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Using the technology available to you, visually and statistically
inspect this data. From this, construct the most appropriate statis-
tical hypotheses.
A.H0:µ˜D=0 HA:µ˜D<0
B.H0:µFirstChild=µSecondChild HA:µFirstChild>µSecondChild
C.H0:µFirstChild=µSecondChild HA:µFirstChil d is not equal to µ
SecondChild D.H0:µ˜D=0 HA:µ˜D>0
E.H0:µ˜FirstChild=µ˜SecondChild HA:µ˜FirstChild>µ˜SecondChild
F.H0:µFirstChild=µSecondChild HA:µFirstChild< µSecondChild
G.H0:µD=0 HA:µD<0
H.H0:µD=0 HA:µD>0
I.H0:µ˜FirstChild=µ˜SecondChild HA:µ˜FirstChild is not equal to µ
S˜econdChild
J.H0:µ˜D=0 HA:µ˜D is not equal to 0.
K.H0:µD=0 HA:µD is not equal to 0.
L.H0:µ˜FirstChild=µ˜SecondChild HA:µ˜FirstChild<µ˜SecondChild
Problem 1. (6 points)
Women are trying measure the difference in oxytocin (love hor-
mone) in the week’s after conception, for two different pregnan-
cies, below you can see the levels. Determine if the oxytocin level
has significantly changed when comparing the first child to the
second.
*Use α=5%
*Use FirstChild=sample1
Download .csv file
Find the p-values of the normality tests, use at least three dec-
imals in your answer.
First child p-value=___.
Second child p-value=___.
Paired difference p-value=___.
(Find the value of the test statistic for this test, use at least one
decimal in your answer.
Test Statistic =
Determine the P-value of your statistical test, and report it to
at least three decimal places.
P=
At α=0.05,this data indicates that you will
[reject/failtoreject]the null hypothesis.
You can say the oxy-tocin level for the first child
[is the same as/more than/less than/is differenr from the second child]
1
Problem 1.csv
"","Woman","first.child","second.child"
"1",1,4.95,1.83
"2",2,6.52,6.43
"3",3,4.22,3.63
"4",4,4.72,4.28
"5",5,4.99,4.62
"6",6,4.56,2.61
"7",7,4.56,5.37
"8",8,4.52,2.33
"9",9,5.24,5.58
"10",10,6.12,5.32
"11",11,5.18,4.86
"12",12,4.15,4.93
"13",13,6.08,1.14
"14",14,5.65,4.19
"15",15,4.27,-0.29
"16",16,5.09,0.0499999999999998
"17",17,5.33,3.17
"18",18,4.92,4.89
"19",19,4.73,3.79
"20",20,5.04,4.73
"21",21,5.58,4.5
"22",22,5.92,3.53
"23",23,4.44,3.1
Problem 2. (6 points)
Does sleep change mental performance? Are creativity and prob-
lem solving linked to adequate sleep? This question was the
subject of research conducted by German scientists at the Univer-
sity of Lubeck (Nature, Jan. 22, 2004).
322 volunteers were divided into two equal-sized groups. Each
volunteer took a math test that involved transforming strings of
eight digits into a new string that fit a set of given rules, as well
as a third, hidden rule. Prior to taking the test, one group received
eight hours of sleep, while the other group stayed awake all night.
The scientists monitored the volunteers to determine whether and
when they figured out the third rule. Of the volunteers who slept,
35 discovered the third rule; of the volunteers who stayed awake
all night, 25 discovered the third rule.
From the study results, what can you infer about the proportions
of volunteers in the two groups who discover the third rule?
(a) Find a 98.19 confidence interval for P(slept)-P(stayedawake)
,the difference in proportions in volunteers in the two groups.
Lower Bound of 98.19%CI=
Upper Bound of 98.19%CI=
(b) The confidence interval found in part (a) indicates the propor-
tion of volunteers in the group who slept and discover the third
rule is [higher/lower/the same] as the proportion of who had no
sleep and discover the third rule.
(c) Based on this data,
•We are
• There is a
• I think maybe that...There is a
• chance
• confident
• likelihood
the [true/estimated/sample] difference in propotion between
these two populations is between the lower and upper interval I
have stated.
2
Problem 3. (6 points)
Religious symbolism in TV commercials. Gonzaga University
professors conducted a study of television commercials and pub-
lished their results in the Journal of Sociology, Social Work and
Social Welfare (Vol. 2, 2008).
The key research question was: ”Do television advertisers use
religious symbolism to sell goods and services?” In a sample of
541 TV commercials collected in 1998, only 16 commercials used
religious symbolism. Of the sample of 1746 TV commercials ex-
amined in the more recent study, 100 commercials used religious
symbolism. Conduct an analysis to determine if the percentage of
TV commercials that use religious symbolism has by more than
0.26
In the context of this question determine the correct Hypothesis
using the recent data as sample 1 (use the 1998 information as
sample 2).
(a) Formulate the appropriate statistical hypotheses.
• A. H0 : ppresent = p1998 HA : ppresent < p1998
• B. H0 : ppresent = p1998 HA : ppresent > p1998
• C. H0 : ppresent = p1998 HA : ppresent 6= p1998
• D. H0 : p̂present = p̂1998 HA : p̂present 6= p̂1998
• E. H0 : p̂present = p̂1998 HA : p̂present < p̂1998
• F. H0 : p̂present = p̂1998 HA : p̂present > p̂1998
(b) If the Test Statistic of this 2-proportion Standard Normal
hypothesis test was found to be 3.0227, what is the P-value Prob-
ability statement? Enter your answers to at least four-decimals.
