Triola Assignment D
Section 8-2 – Basic Skills and Concepts
1) A newspaper article states that “based on a recent survey, it has bee proved that 50% of all
truck drivers smoke.” What is wrong with that statement?
The problem with the statement is the word “proven.” The use of sample statistics and
hypothesis testing helps to provide an indication of the likelihood of an event occurring.
However, these methods do not absolutely prove something. Check: OK
3) What is the difference between a test statistic and a critical value?
The critical value is the boundary between values that would be used to reject the null
hypothesis and those that would fail to reject the null hypothesis. The test statistic is
the value that is the judged against the regions established by the critical value. By
comparing the test statistic to the critical value, one is able to either reject of fail to
reject the null hypothesis. Check: OK In addition, the test statistic is based on the sample
data. The critical value is based on the significance level and the distribution being used.
5) Make a decision about the given claim. (Use the rare event rule to make a subjective estimate
to determine whether the event is likely to occur.) Claim: A coin favors heads when tossed, and
there are 11 heads in 20 tosses.
In this case, the claim that the coin favors heads is not reasonable. One would expect
that, by chance, about 10 heads and 10 tails would occur. The value of 11 is not
significantly different enough to justify the claim of the coin favoring heads. Check: OK
13) Examine the given statement, then express the null hypothesis and the alternative hypothesis
in symbolic form. Be sure to use the correct symbol for the indicated parameter. IQ scores of
college professors have a standard deviation less than 15, which is the standard deviation for the
general population.
Null  H0:  = 15 Alternative  H1:  < 15 Check: OK
17) Find the critical z values. In each case, assume that the normal distribution applies.
Situation: Two-tailed test; α = 0.05
The critical value for this cases would be based on a value of α = 0.025 since the critical
region is split between two tails. Use 1 – 0.025 to find the probability = 0.975. The
critical value is 1.96. Since it is a two-tailed situation, the critical z values are: ±1.96
Check: OK
19) Find the critical z values. In each case, assume that the normal distribution applies.
Situation: Right-tailed test; α = 0.01
The critical value for this cases would be based on a value of α = 0.01 since the critical
region is only in the right tail. Use 1 – 0.01 to find the probability = 0.99. The critical
value is 2.33 Check: OK
27) Find the value of test statistic z. The claim is more than 75% of workers are satisfied with
their job, and sample statistics include 580 employed adults, with 516 of them saying that they
are satisfied with their job.
We need to start by finding the proportion value for the sample:
^
p  516  580  0.8896  0.89
The other values for the formula will be:
p = 0.75
q = 1 – p = 1 – 0.75 = 0.25
n = 580
The final step is to calculate z:
^
z
p p
pq
n
Check: OK

0.89  0.75
0.750.25
580

0.14
0.1875
580

0.14
0.000323275

0.14
 7.786  7.79
0.017979
33) Use the given information to find the P-value. Information: With H1: p > 0.333, the test
statistic is z = 1.50.
The alternative hypothesis indicates that it is a right-tailed test. For a z-score of 1.50,
the probability is 0.9332. Therefore, the P(p > 0.333) = 0.0668. Check: OK
Section 8-3 – Basic Skills and Concepts
1) Assuming that the listed requirements of this section are satisfied, what distribution is used to
test a claim about a population proportion? Why?
The normal distribution as an approximation of a binomial distribution is being used.
The requirements for a binomial distribution include: a fixed number of independent
trials; two categories for outcomes; and the probability of success is the same in all
trials. The normal distribution can be used to approximate because the binomial
distribution as long as there are at least 5 trials that fall into each category. Check: OK
3) American Online conducts a survey in which Internet users are asked to respond to a question.
Among the 96773 responses, there are 76885 responses of “yes.” Is it valid to use these sample
results for testing the claim that the majority of the general population answers “yes”?
The requirements that must be met are: a simple random sample, the conditions for the
binomial distribution are met, and np  5 and nq  5. It is NOT valid to use these
sample results for a hypothesis test because the sample is not a simple random sample.
The sampling was based on voluntary-response sampling. Although the results may in
fact reflect the general population, hypothesis testing based on these data does not
reflect valid statistical procedures. Check: OK
13) Test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, Pvalue or critical value(s), conclusion about the null hypothesis, and final conclusion that
addresses the original claim. Use the P-value method. Technology is dramatically changing the
way we communicate. In 1997, a survey of 880 U.S. households showed that 149 of them use e-
mail. Use those sample results to test the claim that more than 15% of U.S. households use email. Use a 0.05 significance level. Is the conclusion valid today? Why or why not?
 Null hypothesis  H0: p = 0.15
 Alternative hypothesis  H1: p > 0.15
 Test statistic 
^
p  149  880  0.1693  0.17
p  0.15
q  1  p  1  0.15  0.85
n  880
^
z
p p

