
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Lesson activity: Hypothesis testing about a population proportion and difference between two population proportions.: Attem…
A Type II error is
deciding that more than 75% of Canadians age 18 to 24 years would say that the given statement
describes them, when in fact the actual percentage is 75% or less.
thinking that 75% or fewer would respond “yes” when the actual percentage is greater than 75%.
Absolutely correct! Type II error is the error of failing to reject the H0 when H0 is false!
Mark 1.00 out of 1.00
Neither type of error is much more serious than the other, you might choose a value for a of 0.05 (as
opposed to something much smaller or much larger). Hence, choose
Significance level, α=
0.05

Check
Check to make sure that the method selected is appropriate: There are two conditions that need to be met
for the large sample test for a population proportion to be appropiate:
1. The sample size is

enough because n*p=800*
more or equal to

10 successes in the sample AND n*(1-p)=800*(1-
more or equal to

10 failures in the sample.
large
0.75

=
600

] which is
0.75

)=
200

] which is
2. The requirement of a random or representative sample is more difficult. The researchers conducting the
study indicated that they believed that the sample was representative of Canadians in this age group. If
this is the case, it is reasonable.
Calculate
Use the sample data to perform any necessary calculations:
n=
800

a sample proportion,
=0.77
Hence, plugging the above into the test statistic formula gives:
z=[
-
0.75
]/[

0.75

*(1-
0.75
)/

800

]0.5
=[
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Lesson activity: Hypothesis testing about a population proportion and difference between two population proportions.: Attem…
0.77
-

0.75
]/[

0.75
*(1-

0.75
)/

800

]0.5
=
1.306

The associated P-value of the observed sample proportion
=0.77:
This is the
upper-tailed Correct! Because the inequality in the alternative hypothesis >
lower-tailed
two-tailed
Mark 1.00 out of 1.00
test
So, P-value = P[z≥
1.306
when H0 is true]

= area under the z curve to the
right

of this z and is equal to
0.095776

.
Communicate Results
Because the P-value is
greater than or equal
Correct!
less than
Mark 1.00 out of 1.00
to the selected significance level α=
0.05

, you
fail to reject the null hypothesis.
Correct!
reject the null hypothesis.
Mark 1.00 out of 1.00
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Lesson activity: Hypothesis testing about a population proportion and difference between two population proportions.: Attem…
The final conclusion for the test should be stated in context and answer the question posed.
Conclusion:
The sample does not provide convincing evidence that more than 75% of Canadians ages 18 to 24 think
that the statement “When nothing is occupying my attention, the first thing I do is reach for my phone”
describes them. Correct!
The sample provides convincing evidence that more than 75% of Canadians ages 18 to 24 think that the
statement “When nothing is occupying my attention, the first thing I do is reach for my phone” describes
them.
Mark 1.00 out of 1.00
Conclude that based on the sample above, the estimated coefficient,
just appears to be different from 0.75 only because of sampling/chance variation but the true
population proportion may not be "really" different from 0.75
is "real" (that is, true population proportion is likely to be actually different from 0.75),
Mark 1.00 out of 1.00
so based on the hypothesis testing you have performed above,
is not

statistically significant
You may compute the corresponding P-values by dragging the values of lower and upper boundaries of
corresponding intervals along the lines in the applet below:
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Lesson activity: Hypothesis testing about a population proportion and difference between two population proportions.: Attem…
Question 9
Correct
Mark 5.00 out of 5.00
Chapter 11: Asking and Answering Questions About the Difference Between Two Population Proportions
Many studies are carried out in order to estimate the difference between two population proportions. For
example, researchers have studied how people of different ages use mobile technology. As part of a survey
conducted by Pew Research (“Americans’ Views on Mobile Etiquette,” August 26, 2015,
www.pewinternet.org, retrieved December 12, 2016), people in a representative sample of 708 American
adults age 18 to 29 were asked if they thought it was OK to use a cell phone while at a restaurant. The same
question was asked of a representative sample of 1029 adult Americans age 30 to 49.
You might expect the proportion who think it is OK to use a cell phone at a restaurant is higher for the 18
to 29 age group than for the 30 to 49 age group, but how much higher? In this lesson activity, you will learn
how sample data can be used to answer questions like this.
In the applet below, you see:
the red-colored
sampling

samples from Population 1.
the blue-colored sampling

distribution of
1,
distribution of
proportion of successes from random
2,
proportion of successes from random
samples from Population 2
Check the "Difference" box below and the applet will display:
the
green