P (
• X
• Z
• Chi-square
• T
• F
• Alpha
• Half of Alpha
> 3.0227) ∼=
(c) Interpret the meaning/definition of the P-value.
• If the P-value is large, fail to reject the null hypothesis
• If the P-value is large, reject the null hypothesis
• If the P-value is small, fail to reject the null hypothesis
• If the P-value is small, reject the null hypothesis
• If the null hypothesis is true, then the p-value is the probability of any thing more extreme than the test statistic
• P-value is the probability of any thing more extreme than the test statistic
(d) Give the 92.76% confidence bound for, at least, how large the
difference should be.
(e) Does there appear to be a significant increase of more than
0.0026 between the proportions use of Religious symbolism to
sell between 1998 and now? [Yes/No]
3
Problem 4. (6 points)
Classifying air threats with heuristics. The Journal of Behavioral
Decision Making (Jan2007) published a study on the use of
heuristics to classify the threat level of approaching aircraft. Of
special interest was the use of a fast and frugal heuristic compu-
tationally simple procedure for making judgments with limited
information named Take-the-Best-for-Classification (TTB-C).
The subjects were 150 men and women, some from a Canadian
Forces reserve unit, others university students. Each subject was
presented with a radar screen on which simulated approaching
aircraft were identified with asterisks. By using the computer
mouse to click on the asterisk, one could receive further informa-
tion about the aircraft.
The goal was to identify the aircraft as friend or foe as fast as
possible. Half the subjects were given cue-based instructions
for determining the type of aircraft, while the other half were
given pattern-based instructions. The researcher also classified
the heuristic strategy used by the subject as TTB-C, Guess, or
Other.
Data on the two variables Instruction type and Strategy, measured
for each of the 150 subjects, are saved in the Download .csv file.
Do the data provide sufficient evidence at α = 0.04 to indicate
that choice of heuristic strategy depends on type of instruction
provided?
a) Organize the observed data from the file into a table. Fill with
integers.
OBSERVED Cue Pattern total
Guess 28
Other 20 21 41
TTBC 26
total 75
b) If the strategy used were independent of the type of instruc-
tion given, how many of the 150 events would you expect to be
Cued-Based and have a TTB-C Based strategy (use at least
two decimals), and how much would this contribute to the Test
Statistic (use at least four decimals)
c) Calculate the critical value for this test if α= 4% is ___(use at least
five decimals) for testing whether choice of heuristic strategy depends on
type of instruction provided, and report the degrees of freedom___.
d) Give the appropriate conclusion in context of the test men-
tioned, if the Test Statistic is χ2 ∼= 0.043
Based on this data,
•We are
• There is a
• I think maybe that...There is a
• chance
• confident
• likelihood
the two variables Instruction type and Strategy, measured for each
of the 150 subjects provide sufficient evidence at α = 0.04 to in-
dicate that choice of heuristic strategy
• depends on
• is independent of
• confounds
the type of instruction provided.
4
%
Problem 5. (6 points)
Two popular social media websites monitor how long (in seconds)
a user is idle on their site. Facebook and Pinterest both sampled
25 users each to analyze how long a user will wait. Is there sta-
tistical evidence that users are on Pinterest are idle for at most 3
seconds longer than Facebook? In all cases
*Use α = 5*Use Facebook = sample 1
The specifics can be found in the file below.
Download .csv file
Test to see if the distributions appear to be normal.
• A. sample 1 appears to be normal but sample 2 does not
appear to be normal, however, we can use a T test because
the sample sizes appear to be large (¿=25 CLT)
• B. sample 2 appears to be normal but sample 1 does not
appear to be normal, however, we can use a T test because
the sample sizes appear to be large (¿=25 CLT)
• C. neither sample 1 appears to be normal nor sample 2 ap-
pears to be normal, however, we can use a T test because
the sample sizes appear to be large (¿=25 CLT)
• D. Both sample 1 and sample 2 appear to be normal
Form the correct hypothesis
• A. H0 : µFacebook − µPinterest = +3 HA : µFacebook −
µPinterest <+3
• B. H0 : µFacebook − µPinterest = −3 HA : µFacebook −
µPinterest >−3
• C. H0 : µFacebook − µPinterest = +3 HA : µFacebook −
µPinterest >+3
• D. H0 : µFacebook − µPinterest = −3 HA : µFacebook −
µPinterest 6=−3
• E. H0 : µFacebook − µPinterest = −3 HA : µFacebook −
µPinterest <−3
• F. H0 : µFacebook − µPinterest = +3 HA : µFacebook −
µPinterest 6=+3
Report the p-value of Levene’s test of variance, use at least three
decimals in your answer.
P-value =
Using technology available to you, test to see if the variances
are equal or not?
• A. They appear to be equal.
• B. There appears to be more variation in Pinterests time
than in the Facebooks.
C. There appears to be more variation in Facebooks time
than in the Pinterests.
Run the appropriate test to determine that the average time spent
on Pinterest’s website exceeds Facebook’s by more than 3 sec-
onds. Report the test statistic from the test you ran. Use at least
two decimals in your answer.
Report the p-value of the test you ran. Use at least three deci-
mals in your answer.