0.17  0.15

0.02

0.02

0.02
 1.66156  1.66
0.0120368
0.150.85
pq
0.1275
0.000144886
n
880
880
 P-value  P-value = 1 – 0.9515 = 0.0485
 Critical Value  For an α = 0.05 in a right-tailed test, the critical value is: 1.645
 Conclusion About Null  Since 0.0485 < 0.5 (or since 1.66 is in the critical region), the
conclusion is to reject the null hypothesis.
 Final Conclusion  Based on the results of the hypothesis test, we can reject the null
hypothesis and conclude that more than 15% of U.S. households
use e-mail.
 Is the conclusion valid today? Although the conclusion is probably still valid today
since the proliferation of technology is even greater. However, a new sample and a
new hypothesis test may be yield an even greater percentage of households using email.
Check: OK. Although my calculations were done correctly, rounding issues led my z-value
to be a little high (as compared to the answer of 1.60). This, in turn, led my P-value to be off
(the correct P-value was 0.0548). With this value, we would fail to reject the null hypothesis
and this would change the conclusion to state the less than 15% of the population used email.
19) Test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, Pvalue or critical value(s), conclusion about the null hypothesis, and final conclusion that
addresses the original claim. Use the P-value method. A recent Gallup poll of 976 randomly
selected adults showed that 312 of them never drink. Use those survey results to test the claim
that less than 1/3 of all adults never drink. Use a 0.05 significance level. Also, examine the
following wording of the actual questions and determine whether it is likely to elicit honest
responses: “How often, if ever, do you drink alcoholic beverages such as liquor, wine or beer—
every day, a few times a week, about once a week, less than once a week, only on special
occasions, or never?”
 Null hypothesis  H0: p = 0.333
 Alternative hypothesis  H1: p < 0.333
 Test statistic 
^
p  312  976  0.31967  0.320
p  0.333
q  1  p  1  0.333  0.667
n  976
^
z
p p

0.320  0.333

 0.013

 0.013

 0.013
 0.86175  0.86
0.0150855
0.3330.667 
pq
0.222111
0.00022757
n
976
976
 P-value  P-value = 0.1685
 Critical Value  For an α = 0.05 in a left-tailed test, the critical value is: –1.645
 Conclusion About Null  Since 0.1685 > 0.5 (or since –0.86 is not in the critical
region), the conclusion is: fail to reject the null hypothesis.
 Final Conclusion  Based on the results of the hypothesis test, we cannot reject the
null hypothesis. Therefore, it is likely that more than 1/3 of the
population never drinks. However, the null hypothesis does not
state anything about what the actual proportion might be.
 Assess the survey question  The initial part of the question is somewhat leading in
the it asks “how often, if ever…” This might compel participants to answer a certain
way. In addition, there is a big gap between the choices of “less than once a week”
and “only on special occasions.” There might not be sufficient choices for a person to
answer honestly. Check: OK. Although my calculations were done correctly, rounding
issues led my z-value to be a little high (as compared to the answer of –0.91). This, in turn,
led my P-value to be off (the correct P-value was 0.1814). However, we would reach the
same conclusion about the null hypothesis. The main issue associated with the question, as
suggested by the authors, is the sensitivity of the topic. This could influence how people
choose to answer.
Section 8-4 – Basic Skills and Concepts
1) Must you have a sample size of n > 30 in order to use the methods of hypothesis testing
presented in this section? If a simple random sample has fewer than 31 values, what requirement
must be satisfied to justify using the methods of this section?
A sample size of n > 30 provides a relatively larger enough sample size to ensure that
the results will be accurate. If the population is normally distributed, though, you do
not have to a sample that is larger than 30. If a simple random sample has fewer than
31 values, then one must ensure that there are no outliers and that the sample is very
close to being normally distributed. Check: OK
3) You want to test the claim that  < 100 by constructing a confidence interval. If the
hypothesis test is to be conducted with a 0.01 significance level, what confidence level should be
used for the confidence interval?
Because this is a one-tailed test, the confidence level should be 98%. Check: OK
13) Test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, Pvalue or critical value(s), conclusion about the null hypothesis, and final conclusion that
addresses the original claim. Use the P-value method. Randomly selected statistics students of
the author participated in an experiment to test their ability to determine when 1 min (or 60
seconds) has passed. Forty students yielded a sample mean of 58.3 seconds. Assuming that  =
9.5 seconds, use a 0.05 significance level to test the claim that the population mean is equal to 60
seconds. Based on the result, does there appear to be an overall perception of 1 minute that is
reasonably accurate?
 Null hypothesis  H0:  = 60
 Alternative hypothesis  H1:   60
 Test statistic 
x  58.3
 x  60
  9.5
n  40
x   x 58.3  60  1.7
 1.7
 1.7
z




 1.13176  1.13

9.5
9.5
9.5
1.50208188
n
40
40 6.324555
 P-value  P-value = 0.1292 Since it is a two-tailed test, the value is doubled: 0.2584
 Critical Value  For an α = 0.05 in a two-tailed test, the critical value is: 1.96
 Conclusion About Null  Since 0.2584 > 0.5 (or since –1.13 is NOT in the critical
region), the conclusion is to fail to reject the null
hypothesis.
 Final Conclusion  Based on the results of the hypothesis test, we fail to reject the
null hypothesis and conclude that people’s perception of the
length of one minute is near a mean of 60 seconds.
 Is the perception accurate? It does not appear that people do have an accurate
perception of how long 1 minute actually is. However, the standard deviation
indicates that individuals could be far off even though the population mean is near 60
seconds. Check: OK. I had incorrectly written down the formula which messed with my
results. Once I corrected the formula, my calculations were fine.
15) Test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, Pvalue or critical value(s), conclusion about the null hypothesis, and final conclusion that
addresses the original claim. Use the P-value method. When 40 people used the Atkins diet for
one year, their mean weight change was –2.1 lbs. Assume that the standard deviation of all such
weight changes is  = 4.8 lbs and use a 0.05 significance level to test the claim that the man
weight changes is less than 0. Based on these results, does the diet appear to be effective? Does
the mean weight changer appear to be substantial enough to justify the special diet?
 Null hypothesis  H0:  = 0
 Alternative hypothesis  H1:  < 0
 Test statistic 
x  2.1
x  0
  4.8
n  40
z
x  x