-colored sampling distribution of
1-
2, the difference in sample proportions of
successes from independently selected random samples from Population 1 and Population 2. It will help you
to compare proportions from two populations.
Compare the center and spread of
1-
red and blue sampling distributions of
of answers:
The mean/center of
1-
2 sampling
distribution to the centers and spreads of the
1 and
2 by choosing the most appropriate versions
2 sampling
distribution looks
smaller
the means/centers of the red and blue sampling distributions of
The spread/variation of
1-
2 sampling
1 and
distribution looks
the spreads of the red sampling distribution of
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
1
wider
compared to
2.

compared to
and the blue sampling distribution
2.
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Lesson activity: Hypothesis testing about a population proportion and difference between two population proportions.: Attem…
This lesson activity introduces you to a new test, incorporating two populations and comparing
proportions between them. You will consider situations where specific tests apply and learn how to
appropriately select a test.
The chapter introduces a new complexity to the analysis problem by raising the number of groups being
analyzed from one to two. This means there are now two distinct sets of numbers to keep track of and you
can easily become overwhelmed by the number of distinct elements involved in each formula.
It is important to remember the four key questions that should be asked and answered prior to any
statistical analysis to ensure that you consider the context of the problem prior to applying any
procedures.
After completing chapter 11, students should know the following information:
1. Understand the difference between testing two population proportions and testing a single population
proportion
2. Understand what is being tested with a two proportion z-test
3. Understand what assumptions need to be fulfilled to use the z-test for two proportions
4. Be able to formulate hypotheses for this test
5. Be able to compute the z-statistic for this test
6. Understand what to estimate and how to compute the estimate for comparing two
population proportions
7. Understand how the confidence interval is formed and what the interval refers to
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Lesson activity: Hypothesis testing about a population proportion and difference between two population proportions.: Attem…
Question 10
Partially correct
Mark 49.09 out of 50.00
In the video below I explain
the general properties of the sampling distribution of the
difference in sample proportions
for independently selected random samples:
Descriptions of four studies are given below.
In each of the studies, the two populations of interest are the students at a particular university who
live on campus and the students who live off campus. Indicate whether
samples are paired or independently selected in each of the studies below:
Study 1: To determine if there is evidence that the proportion of money spent on housing expenses each
month differs for the two populations, a random sample of 45 KIMEP students who live on campus and a
random sample of 50 KIMEP students who live off campus are selected.
independently selected samples You are right! Two separate random samples were taken.
paired samples
Mark 1.00 out of 1.00
Study 2: To determine if the proportions of time spent studying differs for the two populations, a random
sample of KIMEP students who live on campus is selected. Each student in this sample is asked what
proportion of time he or she spends studying each week. For each of these KIMEP students in the sample
who live on campus, a KIMEP student who lives off campus and who studies the same number of hours
per week is identified and included in the sample of students who live off campus.
independently selected samples
paired samples You are right! Here, only one random sample was taken. The other sample was a
matched pair to the first one.
Mark 1.00 out of 1.00
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Lesson activity: Hypothesis testing about a population proportion and difference between two population proportions.: Attem…
Study 3: To determine if the proportion of time spent surfing the web differs for the two populations, a
random sample of KIMEP students who live on campus and who have a brother or sister who also attends