P-value =
Based on the above calculations, we should [reject/not reject]
the null hypothesis.
5
Problem5.csv
"","Facebook","Pinterest"
"1",21.25,10.56
"2",10.94,16.59
"3",7.94,17.45
"4",2.72,19.88
"5",16.55,15.82
"6",20.06,17.43
"7",20.71,14.72
"8",7.2,12.35
"9",7.19,12.99
"10",10.39,10.99
"11",5.99,16.12
"12",7.63,16.76
"13",16.53,16.07
"14",6.65,10.36
"15",13.33,29.78
"16",7.81,17.29
"17",7.31,26.78
"18",12.06,12.58
"19",9.08,27.06
"20",10.02,15.3
"21",9.45,19.17
"22",8.51,12.99
"23",7.47,30.93
"24",6.34,14.38
"25",15.77,16.09
(d) Based on the above calculations, at a 5% level of significance, it
appear the means are
Problem 6. (6 points)
Bank of America’s Consumer Spending Survey collected data on
annual credit card charges in seven different categories of expen-
ditures: transportation, groceries, dining out, household expenses,
home furnishings, apparel, and entertainment. Using data from
a sample of 19 credit card accounts, assume that each account
was used to identify the annual credit card charges for groceries
(population 1) and the annual credit card charges for dining out
(population 2). Is there reason to believe that the average amount
spent in the ”groceries” and ”dining out” categories is signifi-
cantly different?
Download .csv file
Note: The data appearing in column 2 of the .csv file represent the
credit card number of credit card holder i, i= 1,2, · · · ,19.
Let XGroci denote the money spent on groceries, and XDinOuti be
the money spent on dining out, also let XDi = XGroci −XDinOuti .
(a) Are the samples Normally distributed? Check the Normality
condition of the differences as well, then report the P-values.
P-valueGroc= (use three decimals)
P-valueDinOut= (use three decimals)
P-valueD= (use three decimals)
(b) Choose the correct statistical hypotheses.
• A. H0 : µ˜D = 0, HA : µ˜D 6= 0
• B. H0 : µD = 0, HA : µD < 0
• C. H0 : µ˜D = 0, HA : µ˜D > 0
• D. H0 : µ˜A = µ˜B, HA : µ˜A > µ˜B
• E. H0 : µA = µB, HA : µA 6= µB
• F. H0 : µD = 0, HA : µD > 0
• G. H0 : µ˜A = µ˜B HA : µ˜A 6= µ˜B
• H. H0 : µD = 0 HA : µD 6= 0
• I. H0 : µA = µB, HA : µA > µB
• J. H0 : µ˜D = 0 HA : µ˜D < 0
• K. H0 : µ˜A = µ˜B, HA : µ˜A < µ˜B
• L. H0 : µA = µB, HA : µA < µB
(c) Carry out the appropriate statistical test and find the Test
Statistic and P-value.
Test Statistic= (use three decimals)
P-value= (use Four decimals)
• significantly similar
• not significantly similar
6
Problem 6.csv
"","Credit.card","groceries","dining.out"
"1",4535685852488922,970.4,678.9
"2",4965974712335422,1185.95,974.78
"3",4114418154133804,742.31,785.97
"4",4911261606311460,600.15,507.9
"5",4252989551521860,1478.33,1240.91
"6",4908587231177502,1071.99,952.99
"7",4443724394135217,642.92,580.76
"8",4786379385925066,898.68,899.75
"9",4404233684658931,1140.27,1038.72
"10",4146152199450085,764.09,526.86
"11",4713114799105395,1058.02,1003.58
"12",4533779936405652,1096.18,984.47
"13",4657805568576830,398.06,430.91
"14",4664705194797458,559.9,358.28
"15",4469526045058584,969.7,817.95
"16",4635358247108298,409.11,474.54
"17",4482587495326564,198.17,157.33
"18",4209691446161886,1165.69,1021.99
"19",4592149033199393,881.99,635.55
Problem 7. (6 points)
Suppose you wanted to see if the national support for a particular
product, your company produces, was falling. So you decide to
replicate and then compare a survey conducted last year.
After watching the latest television advertisement the key research
question was: ”Would you buy this product?” In a sample of 1742
potential consumers collected last year, only 85 people committed
to buying the product. Of the more recent sample of 552, 27 said
they would buy. Conduct an analysis to determine if the percent-
age of consumers has dropped since the last year’s study. Use
these results to test this theory α= 0.05.
In the context of this question determine the correct Hypothesis
using the last year’s data as sample 1.
(a) Formulate the appropriate statistical hypotheses.
• A. H0 : plastyear− ppresent = 0 HA : plastyear− ppresent < 0
• B. H0 : plastyear− ppresent = 0 HA : plastyear− ppresent 6= 0
• C. H0 : plastyear− ppresent = 0 HA : plastyear− ppresent > 0
• D. H0 : p̂lastyear− p̂present = 0 HA : p̂lastyear− p̂present < 0
• E. H0 : p̂lastyear− p̂present = 0 HA : p̂lastyear− p̂present < 0
• F. H0 : p̂lastyear− p̂present = 0 HA : p̂lastyear− p̂present 6= 0
(b) If the Test Statistic of this hypothesis test was found to be
-0.0113, what is the P-value probability statement?