 2.1  0  2.1


4.8
4.8
 2.1
 2.1

 2.766992  2.77
4.8
0.7589466
6.324555
n
40
40
 P-value  P-value = 0.0028
 Critical Value  For an α = 0.05 in a left-tailed test, the critical value is: –1.645
 Conclusion About Null  Since 0.0028 < 0.5 (or since –2.77 is in the critical region),
the conclusion is to reject the null hypothesis.
 Final Conclusion  Based on the results of the hypothesis test, we reject the null
hypothesis and conclude that people’s mean weight loss on the
Atkin’s diet is less than zero.
 Is the diet effective? Based on the hypothesis test, the diet does appear to be effective.
Is the diet justified? Having only a have mean weight loss of 2.1 pounds in a year does
not appear to be substantial enough to justify the special diet. Check: OK.
Section 8-5 – Basic Skills and Concepts
1) When using Table A-3 to find critical values, we must use the appropriate number of degrees
of freedom. If a sample consists of five values, what is the appropriate number of degrees of
freedom? If you don’t know any of the five sample values, but you know that their mean is
exactly 20.0, how many values can you make up before the reaming values are determined by the
restriction that the mean is 20.0?
If the sample has five values, the df is 4. This means that you can freely assign the first
four values, but the fifth will be set based on the mean that you are trying to get. Check:
OK
5) Determine whether the hypothesis test involves a sampling distribution of means that is a
normal distribution, Student t distribution, or nether. Claim:  = 2.55. Sample data: n = 7,
x  2.41 , s = 0.66. The sample data appear to come from a normally distributed population with
unknown  and .
Since the population is normally distributed but you do not know the population
standard deviation, you must use the Student t distribution. Check: OK
7) Determine whether the hypothesis test involves a sampling distribution of means that is a
normal distribution, Student t distribution, or nether. Claim:  = 0.0105. Sample data: n = 17,
x  0.0134 , s = 0.0022. The sample data appear to come from a population with a distribution
that is very far from normal, and  is unknown.
Since the population is not normally distributed and you do not know the population
standard deviation, it is neither a normal nor a Student t distribution. Check: OK
13) Find the null hypothesis and alternative hypotheses, test statistic, P-value (or range of Pvalues), critical value(s), and state the final conclusion. Claim: The mean IQ score of statistic
professors is greater than 120. Sample data: n = 12, x  132 , s = 12. The significance level is α
= 0.05.
 Null hypothesis  H0:  = 120
 Alternative hypothesis  H1:  >120
 Test statistic 
x  132
 x  120
s  12
n  12
df  11
x   x 132  120
12



s
12
12
12
12

 3.4641016  3.464
12
3.4641016
n
12
12 3.4641016
 Critical Value  For an α = 0.05 in a one-tailed test, the critical value is: 1.796
 P-value  The range of P-values < 0.005
 Final Conclusion  Based on the results of the hypothesis test, we reject the null
hypothesis and conclude that professors have an IQ score greater than 120. Check:
OK.
t
15) Test the given claim by interpreting the accompanying display of hypothesis test results. Use
a 0.05 significance level to the test the claim that statistics students have a mean IQ greater than
110. The sample data are summarized by the statistics n = 25, x  118.0 , and s = 10.7. See the
accompanying Minitab display.
The critical value for an α = 0.05 and df = 24 is 1.711. Since the test statistic is t = 3.74,
we can reject the null hypothesis and conclude that the mean IQ scores are greater than
110. Check: OK
Section 8-6 – Basic Skills and Concepts
1) What does it mean when we say that the chi-square test of this section is not robust against
departures from normality? How does this affect the conditions that must be satisfied for the chisquare test of this section?
Since the chi-square test is not robust, the test will NOT produce accurate results for
distributions that vary too far from normal. Therefore, the restrictions of using a
simple random sample from a normally distributed population must be adhered to.
Check: OK
5) Calculate the test statistics, then use Table A-4 to find critical values(s) of χ2, then use Table
A-4 to find limits containing the P-value, then determine whether there is sufficient evidence to
support the given alternative hypothesis. H1 =   2.00, α = 0.05, n = 10, s = 3.00
 Test statistic 
n  10
s  3.00
  2.00
df  9
n  1s 2
10  1  3.00 2
9  9 81

 20.25
4
4

2.00
 Critical Values  For an α = 0.05 in a two-tailed test, the critical values are: 2.700
and 19.023
 P-value  The range of P-values is between 0.01 and 0.025. Since it is a two-tailed
test, you must double these values. Therefore, the range is: 0.02 < P-value < 0.05
 Final Conclusion  Due to the fact that the test statistic is in the critical region, we
can reject the null hypothesis and conclude that we can support the alternative
hypothesis. Check: OK. Everything was okay except that I needed to fix my values for the
range of the P-values.
2 
2

2

9) Test the given claim. Assume that a simple random sample is selected from a normally
distributed population. Use the traditional method. A study was conducted of babies born to
mothers who use cocaine during pregnancy, and the following sample data were obtained for
weights at birth: n = 190, x  2700 , s = 645 grams. Use a 0.05 significance level to test the
claim that the standard deviation of birth weights of infants born to mothers who use cocaine is
different from the standard deviation of 696 grams for birth weights of infants born to mothers
who do not use cocaine during pregnancy. (With df = 189, use these critical values:
 L2  152.8222 and  R2  228.9638 ) Based on the result, does cocaine use by mothers appear to
affect the variation of the weights of their babies?
 Null hypothesis  H0:  = 696
 Alternative hypothesis  H1:   696
 Test statistic 
n  190
s  645
  696
df  189
n  1s 2
190  1  645 2
189  416025 78628725