KIMEP but who lives off campus is selected. The sibling who lives on campus is included in the on campus
sample, and the sibling who lives off campus is included in the off-campus sample.
independently selected samples
paired samples You are right! Here, only one random sample was taken. The other sample was
matched to the first one.
Mark 1.00 out of 1.00
Study 4: To determine if the proportion of total budget spent on tuition and textbooks differs for the two
populations, a random sample of KIMEP students who live on campus is selected. A separate random
sample of the same size is selected from the population of students who live off campus.
independently selected samples You are right! Two separate random samples were taken.
paired samples
Mark 1.00 out of 1.00
In the example below you will be applying the general properties of the sampling distribution of the
difference in sample proportions and use data from two independent samples to construct a
confidence interval estimate of a difference between two population proportions:
Cell Phone Etiquette
Let’s return to the example introduced in the Introduction part of this lesson activity to answer the question,
“How much greater is the proportion who think it is OK to use a cell phone in a restaurant for people age 18 to
29 than for those age 30 to 49?”
The study described earlier found that 354 of 708 people in the sample of 18- to 29-yearolds and 412 of the 1029 people in the sample of 30- to 49-year-olds said that they thought
it was OK to use a cell phone in a restaurant.
Based on these sample data, what can you learn about the actual difference in proportions for these two
populations?
Start by answering the four key questions for this problem:
Question type:
Hypothesis testing problem.  Wrong. In this example, you want to estimate the difference in the
proportion of people age 18 to 29 who think it is OK to use a cell phone in a restaurant and this proportion
for those age 30 to 49.
Estimation problem.
Mark 0.00 out of 1.00
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Lesson activity: Hypothesis testing about a population proportion and difference between two population proportions.: Attem…
Study type:
Sample data.
Correct!
Experiment data.
Mark 1.00 out of 1.00
Type of data:
one Correct!There is only one variable with two values - those who think it is OK to use a cell phone in
a restaurant and those who don't think so.
two
Mark 1.00 out of 1.00
AND
categorical Correct! There is one categorical variable with two values - those who think it is OK to use
a cell phone in a restaurant and those who don't think so.
numerical
Mark 1.00 out of 1.00
variable(s).
Number of Samples/Treatments:
one
two
Correct! There are two samples - one from each age group!
more than two
Mark 1.00 out of 1.00
sample(s) or treatment(s).
Now use the five-step process to learn from data:
Estimate
Explain what population characteristic you plan to estimate:
p1-p2, the unknown difference in the p1 =proportion of people age 18 to 29 who think it is OK to use a
cell phone in a restaurant and p2=this proportion for those age 30 to 49. Correct!
I don't know
Mark 1.00 out of 1.00
Method
Select a potential method: Based on the correct answers to the four key questions, Table 7.1 directs you to
the following method: Two-Sample z Confidence Interval for a Difference in Proportions described in chapter 11

Associated with every confidence interval is a
confidence level Correct! A confidence level defines the distance for how close the confidence
interval's limits are to the sample statistic. While significance level measures the probability of rejecting the
null hypothesis when it is true. Significance level is the complement of the confidence level, if a confidence
level is used to make the test. Thus, if the confidence level is chosen as 0.95, the significance level is 0.05
significance level
Mark 1.00 out of 1.00
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Lesson activity: Hypothesis testing about a population proportion and difference between two population proportions.: Attem…
. You also need to specify it. For this example,
confidence level

of 90% will be used.
Check
Check to make sure that the method selected is appropriate:
There are two conditions that need to be met in order to use the large sample confidence interval for a
difference in populations proportions:
1. The large-samples condition is
satisfied

because
with “success” defined as "a person who thinks it is OK to use a cell phone in a restaurant",
there are
354