P (
• X
• Z
• Chi-square
• T
• F
• Alpha
• Half of Alpha
> -0.0113) = (use four decimals in your answer)
(c) Interpret the meaning/definition (not simply the result) of
the P-value.
• If the P-value is large, fail to reject the null hypothesis
• If the P-value is large, reject the null hypothesis
• If the P-value is small, fail to reject the null hypothesis
• If the P-value is small, reject the null hypothesis
• If the null hypothesis is true, then the p-value is the probability
of any thing more extreme than the test statistic
• P-value is the probability of any thing more extreme than the
test statistic
(d) Does there appear to be a significant increase between the per-
centages who would buy the product then versus now?[Yes/No]
7
Problem 8. (6 points)
What is the short-term effect of caffeine on your heart rate? After
measuring the heart rate of each of n= 36 healthy participants all
of whom were under the age of 30, researchers randomly divided
them into two groups. nCa f = 16 were given a 175 millilitres of
caffeinated coffee and nDeCa f = 20 were given 175 millilitres of
decaffeinated coffee. After 10 minutes, each persons heart rate
was measured again and the before and after change in heart rate
was measured. The data appears in the .csv file. Download this
file, then copy−and−paste into the statistical software.
(a) Let µCa f represent the mean change in the heart rate of those
who consumed caffeinated coffee and µDeCa f be the mean change
in the heart rate of those who consumed decaffeinated coffee. You
wish to investigate the impact caffeine has on ones heart rate,
specifically if caffeine increases one’s heart rate. Select the ap-
propriate statistical hypotheses. Define change=after-before.
• A. H0 : XCa f ≥ XDeCa f HA : XCa f < XDeCa f
• B. H0 : µCa f = µDeCa f HA : µCa f 6= µDeCa f
• C. H0 : µCa f > µDeCa f HA : µCa f ≤ µDeCa f
• D. H0 : XCa f = XDeCa f HA : XCa f > XDeCa f
• E. H0 : µCa f = µDeCa f HA : µCa f > µDeCa f
• F. H0 : XCa f > XDeCa f HA : XCa f ≤ XDeCa f
• G. H0 : XCa f = XDeCa f HA : XCa f 6= XDeCa f
• H. H0 : µCa f ≥ µDeCa f HA : µCa f < µDeCa f
(b) Use R-Studio assess the Normalilty for the caffeine data and
the decaffeinated data. Use these to correctly complete the state-
ments (i) and (ii) below. If necessarily use α= 0.05.
Note: You can either leave the data in the current stacked form,
or copy−and−paste the Caf data and the Decaf data into separate
columns.
(i) One can conclude that the change in heart rate for those who
consumed caffeinated coffee
• is Normally distributed
• is not Normally distributed
because the P−value is . (use three decimals).
(ii) One can conclude that the change in heart rate for those who
consumed decaffeinated coffee
• is Normally distributed
• is not Normally distributed
because the P−value is . (use three decimals).
(c) Side−by−side boxplots of these data appear below.
This type of visual inspection of these data is primarily used to
determine if the
• ?
• heart rate change is Normally distributed
• mean heart rate change
• median heart rate change
• standard deviation of the heart rate change
for persons who consumed caffeinated coffee is
• ?
• greater than
• less than
• similar
when compared to those persons who consumed decaffeinated
coffee.
(d) Levenes Test was applied to the data that is summarized in the boxplots. This produced a P−value of 0.905115972877851. Given
what you know about these data, compute the test statistic used to test the null hypothesis in part (a) .
Test Statistic = (Use three decimals in your answer)
(e) Compute the P−value of the result in part (a).
P−value = (Use five decimals in your answer)
(f) Correctly complete the concluding statement. Use α= 0.05.
From these data, I would [?/fail to reject/reject] the null hypothesis. From these data, I can infer that the
• ?
• median
• mean
• standard deviation
8
This type of visual inspection of these data is primarily
used to determine if the
[heart rate change is normally distributed/
mean heart rate change/
median heart rate change/
standard deviation of the heart rate change]
for persons who consumed caffeinated coffee is
[greater than/
less than/
similar]
when compared to those persons who consumed
decaffeinated coffee.
(d) Levenes Test was applied to the data that is
summarized in the boxplots. This produced a
P-value of 0.905115972877851. Given what you know
about these data, compute the test statistic used to test the
null hypothesis in part (a) .
Test statistic=___. (Use three decimals in your answer)
(e) Compute the P-value of the result in part (a).
P-value=___. (Use five decimals in your answer)
(f) Correctly complete the concluding statement. Use
α=0.05.
From these data, I would [reject/fail to reject] the Null
Hypothesis.
From these data, I can infer that the [median/mean/
standard deviation] change in the heart rate
between those who consumed caffeinated coffee and those
who consumed decaffeinated coffee is statistically
[significant/insignificant]
(g) Complete a two.sided 95% confidence interval estimate for
the difference in the mean heart rate change between caffeinated
and decaffeinated coffee drinkers. Ensure you use three decimals
in each of your answers.
___<µ(caf)-µ(decaf)<___
< µCa f −µDeCa f <
9
Problem 8.csv
"","CafData","DecafData"
"1","3",8
"2","4",3
"3","5",2
"4","6",9
"5","1",13
"6","7",8
"7","12",7
"8","2",13
"9","2",-1
"10","7",3
"11","1",10
"12","8",3
"13","4",9
"14","12",7
"15","8",9
"16","14",3
"17","",12
"18","",2
"19","",13
"20","",8
inspect this data. From this, construct the most appropriate statis-
tical hypotheses.