 162.316531  162.3165
484416
484416

696
 Critical Values  For an α = 0.05 in a two-tailed test, the critical values are:
 L2  152.8222 and  R2  228.9638
 Final Conclusion  Due to the fact that the test statistic is NOT in the critical region,
we fail to reject the null hypothesis and conclude that cocaine use does not affect the
variation of birth weights. Check: OK.
2 
2

2

Unit 8 – Review Exercises
1) Based on the given conditions, identify the alternative hypothesis. Also, either identify the
sampling distribution (normal, t, chi-square) of the test statistic, or state that the methods of this
chapter should not be used.
a) Claim: The mean annual income of full-time college students is below $10000. Sample
data: For 750 randomly selected college students, the mean is $3662 and the standard
deviation is $2996. The sample data suggest that the population has a distribution that is not
normal.
H1:  < 10000
Sampling Distribution: Student t (even though not normal, sample is > 30) Check:
OK
b) Claim: The majority of college students have at least one credit card. Sample data: Of 500
randomly selected college students, 82% have at least one credit card.
H1: p > .50
Sampling Distribution: Normal Check: OK
c) Claim: The mean IQ of adults is equal to 100. Sample data: n = 150 and x  98.8 . It is
reasonable to assume that  = 15.
H1:   100
Sampling Distribution: Normal Check: OK
d) Claim: With manual assembly of telephone parts, the assembly times vary more than the
times for automated assembly, which are know to have a mean of 27.6 seconds and a
standard deviation of 1.8 seconds. Sample data: For 24 randomly selected times, the mean is
25.4 seconds, the standard deviation is 2.5 seconds, and the sample data suggest that the
population has a distribution that is far from normal.
H1:  > 1.8
Sampling Distribution: Due to the fact that the distribution is not normal, we cannot
use the techniques from this chapter. Check: OK
e) Claim: Statistics professors have IQ scores that vary less than IQ scores from the adult
population, which has standard deviation of 15. Sample data: For 24 randomly selected
statistics professors, the standard deviation is 10.2, and the sample data suggest that the
population has a distribution that is very close to a normal distribution.
H1:  < 15
Sampling Distribution: Chi-Square Check: OK
3) Glamour magazine sponsored a survey of 2500 prospective brides and found that 60% of them
spent less that $750 of their wedding gown. Use a 0.01 significance level to test the claim that
less than 62% of brides spend less than $750 of their wedding gown. How are the results
affected if it is learned the responses were obtained from magazine readers who decided to
responded to the survey through an Internet web site?
 Null hypothesis  H0: p = 0.62
 Alternative hypothesis  H1: p < 0.62
 Test statistic 
^
p  0.60
p  0.62
q  1  p  1  0.62  0.38
n  2500
^
z
p p
pq
n

0.60  0.62
0.620.38

 0.02
0.2356
2500
2500

 0.02
0.00009424

 0.02
 2.0602  2.06
0.0097077288
 P-value  P-value = 0.0197
 Critical Value  For an α = 0.01 in a left-tailed test, the critical value is: –2.33
 Conclusion About Null  Since 0.0197 > 0.01 (or since –2.06 is NOT in the critical
region), the conclusion is to fail to reject the null
hypothesis.
 Final Conclusion  Based on the results of the hypothesis test, we fail to reject the
null hypothesis and cannot conclude less than 62% of women
spend less than $750 on their wedding gowns.
 What would happen if data was obtained from Internet poll? If the data were
gathered via a self-response poll, the integrity of the results would be in question. In
addition, this type of sample is not simple random sampling, therefore the techniques
used in this chapter would not legitimately apply. Check: OK.
7) For a simple random sample of adults, IQ scores are normally distributed with a mean of 100
and a standard deviation of 15. A simple random sample of 13 statistics professors yields a
standard deviation of s = 7.2. A psychologist is quite sure that statistics professors have IQ
scores that have a mean greater than 100. He doesn’t understand the concept of standard
deviation very well and does not realize that the standard deviation should be lower than 15.
Instead, he claims that statistics professors have IQ scores with a standard deviation equal to 15,
the same standard deviation for the general population. Assume that IQ scores of statistics
professors are normally distributed and use a 0.05 significance level to test the claim that  = 15.
Based on the result, what do you conclude about the standard deviation of IQ scores for statistics
professors?
 Null hypothesis  H0:  = 15
 Alternative hypothesis  H1:   15
 Test statistic 
n  13
s  7.2
  15
df  18

n  1s 2


13  1  7.2 2

12  51.84 622.08

 2.7648  2.7648
225
225

15
 Critical Values  For an α = 0.05 in a two-tailed test, the critical values are:
 L2  4.404 and  R2  23.337
 Final Conclusion  Due to the fact that the test statistic is in the critical region
(2.7648 < 4.404), we reject the null hypothesis and conclude that the IQ scores of
statistics professors varies less than the general population. Check: OK.