“successes” in the sample of 18- to 29-year-olds which is
more than or equal to

10 and
more than or equal to

10.
412

“successes” in the sample of 30- to 49-year-olds which is
and there are
354

“failures” in the sample of 18- to 29-year-olds which is
more than or equal to

10 and
617

“failures” in the sample of 30- to 49-year-olds which is
more than or equal to

10.
2. The requirement of independent random samples or samples that are representative of the
corresponding populations is more difficult. The researchers who collected these data indicate in the
report that they believe that the samples are representative of the two populations of interest.
Calculate
Use the sample data to perform any necessary calculations:
n18-29 =
708

and n30-49 =
1029

18-29=
354

/
708

=
0.5

30-49=
412

/
1029

=
0.4

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Lesson activity: Hypothesis testing about a population proportion and difference between two population proportions.: Attem…
Use the applet below to find the appropriate z-critical value for a confidence level of
90%: z90%critical value =
1.645

**
**Use the applet below to check your answer by dragging the lower boundary of z to the -z-critical value in your answer
and the upper boundary of z to the +z-critical value in your answer. Verify that the area between the -z-critical
value and +z-critical value is 0.9 or 90%!
Now use the formula below, to establish a confidence interval for the
the difference in sample proportions by plugging in the values of n18-29, n30-49, 18-29 and
30-49:
Confidence interval is:
18-29 -
30-49±z90%critical value
*[
18-29*(1-
18-29)/n18-29+
30-49*(1-
0.5
30-49)/n30-49] =
=
0.5

-
0.4

1.645

*[
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Lesson activity: Hypothesis testing about a population proportion and difference between two population proportions.: Attem…
0.5
*(1-

0.5
)/

708
+

0.4
*(1-

0.4
)/

1029
 ]0.5=
=
0.1

0.0398
=

=(
0.0602

,
0.1398

)
Interpreting Confidence Intervals for a Difference:
Communicate Results
Confidence interval: Assuming that the samples were selected in a reasonable way, you can
be
90
% confident that the actual difference in the proportion of people who think it is OK to use
a cell phone in a restaurant for 18- to 29-year-olds and for 30- to 49-year-olds is between

0.0602

and
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Lesson activity: Hypothesis testing about a population proportion and difference between two population proportions.: Attem…
0.1398

. This means that you can be confident that the percentage is higher for
than for

30- to 49-year-olds

18- to 29-year-olds
by somewhere between
6

% and
14

%.
Confidence

level: The method used to construct this interval estimate will be successful in capturing
the actual value of the difference in population proportions about
90

% of the time.
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Lesson activity: Hypothesis testing about a population proportion and difference between two population proportions.: Attem…
Question 11
Partially correct
Mark 49.02 out of 50.00
Working Parents
The report “Raising Kids and Running a Household: How Working Parents Share the Load”
described a survey of parents of children under the age of 18. Each person in a representative
sample of 825 working fathers and a sample of 586 working mothers was asked if balancing
the responsibilities of a job and a family was difficult. It was reported that 429 (52%) of the
fathers surveyed and 352 (60%) of the mothers surveyed said that it was difficult. The two
samples were independently selected and were thought to be representative of working fathers
and mothers of children under 18 years old.
This information can be used to estimate the difference between the proportion of working
fathers who find balancing work and family difficult, pf, and this proportion for working
mothers, pm.
Start by answering the four key questions for this problem:
Question type:
Hypothesis testing problem.
Estimation problem. Correct! In this example, you want to estimate the difference
between the proportion of working fathers who find balancing work and family difficult and this
proportion of working mothers.
Mark 1.00 out of 1.00
Study type:
Sample data.
Correct!
Experiment data.
Mark 1.00 out of 1.00
Type of data:
one Correct!There is only one variable with two values - difficult an not difficult
two
Mark 1.00 out of 1.00
AND
categorical
Correct! There is one categorical variable with two values - difficult and not difficult
numerical
Mark 1.00 out of 1.00
variable(s).
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Lesson activity: Hypothesis testing about a population proportion and difference between two population proportions.: Attem…
Number of Samples/Treatments:
one
two Correct!There are two samples - one from population of fathers and one from the population of
mothers
more than two
Mark 1.00 out of 1.00
sample(s) or treatment(s).
Now use the five-step process to learn from data:
Estimate
Explain what population characteristic you plan to estimate:
pf-pm, the difference in the pf =proportion of working fathers who find balancing work and family
difficult and pm=proportion of working mothers who find balancing work and family difficult
Correct!
I don't know
Mark 1.00 out of 1.00
Method
Select a potential method: Based on the correct answers to the four key questions, Table 7.1 directs you to
the following method: Two-Sample z Confidence Interval for a Difference in Proportions described in chapter 11