A.H0:µ˜D=0 HA:µ˜D<0
B.H0:µFirstChild=µSecondChild HA:µFirstChild>µSecondChild
C.H0:µFirstChild=µSecondChild HA:µFirstChil d is not equal to µ
SecondChild D.H0:µ˜D=0 HA:µ˜D>0
E.H0:µ˜FirstChild=µ˜SecondChild HA:µ˜FirstChild>µ˜SecondChild
F.H0:µFirstChild=µSecondChild HA:µFirstChild< µSecondChild
G.H0:µD=0 HA:µD<0
H.H0:µD=0 HA:µD>0
I.H0:µ˜FirstChild=µ˜SecondChild HA:µ˜FirstChild is not equal to µ
S˜econdChild
J.H0:µ˜D=0 HA:µ˜D is not equal to 0.
K.H0:µD=0 HA:µD is not equal to 0.
L.H0:µ˜FirstChild=µ˜SecondChild HA:µ˜FirstChild<µ˜SecondChild
Problem 1. (6 points)
Women are trying measure the difference in oxytocin (love hor-
mone) in the week’s after conception, for two different pregnan-
cies, below you can see the levels. Determine if the oxytocin level
has significantly changed when comparing the first child to the
second.
*Use α=5%
*Use FirstChild=sample1
Download .csv file
Find the p-values of the normality tests, use at least three dec-
imals in your answer.
First child p-value=___.
Second child p-value=___.
Paired difference p-value=___.
(Find the value of the test statistic for this test, use at least one
decimal in your answer.
Test Statistic =
Determine the P-value of your statistical test, and report it to
at least three decimal places.
P=
At α=0.05,this data indicates that you will
[reject/failtoreject]the null hypothesis.
You can say the oxy-tocin level for the first child
[is the same as/more than/less than/is differenr from the second child]
1
Problem 1.csv
"","Woman","first.child","second.child"
"1",1,4.95,1.83
"2",2,6.52,6.43
"3",3,4.22,3.63
"4",4,4.72,4.28
"5",5,4.99,4.62
"6",6,4.56,2.61
"7",7,4.56,5.37
"8",8,4.52,2.33
"9",9,5.24,5.58
"10",10,6.12,5.32
"11",11,5.18,4.86
"12",12,4.15,4.93
"13",13,6.08,1.14
"14",14,5.65,4.19
"15",15,4.27,-0.29
"16",16,5.09,0.0499999999999998
"17",17,5.33,3.17
"18",18,4.92,4.89
"19",19,4.73,3.79
"20",20,5.04,4.73
"21",21,5.58,4.5
"22",22,5.92,3.53
"23",23,4.44,3.1
Problem 2. (6 points)
Does sleep change mental performance? Are creativity and prob-
lem solving linked to adequate sleep? This question was the
subject of research conducted by German scientists at the Univer-
sity of Lubeck (Nature, Jan. 22, 2004).
322 volunteers were divided into two equal-sized groups. Each
volunteer took a math test that involved transforming strings of
eight digits into a new string that fit a set of given rules, as well
as a third, hidden rule. Prior to taking the test, one group received
eight hours of sleep, while the other group stayed awake all night.
The scientists monitored the volunteers to determine whether and
when they figured out the third rule. Of the volunteers who slept,
35 discovered the third rule; of the volunteers who stayed awake
all night, 25 discovered the third rule.
From the study results, what can you infer about the proportions
of volunteers in the two groups who discover the third rule?
(a) Find a 98.19 confidence interval for P(slept)-P(stayedawake)
,the difference in proportions in volunteers in the two groups.
Lower Bound of 98.19%CI=
Upper Bound of 98.19%CI=
(b) The confidence interval found in part (a) indicates the propor-
tion of volunteers in the group who slept and discover the third
rule is [higher/lower/the same] as the proportion of who had no
sleep and discover the third rule.
(c) Based on this data,
•We are
• There is a
• I think maybe that...There is a
• chance
• confident
• likelihood
the [true/estimated/sample] difference in propotion between
these two populations is between the lower and upper interval I
have stated.
2
Problem 3. (6 points)
Religious symbolism in TV commercials. Gonzaga University
professors conducted a study of television commercials and pub-
lished their results in the Journal of Sociology, Social Work and
Social Welfare (Vol. 2, 2008).
The key research question was: ”Do television advertisers use
religious symbolism to sell goods and services?” In a sample of
541 TV commercials collected in 1998, only 16 commercials used
religious symbolism. Of the sample of 1746 TV commercials ex-
amined in the more recent study, 100 commercials used religious
symbolism. Conduct an analysis to determine if the percentage of
TV commercials that use religious symbolism has by more than
0.26
In the context of this question determine the correct Hypothesis
using the recent data as sample 1 (use the 1998 information as
sample 2).
(a) Formulate the appropriate statistical hypotheses.
• A. H0 : ppresent = p1998 HA : ppresent < p1998
• B. H0 : ppresent = p1998 HA : ppresent > p1998
• C. H0 : ppresent = p1998 HA : ppresent 6= p1998
• D. H0 : p̂present = p̂1998 HA : p̂present 6= p̂1998
• E. H0 : p̂present = p̂1998 HA : p̂present < p̂1998
• F. H0 : p̂present = p̂1998 HA : p̂present > p̂1998
(b) If the Test Statistic of this 2-proportion Standard Normal
hypothesis test was found to be 3.0227, what is the P-value Prob-
ability statement? Enter your answers to at least four-decimals.