2
2
2

Unit 8 – Cumulative Review Exercises
1) Listed below are the heights (in inches) of the Presidents who served in the 20th century.
Assume that these values are sample data from some larger population:
67 70 72 71 72 70 71 74 69 70.5 72 75 71.5 69.5 73 74 74.5
a) Mean:
b) Median:
c) SD:
67  70  72  ...  73  74  74.5 1216

 71.529411  71.53 Check: OK
17
17
67 69 69.5 70 70 70.5 71 71 71.5 72 72 72 73 74 74 74.5 75
The median is: 71.5 Check: OK
 x   67  69  69.5  ...  74  74.5  75  87053
 x   (67  69  69.5  ...  74  74.5  75)  1216  1478656
n x    x 
17(87053)  1478656
1479901  1478656
s



2
2
2
2
2
2
2
2
2
2
2
2
nn  1
1717  1
d) Variance:
4.57720588235  2.13944  2.14
Check: OK
s 2  2.13944 2  4.57720588235  4.58 Check: OK
e) Range:
75  67  8 Check: OK
1716
1245

272
f) Assuming that the values are sample data, construct a 95% confidence interval estimate of
the population mean.
  1  0.95  0.05
t  2.120
s  2.14
n  17
s
2.14
2.14
E  t / 2 
 2.120 
 2.120 
 2.120  0.51889  1.1000  1.10
4.1231056
n
17
x  71.53
x  E    x  E  71.53  1.10    71.53  1.10  70.43    72.63
Check: OK
g) The mean height of men is 69.0 inches. Use a 0.05 significance level to test the claim that
this sample comes from a population with a mean greater than 69.0 inches. Do Presidents
appear to be taller than the typical man?
 Null hypothesis  H0:  = 69.0
 Alternative hypothesis  H1:  > 69.0
 Test statistic 
x  71.53
 x  69.0
s  2.14
n  17
df  16
t
x   x 71.53  69.0 2.53



s
2.14
2.14
n
17
17
2.53
2.53

 4.8745127  4.875
2.14
0.519026
4.1231056
 Critical Value  For an α = 0.05 in a one-tailed test with df = 16, the critical value is:
1.746
 Final Conclusion  Based on the results of the hypothesis test, we reject the null
hypothesis and conclude that Presidents come from a population of taller men. Is
does appear that Presidents tend to be taller than the typical man. Check: OK.
3) The math SAT scores for women are normally distributed with a mean of 496 and standard
deviation of 108.
a) If a woman who takes the math portion of the SAT is randomly selected, find the
probability that her score is above 500.
x   500  496
4


 0.037  0.04
 Convert 500 to a z-score: z 

108
108
 Find the probability of a score greater than 500: 1 – 0.5160 = 0.484. Check: OK
b) If five math SAT scores are randomly selected from the population of women who take
the test, find the probability that all five of the scores are above 500.
x   500  496
4


 0.037  0.04
 Convert 500 to a z-score: z 

108
108
 Find the probability of all five: 0.484·0.484·0.484·0.484·0.484 = 0.0265599 = 0.027.
Check: OK
c) If five women who take the math portion of the SAT are randomly selected, find the
probability that their mean is above 500.
x   x 500  496
4
4



 0.0828  0.08
 Convert 500 to a z-score: z 

108
108
48.299
2.236
n
5
 Find the probability of a score greater than 500: 1 – 0.5319 = 0.4681. Check: OK
d) Find P90, the score separating the bottom 90% from the top 10%.
 The z-score corresponding to 0.90 is: 1.28
 Convert 1.28 to raw score:
x  z      1.28  108  496  138.24  496  634.24  634 Check: OK
Section 9-2 – Basic Skills and Concepts
1) What is the pooled sample proportion and what is the symbol that represents it? Is it used for
hypothesis tests and confidence intervals?
The symbol for the pooled sample proportion is: p . The pooled sample proportion is
the combination of the number of “successes” (in proportion, decimal, or percentage
form) from both samples. It is calculated by dividing the sum of the successes divided
by the sum of the sample sizes. It is used to calculate the test statistic for hypothesis
testing, but it is not used for constructing confidence intervals. Check: OK
5) Find the number of successes x suggested by the given statement. From a New York Times
article: Among 3250 pedestrian walk buttons in New York City, 23% of the work.
Multiply the percent by the number of trials: .23  3250  747.5  748 Check: OK
9) Assume that you plan to use a significance level of α = 0.05 to test the claim that p1 = p2.
Use the given sample sizes and numbers of successes to find (a) the pooled estimate p , (b) the z
test statistic, (c) the critical z values, and (d) the P-value.
Treatment Group
n1 = 500
x1 = 100
a) p 
Placebo Group
n2 = 400
x2 = 50
x1  x2
100  50 150


 0.16  0.167
n1  n2 500  400 900
b) p  0.167
q  1  p  1  0.167  0.833
^
^
x
x
50
100
 0.125
p1  1 
 0.2 p2  2 
n2 400
n1 500
^
 ^

 p1  p 2    p1  p 2 

z

pq pq

n1
n2
0.075
0.2  0.125  0
0.167  0.833 0.167  0.833

500
400
0.075
0.13911 0.13911

500
400

0.075
 2.9976  3.00
0.00027822  0.00034777
0.0006259995 0.02501998
c) The critical values for α = 0.05 in a two-tailed test are: 1.96
d) The P-value is: 1 – 0.9987 = 0.0013. This needs to be doubled for a two-tailed test:
0.0026. Check: OK