Associated with every confidence interval is a
significance level
confidence level Correct! A confidence level defines the distance for how close the confidence
interval's limits are to the sample statistic. While significance level measures the probability of rejecting the
null hypothesis when it is true. Significance level is the complement of the confidence level, if a confidence
level is used to make the test. Thus, if the confidence level is chosen as 0.95, the significance level is 0.05
Mark 1.00 out of 1.00
You also need to specify it. For this example,
confidence level
Correct!
significance level
Mark 1.00 out of 1.00
of 95% will be used.
Check
Check to make sure that the method selected is appropriate:
There are two conditions that need to be met in order to use the large sample confidence interval for a
difference in populations proportions:
1. The large-samples condition is
satisfied

because:
with “success” defined as "someone who finds balancing work and family difficult", there are
429

“successes” in the sample of working fathers which is
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more than or equal to

10 and
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Lesson activity: Hypothesis testing about a population proportion and difference between two population proportions.: Attem…
352

“successes” in the sample of working mothers which is
more than or equal to

10.
and there are
396

“failures” in the sample of working fathers which is
more than or equal to

10 and
234

“failures” in the sample of working mothers which is
more than or equal to

10.
2. The example specified that the samples were independently selected and that the samples could be
considered as representative of the populations.
Calculate
Use the sample data to perform any necessary calculations:
nf =
825

and nm =
586

f=
429

/
825

=
0.52

m=
352

/
586

=
0.6

Use the applet below to find the appropriate zcritical value for a confidence level of
95%: z95%critical value =
1.959

**
**Use the applet below to check your answer by dragging the lower boundary of z to the -z-critical value in your
answer and the upper boundary of z to the +z-critical value in your answer. Verify that the area between the -zcritical value and +z-critical value is 0.95 or 95%!
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Lesson activity: Hypothesis testing about a population proportion and difference between two population proportions.: Attem…
Now use the formula below, to establish a confidence interval for the the difference in sample
proportions by plugging in the values of nf, nm,
f and
m:
Confidence interval is:
f
-
m±z95%critical value*[
f*(1-
f
)/nf+
m*(1-
0.5
m)/nm] =
=
0.52

-
0.6

1.959

*[
0.52

*(1-
0.52

)/
825

+
0.6
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Lesson activity: Hypothesis testing about a population proportion and difference between two population proportions.: Attem…
*(1-

0.6
)/

586
 ]0.5=
=
-0.08

0.052
=

=(
-0.132

,
-0.028

)
Endpoints of the confidence interval you have obtained above are
both negative

Communicate Results
Confidence interval: You can be
95
% confident that the actual difference in the proportions of working fathers and working
mothers who find balancing work and family difficult is somewhere in between

-0.132

and
-0.028
. This means that you think that the proportion of working fathers who find balancing
work and family difficult is less
 than the proportion of working mothers somewhere

between (enter from smaller to larger value)
2.8

% and
13.2

%.
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Confidence
Lesson activity: Hypothesis testing about a population proportion and difference between two population proportions.: Attem…

level: The method used to construct this interval estimate will be successful in capturing
the actual value of the difference in population proportions about
95