P (
• X
• Z
• Chi-square
• T
• F
• Alpha
• Half of Alpha
> 3.0227) ∼=
(c) Interpret the meaning/definition of the P-value.
• If the P-value is large, fail to reject the null hypothesis
• If the P-value is large, reject the null hypothesis
• If the P-value is small, fail to reject the null hypothesis
• If the P-value is small, reject the null hypothesis
• If the null hypothesis is true, then the p-value is the probability of any thing more extreme than the test statistic
• P-value is the probability of any thing more extreme than the test statistic
(d) Give the 92.76% confidence bound for, at least, how large the
difference should be.
(e) Does there appear to be a significant increase of more than
0.0026 between the proportions use of Religious symbolism to
sell between 1998 and now? [Yes/No]
3
Problem 4. (6 points)
Classifying air threats with heuristics. The Journal of Behavioral
Decision Making (Jan2007) published a study on the use of
heuristics to classify the threat level of approaching aircraft. Of
special interest was the use of a fast and frugal heuristic compu-
tationally simple procedure for making judgments with limited
information named Take-the-Best-for-Classification (TTB-C).
The subjects were 150 men and women, some from a Canadian
Forces reserve unit, others university students. Each subject was
presented with a radar screen on which simulated approaching
aircraft were identified with asterisks. By using the computer
mouse to click on the asterisk, one could receive further informa-
tion about the aircraft.
The goal was to identify the aircraft as friend or foe as fast as
possible. Half the subjects were given cue-based instructions
for determining the type of aircraft, while the other half were
given pattern-based instructions. The researcher also classified
the heuristic strategy used by the subject as TTB-C, Guess, or
Other.
Data on the two variables Instruction type and Strategy, measured
for each of the 150 subjects, are saved in the Download .csv file.
Do the data provide sufficient evidence at α = 0.04 to indicate
that choice of heuristic strategy depends on type of instruction
provided?
a) Organize the observed data from the file into a table. Fill with
integers.
OBSERVED Cue Pattern total
Guess 28
Other 20 21 41
TTBC 26
total 75
b) If the strategy used were independent of the type of instruc-
tion given, how many of the 150 events would you expect to be
Cued-Based and have a TTB-C Based strategy (use at least
two decimals), and how much would this contribute to the Test
Statistic (use at least four decimals)
c) Calculate the critical value for this test if α= 4% is ___(use at least
five decimals) for testing whether choice of heuristic strategy depends on
type of instruction provided, and report the degrees of freedom___.
d) Give the appropriate conclusion in context of the test men-
tioned, if the Test Statistic is χ2 ∼= 0.043
Based on this data,
•We are
• There is a
• I think maybe that...There is a
• chance
• confident
• likelihood
the two variables Instruction type and Strategy, measured for each
of the 150 subjects provide sufficient evidence at α = 0.04 to in-
dicate that choice of heuristic strategy
• depends on
• is independent of
• confounds
the type of instruction provided.
4
%
Problem 5. (6 points)
Two popular social media websites monitor how long (in seconds)
a user is idle on their site. Facebook and Pinterest both sampled
25 users each to analyze how long a user will wait. Is there sta-
tistical evidence that users are on Pinterest are idle for at most 3
seconds longer than Facebook? In all cases
*Use α = 5*Use Facebook = sample 1
The specifics can be found in the file below.
Download .csv file
Test to see if the distributions appear to be normal.
• A. sample 1 appears to be normal but sample 2 does not
appear to be normal, however, we can use a T test because
the sample sizes appear to be large (¿=25 CLT)
• B. sample 2 appears to be normal but sample 1 does not
appear to be normal, however, we can use a T test because
the sample sizes appear to be large (¿=25 CLT)
• C. neither sample 1 appears to be normal nor sample 2 ap-
pears to be normal, however, we can use a T test because
the sample sizes appear to be large (¿=25 CLT)
• D. Both sample 1 and sample 2 appear to be normal
Form the correct hypothesis
• A. H0 : µFacebook − µPinterest = +3 HA : µFacebook −
µPinterest <+3
• B. H0 : µFacebook − µPinterest = −3 HA : µFacebook −
µPinterest >−3
• C. H0 : µFacebook − µPinterest = +3 HA : µFacebook −
µPinterest >+3
• D. H0 : µFacebook − µPinterest = −3 HA : µFacebook −
µPinterest 6=−3
• E. H0 : µFacebook − µPinterest = −3 HA : µFacebook −
µPinterest <−3
• F. H0 : µFacebook − µPinterest = +3 HA : µFacebook −
µPinterest 6=+3
Report the p-value of Levene’s test of variance, use at least three
decimals in your answer.
P-value =
Using technology available to you, test to see if the variances
are equal or not?
• A. They appear to be equal.
• B. There appears to be more variation in Pinterests time
than in the Facebooks.
C. There appears to be more variation in Facebooks time
than in the Pinterests.
Run the appropriate test to determine that the average time spent
on Pinterest’s website exceeds Facebook’s by more than 3 sec-
onds. Report the test statistic from the test you ran. Use at least
two decimals in your answer.
Report the p-value of the test you ran. Use at least three deci-
mals in your answer.