0.075


Section 9-3 – Basic Skills and Concepts
1) See the age data used in the hypothesis testing and confidence interval examples of this
section. If we include one more age of 80 years for an unsuccessful applicant, are the
requirements given in Part 1 satisfied? Why or why not?
If an age of 80 is included in the list for unsuccessful applicants, the requirements for
the test will not be met. Since the populations for each sample are less than 30, the
samples must be near normal and have no outliers. 80 is an outlier which will affect the
calculations and yield less than reliable results. Check: OK
5) Determine whether the samples are independent or consist of matched pairs: The
effectiveness of Prilosec for treating heartburn is tested by measuring gastric acid secretion in a
group of patients treated with Prilosec and another group of patients given a placebo.
Independent. Check: OK
9) Assume that the two samples are independent simple random samples selected from normally
distributed populations. Do not assume that the population standard deviations are equal. In a
randomized, double-blind, placebo-controlled trial of children, echinacea was tested as a
treatment for upper respiratory infections in children. “Days of fever” was on criterion used to
measure effects. Among 337 children treated with Echinacea, the mean number of days with
fever was 0.81, with a standard deviation of 1.50 days. Among 370 children given a placebo, the
mean number of days with fever was 0.64 with standard deviation of 1.16 days. Use a 0.05
significance level to test the claim that echinacea affects the number of days with fever. Based
on these results, does echinacea appear to be effective?
 Null hypothesis  H0: 1 = 2
 Alternative hypothesis  H1: 1  2
 Test statistic 
x  x2   1   2   0.81  0.64  0 
0.17
0.17
t 1

2.25 1.3456
0.006676  0.003636
s12 s 22
1.50 2 1.16 2



337
370
337
370
n1 n2

0.17

0.17
 1.673978  1.674
0.101554490
0.010313314621
 Critical Value 
df  337  1  336 or df  370  1  369 Use 336 as the df since it is smaller
For an α = 0.05 in a two-tailed test, the critical value is: 1.968
 Conclusion About Null  Since 1.67 is NOT in the critical region, the conclusion is to
fail to reject the null hypothesis.
 Final Conclusion  Since we failed to reject the null hypothesis, is appears that
echinacea is not effective in treating the number of “days of fever.” Check: OK.
13) Assume that the two samples are independent simple random samples selected from
normally distributed populations. Do not assume that the population standard deviations are
equal. Many studies have been conducted to test the effects of marijuana use on mental abilities.
IN one such study, groups of light and heavy users of marijuana in college were tested for
memory recall, with the results given below. Use a 0.01 significance level to test the lcaim that
the population of heavy marijuana users has a lower mean than the light users? Should
marijuana use be of concern to college students?
1) Light users:
2) Heavy users:
n = 64, x  53.3 , s = 3.6
n = 65, x  51.3 , s = 4.5
 Null hypothesis  H0: 1 = 2
 Alternative hypothesis  H1: 1 > 2
 Test statistic 
x  x2   1   2   53.3  51.3  0 
2.00
t 1

2
2
2
2
12.96 20.25
s1 s 2
3.6
4.5



64
65
64
65
n1 n2

2.00

2.00
0.2025  0.3115384
2.00
 2.78953739  2.790
0.716964756
0.51403846
 Critical Value 
df  64  1  63 or df  65  1  64 Use 64 as the df since it is smaller
For an α = 0.01 in a right-tailed test, the critical value is: 2.390
 Conclusion About Null  Since 2.790 is in the critical region, the conclusion is to
reject the null hypothesis.
 Final Conclusion  Since we rejected the null hypothesis, it appears that marijuana
does have a negative effective on memory, particularly for heavy users. Check: OK.
17) Assume that the two samples are independent simple random samples selected from
normally distributed populations. Do not assume that the population standard deviations are
equal. People send huge sums of money for the purchase of magnets used to treat a wide variety
of pains. Researchers conducted a study to determine whether magnets are effective in treating
back pain. Pain was measured using the visual analog scale, and the results given below are
among the results obtained in the study. Use a 0.05 significance level to test the claim that those
treated with magnets have a greater reduction in pain than those given a sham treatment. Does it
appear that magnets are effective in treating back pain? Is it valid to argue that magnets might
appear to be effective if the sample sizes are larger?
1) Magnet Treatment:
2) Sham Treatment:
n = 20, x  0.49 , s = 0.96
n = 20, x  0.44 , s = 1.4
 Null hypothesis  H0: 1 = 2
 Alternative hypothesis  H1: 1 > 2
 Test statistic 
x  x2   1   2   0.49  0.44  0 
t 1
s12 s 22
0.96 2 1.40 2


20
20
n1 n2
0.9216 1.96

20
20

0.05
0.04608  0.098
0.05
 0.1317249  0.132
0.14408 0.3795787
 Critical Value 
df  20  1  19
For an α = 0.05 in a two-tailed test, the critical value is: 1.729
 Conclusion About Null  Since 0.132 is NOT in the critical region, the conclusion is
to fail to reject the null hypothesis.

0.05
0.05

 Final Conclusion  Since we failed to reject the null hypothesis, is appears that
magnets are not effective in treating back pain. If a larger sample size was used, the
results may be different. However, there is no guarantee. Check: OK.
Section 9-4 – Basic Skills and Concepts
1) Sample data consist of heights and weights of 12 randomly selected bears. When analyzing
the relationship between height and weight, do the methods of this section apply? Why or why
not?
Although the samples do consist of matched pairs (i.e. both measurements coming from
the same bear), we are not comparing the same type of data. Therefore, the methods of
this section do not really work. This type of situation would involve correlating two
different variables rather than comparing means from different samples. Check: OK
3) What do the symbols d and sd denote?
d is the mean of the differences of the matched pairs from each sample. sd is the
standard deviation of these differences. Check: OK
15) The article “An SAT Coaching Program That Works,” by Kaplan included a graph depicting
SAT scores for 50 subjects in a control group. Nine of the 50 points were randomly selected,
with each point representing the score on the SAT test taken the first time and the score on the
SAT test taken a second time, with no preparatory course taken between the two tests. The
graph was used to identify the scores below. Test the claim that the differences have a mean of
0. What do the results suggest?
Student
First
Second
A
480
460
B
510
500
C
530
530
D
540
520
E
550
580
F
560
580
G
600
560
H
620
640
I
660
690
C
530
530
0
D
540
520
20
E
550
580
–30
F
560
580
–20
G
600
560
40
H
620
640
–20
I
660
690
–30
 Calculate values for d and sd
Student
First
Second
Difference
d
A
480
460
20
B
510
500
10
20  10  0  20  30  20  40  20  30  10