% of the time.
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Lesson activity: Hypothesis testing about a population proportion and difference between two population proportions.: Attem…
Question 12
Correct
Mark 59.00 out of 59.00
Testing Hypothesis About the Difference between Two Population Proportions
According to the video above, p1=p2 is equivalent to
p1 – p2 = 0 Correct!
p1 –p2>0
p1 –p2<0
Mark 1.00 out of 1.00
Similarly, p1>p2 is equivalent to
p1 – p2 > 0
Correct!
p1 –p2=0
p1 –p2<0
Mark 1.00 out of 1.00
and
p1<p2 is equivalent to
p1 – p2 < 0 Correct!
p1 –p2=0
p1 –p2>0
Mark 1.00 out of 1.00
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Lesson activity: Hypothesis testing about a population proportion and difference between two population proportions.: Attem…
Detecting Plagiarism
In the example below you will apply the general properties of the sampling distribution of the
difference in sample proportions and use data from two independent samples to test hypotheses
about the difference between two population proportions:
The report “The 2016 Inside Higher Ed Survey of Faculty Attitudes on Technology” describes the results of a
survey of 1129 full-time college faculty and 293 part-time college faculty. Survey participants were
asked if they require undergraduate students to submit papers through plagiarism-detection
software; 40% of the full-time faculty and 38% of the part-time faculty said “yes.”
One question of interest is whether the proportion who require students to submit papers through
plagiarism detection software is different for full-time faculty and part-time faculty.
The following table summarizes what you know so far.
Notice that the two sample proportions are not equal. But even if the two population
proportions were equal, the two sample proportions wouldn’t necessarily be equal because of
sampling
 variability—differences that occur from one sample to another just by chance.
The important question is whether chance is a believable explanation for the observed difference in the two
sample proportions or whether this difference is large enough that it is unlikely to have occurred just by
chance. A hypothesis test will help you to make this determination.
Start by answering the four key questions for this problem:
Question type:
Hypothesis testing problem. Correct! In this example, you want to test the claim about the difference
in proportions for the two faculty groups.
Estimation problem.
Mark 1.00 out of 1.00
Study type:
Sample data.
Correct!
Experiment data.
Mark 1.00 out of 1.00
Type of data:
one Correct!There is only one variable with two values - those who use plagiarism detecting software
and those who don't use
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Lesson activity: Hypothesis testing about a population proportion and difference between two population proportions.: Attem…
two
Mark 1.00 out of 1.00
AND
categorical Correct! There is only one categorical variable with two values - those who use plagiarism
detecting software and those who don't use
numerical
Mark 1.00 out of 1.00
variable(s).
Number of Samples/Treatments:
one
two
Correct! There are two samples - one from full-time faculty and another from part-time
more than two
Mark 1.00 out of 1.00
sample(s) or treatment(s).
Now use the five-step process for hypothesis testing:
H Hypotheses
The claim is about the difference in proportions for the two faculty groups. Use the following notation:
pft =proportion of full-time college faculty who require students to submit papers through plagiarismdetection software
ppt= proportion of part-time college faculty who require students to submit papers through plagiarismdetection software
The question of interest (are the population proportions different?) translates into the following alternative
and null hypothesis:
H0: pft-ppt=
0

no difference in population proportions Correct! Under the null: pft=ppt or pft-ppt=0.
Although we observe some difference between sample proportions but it is possible that
population proportions are still the same.
there is a difference in population proportions
Mark 1.00 out of 1.00
Ha: pft-ppt≠
0

no difference in population proportions
there is a difference in population proportions
Correct!
Mark 1.00 out of 1.00
Method
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Lesson activity: Hypothesis testing about a population proportion and difference between two population proportions.: Attem…
Select a potential method: Based on the correct answers to the four key questions, Table 7.1 directs you to
the following method: Two-Sample z Test for a Difference in Proportions described in chapter 11