P-value =
Based on the above calculations, we should [reject/not reject]
the null hypothesis.
5
Problem5.csv
"","Facebook","Pinterest"
"1",21.25,10.56
"2",10.94,16.59
"3",7.94,17.45
"4",2.72,19.88
"5",16.55,15.82
"6",20.06,17.43
"7",20.71,14.72
"8",7.2,12.35
"9",7.19,12.99
"10",10.39,10.99
"11",5.99,16.12
"12",7.63,16.76
"13",16.53,16.07
"14",6.65,10.36
"15",13.33,29.78
"16",7.81,17.29
"17",7.31,26.78
"18",12.06,12.58
"19",9.08,27.06
"20",10.02,15.3
"21",9.45,19.17
"22",8.51,12.99
"23",7.47,30.93
"24",6.34,14.38
"25",15.77,16.09
(d) Based on the above calculations, at a 5% level of significance, it
appear the means are
Problem 6. (6 points)
Bank of America’s Consumer Spending Survey collected data on
annual credit card charges in seven different categories of expen-
ditures: transportation, groceries, dining out, household expenses,
home furnishings, apparel, and entertainment. Using data from
a sample of 19 credit card accounts, assume that each account
was used to identify the annual credit card charges for groceries
(population 1) and the annual credit card charges for dining out
(population 2). Is there reason to believe that the average amount
spent in the ”groceries” and ”dining out” categories is signifi-
cantly different?
Download .csv file
Note: The data appearing in column 2 of the .csv file represent the
credit card number of credit card holder i, i= 1,2, · · · ,19.
Let XGroci denote the money spent on groceries, and XDinOuti be
the money spent on dining out, also let XDi = XGroci −XDinOuti .
(a) Are the samples Normally distributed? Check the Normality
condition of the differences as well, then report the P-values.
P-valueGroc= (use three decimals)
P-valueDinOut= (use three decimals)
P-valueD= (use three decimals)
(b) Choose the correct statistical hypotheses.
• A. H0 : µ˜D = 0, HA : µ˜D 6= 0
• B. H0 : µD = 0, HA : µD < 0
• C. H0 : µ˜D = 0, HA : µ˜D > 0
• D. H0 : µ˜A = µ˜B, HA : µ˜A > µ˜B
• E. H0 : µA = µB, HA : µA 6= µB
• F. H0 : µD = 0, HA : µD > 0
• G. H0 : µ˜A = µ˜B HA : µ˜A 6= µ˜B
• H. H0 : µD = 0 HA : µD 6= 0
• I. H0 : µA = µB, HA : µA > µB
• J. H0 : µ˜D = 0 HA : µ˜D < 0
• K. H0 : µ˜A = µ˜B, HA : µ˜A < µ˜B
• L. H0 : µA = µB, HA : µA < µB
(c) Carry out the appropriate statistical test and find the Test
Statistic and P-value.
Test Statistic= (use three decimals)
P-value= (use Four decimals)
• significantly similar
• not significantly similar
6
Problem 6.csv
"","Credit.card","groceries","dining.out"
"1",4535685852488922,970.4,678.9
"2",4965974712335422,1185.95,974.78
"3",4114418154133804,742.31,785.97
"4",4911261606311460,600.15,507.9
"5",4252989551521860,1478.33,1240.91
"6",4908587231177502,1071.99,952.99
"7",4443724394135217,642.92,580.76
"8",4786379385925066,898.68,899.75
"9",4404233684658931,1140.27,1038.72
"10",4146152199450085,764.09,526.86
"11",4713114799105395,1058.02,1003.58
"12",4533779936405652,1096.18,984.47
"13",4657805568576830,398.06,430.91
"14",4664705194797458,559.9,358.28
"15",4469526045058584,969.7,817.95
"16",4635358247108298,409.11,474.54
"17",4482587495326564,198.17,157.33
"18",4209691446161886,1165.69,1021.99
"19",4592149033199393,881.99,635.55
Problem 7. (6 points)
Suppose you wanted to see if the national support for a particular
product, your company produces, was falling. So you decide to
replicate and then compare a survey conducted last year.
After watching the latest television advertisement the key research
question was: ”Would you buy this product?” In a sample of 1742
potential consumers collected last year, only 85 people committed
to buying the product. Of the more recent sample of 552, 27 said
they would buy. Conduct an analysis to determine if the percent-
age of consumers has dropped since the last year’s study. Use
these results to test this theory α= 0.05.
In the context of this question determine the correct Hypothesis
using the last year’s data as sample 1.
(a) Formulate the appropriate statistical hypotheses.
• A. H0 : plastyear− ppresent = 0 HA : plastyear− ppresent < 0
• B. H0 : plastyear− ppresent = 0 HA : plastyear− ppresent 6= 0
• C. H0 : plastyear− ppresent = 0 HA : plastyear− ppresent > 0
• D. H0 : p̂lastyear− p̂present = 0 HA : p̂lastyear− p̂present < 0
• E. H0 : p̂lastyear− p̂present = 0 HA : p̂lastyear− p̂present < 0
• F. H0 : p̂lastyear− p̂present = 0 HA : p̂lastyear− p̂present 6= 0
(b) If the Test Statistic of this hypothesis test was found to be
-0.0113, what is the P-value probability statement?
P (
• X
• Z
• Chi-square
• T
• F
• Alpha
• Half of Alpha
> -0.0113) = (use four decimals in your answer)
(c) Interpret the meaning/definition (not simply the result) of
the P-value.