 1.111111  1.11
9
9
 x   20  10  0  ...  40  (20)  (30)  5100
 x   (20  10  0  ...  40  20  30)   10  100
n x    x 
9(5100)  100
45900  100
2
2
2
2
2
2
2
2
2
sd 
2
nn  1
2
2

99  1
636.1111111  25.221243  25.221

98

45800

72
 Null hypothesis  H0: d = 0
 Alternative hypothesis  H1: d  0
 Test statistic 
d   d  1.111  0  1.111  1.111
 1.111
t




 0.1321637  0.132
sd
25.221
25.221 25.221 8.40708108
3
9
9
n
 Critical Value 
df  9  1  8
For an α = 0.05 in a two-tailed test, the critical value is: 2.306
 Conclusion About Null  Since –0.132 is NOT in the critical region, the conclusion is
to fail to reject the null hypothesis.
 Final Conclusion  Based on the results of the hypothesis test, we fail to reject the
null hypothesis and conclude that the mean of the differences
between the first and second SAT scores is zero.
 What do the results suggest? This suggest that when taking the SAT a second time
without any special preparation, the second score is not really influenced by what was
learned from taking it the first time. In other words, the first attempt on the test does
not appear to influence performance on the second attempt.
Check: OK. I did have a minor calculation error in figuring the standard deviation,
otherwise things were fine.
Section 9-5 – Basic Skills and Concepts
1) What is the major weakness of the F test described in this section? Name two alternatives that
do a better job of overcoming that weakness.
The major weakness is that the two populations must be distributed normally.
Deviations from normality have a significant impact on the results of the test. Two
alternatives that are more robust (i.e. they handle departures from the normality
better) are the count five method and the modified Levene’s test. Check: OK
3) In the context of this section, what is the F-distribution?
The F-distribution is based on the test-statistic for the F-test. If the samples are
continually selected from a normally distribution population and the variances are used
to calculate the test statistic for the F-test, the shape of the results will tend toward the
F-distribution. Check: OK
14) Students at the author’s college randomly selected 217 student cars and foudnt hat they had
ages with a mean of 7.89 years and a standard deviation of 3.67 years. They also randomly
selected 152 faculty cars and found that they had ages with a mean of 5.99 years and a standard
deviation of 3.65 years. Is there sufficient evidence to support the claim that the ages of faculty
cars vary less than the ages of student cars?
 The values for the test are:
Students have larger variance: s12  3.67 2  13.4689
Faculty has smaller variance: s 22  3.65 2  13.3225
 Null hypothesis  H0: 12 = 22
 Alternative hypothesis  H1: 12  22
 Test statistic 
s 2 13.4689
F  12 
 1.0109889  1.0110
s 2 13.3225
 Critical Value 
Numerator: df  217  1  216
Denominator: df  152  1  151
For an α = 0.05 in a two-tailed test (0.025 in right tail), the critical value is between:
1.0000 and 1.4327
 Conclusion About Null  Since 1.0110 is in the critical region, the conclusion is to
reject the null hypothesis.
 Final Conclusion  Although the test results yield a conclusion to reject the null
hypothesis, this conclusion must be taken with a grain of salt. First, because the dfs
used exceeded the highest values in the available table, the range for the critical value
is wide. Second, it is unlikely that the two variances are exactly the same; however,
examination of the test statistic indicates that the variances are pretty close. A
different conclusion might be reached if software was used to determine the critical
value for the test. Therefore, there is not absolute proof that the variance of the car
ages for faculty is less than the cars ages for the students. Check: OK.
17) Many students have had the unpleasant experience of panicking on a test because the first
question was exceptionally difficult. The arrangement of test items was studied for its effect on
anxiety. Sample values consisting of measure of “debilitating test anxiety” are obtained for a
group of subjects with test questions arranged from easy to difficult, and another group with test
questions arranged from difficult to easy. Test the claim that the two samples come from
populations with the same variance. Here are the summary statistics:
1) Easy-to-Difficult: n = 25, x  27.115 , s2 = 47.020
2) Difficult-to-Easy: n = 16, x  31.728 , s2 = 18.150
 Null hypothesis  H0: 12 = 22
 Alternative hypothesis  H1: 12  22
 Test statistic 
s 2 47.020
F  12 
 2.5906336  2.5906
s 2 18.150
 Critical Value 
Numerator: df  25  1  24
Denominator: df  16  1  15
For an α = 0.05 in a two-tailed test (0.025 in right tail), the critical value is 2.7006
 Conclusion About Null  Since 2.5906 is NOT in the critical region, the conclusion is
to fail to reject the null hypothesis.
 Final Conclusion  There is not enough evidence to reject the claim that the samples
comes from populations with different variances. Check: OK.
Unit 9 – Review Exercises
1) Racial profiling is the controversial practice of targeting someone for suspicion of criminal
behavior on the basis of race, national origin, or ethnicity. The table below includes data from
randomly selected drivers stopped by police in a recent year:
Drivers Stopped by Police
Total # of Observed Drivers
Race and Ethnicity
Black and Non-Hispanic White and Non-Hispanic
24
147
200
1400
a) Use a 0.05 significance level to test the claim that the proportion of blacks stopped by
police is significantly greater than the proportion of whites.
 Null hypothesis  H0: p1 = p2
 Alternative hypothesis  H1: p1 > p2
 Test statistic 
x  x2
24  147
171
p 1