Mark 3.00 out of 3.00
The test statistic for this test is:
z =[
ft -
pt]/[
c*(1-
c)/nft+
c*(1-
0.5
c)/npt]
You also need to select a significance level for the test. Because no significance level was
specified, you should choose a significance level based on a consideration of the
consequences of Type I and Type II errors.
In this example, a Type I error would be
deciding that the proportions were not the same for the two faculty groups, when in fact
the actual proportions were equal.  Correct! Type I error is the error of rejecting the H0
when H0 is true!
not thinking that there is a difference in the population proportions, when in fact the two
proportions were not the same.
Mark 1.00 out of 1.00
A Type II error is
deciding that the proportions were not the same for the two faculty groups, when in fact
the actual proportions were equal.
not thinking that there is a difference in the population proportions, when in fact the two
proportions were not the same. Correct! Type II error is the error of failing to reject the
H0 when H0 is false!
Mark 1.00 out of 1.00
Neither type of error is much more serious than the other, you might choose a value for a of
0.05 (as opposed to something much smaller or much larger).
Hence, chosen significance level, α=
0.05

Check
There are two conditions that need to be met in order for the large-sample test for a difference in
population proportions to be appropriate.
The large samples condition is easily verified.
1. The large-samples condition is
nft*
satisfied

because
ft =1129*
0.4
which is

(1-
more or equal to

10 successes in the sample of full-time college faculty AND nft*
ft)=1129*(1-
0.4

) which is
npt*
more or equal to

10 failures in the sample of full-time college faculty AND
pt=293*
0.38
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Lesson activity: Hypothesis testing about a population proportion and difference between two population proportions.: Attem…
which is

more or equal to

10 successes in the sample of part-time college faculty AND npt*(1-
pt)=293*(10.38

) which is
more or equal to

10 failures in the sample of part-time college faculty.
2. From the study description, it is known that the samples were independently selected. The report states
that the responses are from a survey sent to faculty members by e-mail. The survey had a large non-response
rate, so it is possible that the samples are not representative of the populations of interest. You should keep
this possibility in mind when you interpret the outcome of the hypothesis test.
Calculate
To calculate the value of the test statistic, you need to find the values of the sample proportions and the
value of
c, the combined estimate of the common population proportion.
nft =
1129
and npt =

293

ft=
0.4

and
pt=
0.38

c =(nft
*
ft+npt
*
pt)/[nft+npt]=
=(
1129

*
0.4

+
293

*
0.38

)/[
1129

+
293

]=
0.396

Hence, the value of test statistic is:
z =[
ft -
pt]/[
c*(1-
c)/nft+
c*(1-
0.5
c)/npt]
=[
0.4

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Lesson activity: Hypothesis testing about a population proportion and difference between two population proportions.: Attem…
0.38
)]/[

0.396

*(1-
0.396

)/
1129

+{1:NM:=0.396:0.01}*(1-
0.396

)/
293

]0.5 =
0.62

A test statistic is {1:MRS:=a measure of how much the sample estimate differs from the null
hypothesis.#Right!~a measure of how much the sample estimate differs from the alternative hypothesis.#
}
This is the
lower-tailed
two-tailed
Correct! Indeed, the inequality in the alternative hypothesis is ≠
upper-tailed
Mark 1.00 out of 1.00
test
So, in this example the p-value =
=likelihood of the observed difference in the sample proportions
ftpt =0.02 or even larger happening
if the null about actual populations proportions H0:pft-ppt=0 were true)=
=
2
2


*area under the z curve to the
*P[z
>
right

of the calculated value of z and is equal to =

0.62
if H0 were true]

=
0.5352

.
Communicate Results
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Lesson activity: Hypothesis testing about a population proportion and difference between two population proportions.: Attem…
This p-value you have computed above implies that the z-statistic you have computed would be
quite likely
 if the H0 is true, and, hence, the sample data can be regarded as
inconsistent
consistent
Yes!
Mark 1.00 out of 1.00
with the null hypothesis H0
Because the p-value is
greater than or equal to
Correct
less than
Mark 1.00 out of 1.00
the selected significance level α=
0.05