• If the P-value is large, fail to reject the null hypothesis
• If the P-value is large, reject the null hypothesis
• If the P-value is small, fail to reject the null hypothesis
• If the P-value is small, reject the null hypothesis
• If the null hypothesis is true, then the p-value is the probability
of any thing more extreme than the test statistic
• P-value is the probability of any thing more extreme than the
test statistic
(d) Does there appear to be a significant increase between the per-
centages who would buy the product then versus now?[Yes/No]
7
Problem 8. (6 points)
What is the short-term effect of caffeine on your heart rate? After
measuring the heart rate of each of n= 36 healthy participants all
of whom were under the age of 30, researchers randomly divided
them into two groups. nCa f = 16 were given a 175 millilitres of
caffeinated coffee and nDeCa f = 20 were given 175 millilitres of
decaffeinated coffee. After 10 minutes, each persons heart rate
was measured again and the before and after change in heart rate
was measured. The data appears in the .csv file. Download this
file, then copy−and−paste into the statistical software.
(a) Let µCa f represent the mean change in the heart rate of those
who consumed caffeinated coffee and µDeCa f be the mean change
in the heart rate of those who consumed decaffeinated coffee. You
wish to investigate the impact caffeine has on ones heart rate,
specifically if caffeine increases one’s heart rate. Select the ap-
propriate statistical hypotheses. Define change=after-before.
• A. H0 : XCa f ≥ XDeCa f HA : XCa f < XDeCa f
• B. H0 : µCa f = µDeCa f HA : µCa f 6= µDeCa f
• C. H0 : µCa f > µDeCa f HA : µCa f ≤ µDeCa f
• D. H0 : XCa f = XDeCa f HA : XCa f > XDeCa f
• E. H0 : µCa f = µDeCa f HA : µCa f > µDeCa f
• F. H0 : XCa f > XDeCa f HA : XCa f ≤ XDeCa f
• G. H0 : XCa f = XDeCa f HA : XCa f 6= XDeCa f
• H. H0 : µCa f ≥ µDeCa f HA : µCa f < µDeCa f
(b) Use R-Studio assess the Normalilty for the caffeine data and
the decaffeinated data. Use these to correctly complete the state-
ments (i) and (ii) below. If necessarily use α= 0.05.
Note: You can either leave the data in the current stacked form,
or copy−and−paste the Caf data and the Decaf data into separate
columns.
(i) One can conclude that the change in heart rate for those who
consumed caffeinated coffee
• is Normally distributed
• is not Normally distributed
because the P−value is . (use three decimals).
(ii) One can conclude that the change in heart rate for those who
consumed decaffeinated coffee
• is Normally distributed
• is not Normally distributed
because the P−value is . (use three decimals).
(c) Side−by−side boxplots of these data appear below.
This type of visual inspection of these data is primarily used to
determine if the
• ?
• heart rate change is Normally distributed
• mean heart rate change
• median heart rate change
• standard deviation of the heart rate change
for persons who consumed caffeinated coffee is
• ?
• greater than
• less than
• similar
when compared to those persons who consumed decaffeinated
coffee.
(d) Levenes Test was applied to the data that is summarized in the boxplots. This produced a P−value of 0.905115972877851. Given
what you know about these data, compute the test statistic used to test the null hypothesis in part (a) .
Test Statistic = (Use three decimals in your answer)
(e) Compute the P−value of the result in part (a).
P−value = (Use five decimals in your answer)
(f) Correctly complete the concluding statement. Use α= 0.05.
From these data, I would [?/fail to reject/reject] the null hypothesis. From these data, I can infer that the
• ?
• median
• mean
• standard deviation
8
This type of visual inspection of these data is primarily
used to determine if the
[heart rate change is normally distributed/
mean heart rate change/
median heart rate change/
standard deviation of the heart rate change]
for persons who consumed caffeinated coffee is
[greater than/
less than/
similar]
when compared to those persons who consumed
decaffeinated coffee.
(d) Levenes Test was applied to the data that is
summarized in the boxplots. This produced a
P-value of 0.905115972877851. Given what you know
about these data, compute the test statistic used to test the
null hypothesis in part (a) .
Test statistic=___. (Use three decimals in your answer)
(e) Compute the P-value of the result in part (a).
P-value=___. (Use five decimals in your answer)
(f) Correctly complete the concluding statement. Use
α=0.05.
From these data, I would [reject/fail to reject] the Null
Hypothesis.
From these data, I can infer that the [median/mean/
standard deviation] change in the heart rate
between those who consumed caffeinated coffee and those
who consumed decaffeinated coffee is statistically
[significant/insignificant]
(g) Complete a two.sided 95% confidence interval estimate for
the difference in the mean heart rate change between caffeinated
and decaffeinated coffee drinkers. Ensure you use three decimals
in each of your answers.
___<µ(caf)-µ(decaf)<___
< µCa f −µDeCa f <
9
Problem 8.csv
"","CafData","DecafData"
"1","3",8
"2","4",3
"3","5",2
"4","6",9
"5","1",13
"6","7",8
"7","12",7
"8","2",13
"9","2",-1
"10","7",3
"11","1",10
"12","8",3
"13","4",9
"14","12",7
"15","8",9
"16","14",3
"17","",12
"18","",2
"19","",13
"20","",8