, you
fail to reject the null hypothesis.
Correct!
reject the null hypothesis.
Mark 1.00 out of 1.00
of no difference between the proportions.
Conclusion:
Assuming that the samples are representative of the populations of interest, you would conclude that
there is not convincing evidence that the proportion who require students to turn in papers
using plagiarism detection software differs for full-time and part-time college faculty
members. Correct!
there is convincing evidence that the proportion who require students to turn in papers
using plagiarism detection software differs for full-time and part-time college faculty
members.
Mark 1.00 out of 1.00
You may compute the corresponding P-values by dragging the values of lower and upper boundaries of
corresponding intervals along the lines in the applet below:
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Lesson activity: Hypothesis testing about a population proportion and difference between two population proportions.: Attem…
Question 13
Correct
Mark 22.00 out of 22.00
How to Read Research Papers:
Randomized Controlled Trials (RCTs)
The West Point study of the effects of classroom electronics featured in this part of the
lesson activity was an experiment
 . In this kind of study, researchers will manipulate
the environment in some way to observe the response of the objects of interest (people,
mice, ball bearings, etc.). When the objects of interest in an experiment are people, we refer
to them as subjects; otherwise, we call them experimental units. In this case, the researchers
randomly assigned the subjects—the students—to classes subject to three treatments:
laptops and tablest use allowed in class, only tablets face-up use allowed in class and no
tablets neither laptops allowed in class.
The treatment group in the West Point study is the group that is
allowed to use electronics in the classroom.
not allowed to use electronics in the classroom.
attending West Point.
not attending West Point.
Mark 1.00 out of 1.00
Which treatment effect below would be considered statistically significant?
Treatment effect: 0.3, standard error: 0.2
Treatment effect: 5, standard error: 4
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Lesson activity: Hypothesis testing about a population proportion and difference between two population proportions.: Attem…
Treatment effect: -2, standard error: 0.8 You are right! Because treatment effect is
more than two standard errors below zero effect! (=-2/0.8=2.5 standard errors!). Hence, it
is statistically different from zero effect.
Treatment effect: -0.4, standard error: 0.3
Mark 1.00 out of 1.00
The table below (‘Table 2’) presents summary statistics from the West Point study of the effects of classroom
electronics featured in this video and is referenced in the first two questions below:
What proportion of Treatment Group 1 were Division I athletes reported in the paper's summary statistics?
40%
 or just 0.40

The standard deviation for the proportion of Treatment Group 1 who were Division I athletes
is not

reported next to the sample proportion inside the square brackets in the paper's summary.
The standard deviation for the proportion of Treatment Group 1 who were Division I athletes is calculated
as [
*(1)/n]0.5 =[
0.4

*(1-
0.4

)/
248
 ]0.5=
=
0.0311

Column 4 (“Both treatments vs control”) compares a sample containing both treatment groups with the
control group. Which of the following is true?
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Lesson activity: Hypothesis testing about a population proportion and difference between two population proportions.: Attem…
(a) The proportion of black students in the treatment groups is significantly below that in the control
group.
(b) Average computer use is significantly higher in treatment groups than in the control group.
Right! Indeed, the difference between the two groups is 0.42 and it is statistically different from zero
because standard error of that difference estimate is 0.02 (inside parenthesis). So 0.42/0.02=21 which is
much greater than 2 in absolute value! Also see *** which indicates that the estimate of the difference in
computer use is statistically significant at 1% significance level! And there is at least one more version of
the answer which is also correct!
(c) %50%White participants make up a larger proportion of the treatment groups than the control
groups, but this difference is not statistically significant.  Right! The column (4) does shows that the
proportion of white students is 2% higher in treatment groups compared to the control group but it is
not statistically significant because the standard error of the difference is 0.04 (inside parenthesis). So
0.02/0.04=0.5 which is less than 2 in absolute value! And there is at least one more version of the answer
which is also correct!
(d) b and c are both correct!
Awesome!
All of the above
Mark 1.00 out of 1.00
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