 0.106875  0.1069
n1  n2 200  1400 1600
p  0.1069
q  1  p  1  0.1069  0.8931
^
^
x
x
24
147
p1  1 
 0.12
p2  2 
 0.105
n1 200
n2 1400
^
 ^

 p1  p 2    p1  p 2 

z

pq pq

n1
n2
0.015
0.00047736  0.000068194
0.12  0.105  0
0.1069  0.8931 0.1069  0.8931

200
1400

0.015
0.0005455565


0.015
0.09547239 0.09547239

200
1400
0.015
 0.6422016  0.642
0.02335715
 Critical Value  For an α = 0.05 in a right-tailed test, the critical value is: 1.645
 Conclusion About Null  Since 0.642 is NOT in the critical region, the conclusion
is to fail to reject the null hypothesis.
 Final Conclusion  Based on the results of the hypothesis test, we cannot support
the claim that blacks are stopped more often than whites. Check: OK.
b) Construct a confidence interval that could be used to test the claim in part (a). Be sure to
use the correct level of significance. What do you conclude based on the confidence
interval?
^
^
x
x
24
147
p1  1 
 0.12
p2  2 
 0.105
n1 200
n2 1400
For a significance level of 0.05 and a right-tailed test, use a confidence level of 90%.
This yields a zα/2 value of 1.645.

^
E  z / 2 
^
^
^
p1 q1 p 2 q 2
0.12  0.88 0.105  0.895
0.1056 0.093975

 1.645 

 1.645 


n1
n2
200
1400
200
1400
1.645  0.000528  0.000067125  1.645  0.000595125  1.645  0.024395  0.04013  0.0401
Construct the confidence interval:
^
^
 ^

 ^

 p1  p 2   E   p1  p 2    p1  p 2   E  0.015  0.040   p1  p 2   0.015  0.040 




 0.025   p1  p 2   0.055
Due to the fact that zero in the confidence interval, we can conclude that there is no
difference between the proportions of black and whites that are stopped.
Check: OK
3) Listed below are Flesch Reading Ease scores taken from randomly selected pages in Harry
Potter and the Sorcerer’s Stone and War and Peace. Use a 0.05 significance level to test the
claim that Harry Potter is easier to read than War and Peace. Is the result expected?
Potter:
War Peace:
85.3 84.3 79.5 82.5 80.2 84.6 79.2 70.9 78.6 86.2 74.0 83.7
69.4 64.2 71.4 71.6 68.5 51.9 72.2 74.4 52.8 58.4 65.4 73.6
85.3  84.3  ...  74.0  83.7 969

 80.75
12
12
x 2  85.32  84.32 ...  74.0 2  83.7 2  78487.82
x1 
 
 x   (85.3  84.3  ...  74.0  83.7)  969  938961
n x    x 
12(78487.82)  938961
941853.84  938961
s 



2
2
2
2
nn  1
1
2
1212  1
1211
2892.84

132
21.91545454545  4.6813945  4.681
69.4  64.2  ...  65.4  73.6 793.8

 66.15
12
12
x 2  69.4 2  64.2 2 ...  65.4 2  73.6 2  53189.1
x2 
 
 x   (69.4  64.2  ...  65.4  73.6)  793.8  630118.44
n x    x 
12(53189.1)  630118.44
638269.2  630118.44
s 



2
2
2
2
nn  1
2
2
1212  1
61.74818181818  7.85800113376  7.858
 Null hypothesis  H0: 1 = 2
 Alternative hypothesis  H1: 1 > 2
 Test statistic 
1211
8150.76

132
t
x1  x2   1   2   80.75  66.15  0 
2
1
2
2
2
s
s

n1 n2

14.6
6.9716604

4.681
7.858

12
12
2
14.6

21.911761 61.748164

12
12
14.6
1.8259  5.1456
14.6
 5.5294857522  5.529
2.640390
 Critical Value 
df  12  1  11
For an α = 0.05 in a right-tailed test, the critical value is: 1.796
 Conclusion About Null  Since 5.529 is in the critical region, the conclusion is to
reject the null hypothesis.
 Final Conclusion  Since we rejected the null hypothesis, it appears that Harry
Potter has an easier reading level that War and Peace. This would be expected since
Harry Potter is intended for younger readers and War and Peace is intended for adult
readers. Check: OK.
5) An article published in USA Today stated that “in a study of 200 colorectal surgery patients,
104 were kept warm with blankets and intravenous fluids; 96 were kept cool. The results show:
Only 6 of those warmed developed wound infections versus 18 who were kept cool.”
a) Use a 0.05 significance level to test the claim of the article’s headline: “Warmer surgical
patients recover better.” If these results are verified, should surgical patients be routinely
warmed?
 Null hypothesis  H0: p1 = p2
 Alternative hypothesis  H1: p1 > p2
 Test statistic 
x  x2
98  78 176
p 1


 0.88
n1  n2 104  96 200
p  0.88
q  1  p  1  0.88  0.12
^
x
98
p1  1 
 0.9423
n1 104
^
 ^

 p1  p 2    p1  p 2 

z

pq pq

n1
n2
0.12980769
0.001015  .0011

^
p2 
x 2 78

 0.8125
n2 96
0.9423  0.8125  0
0.88  0.12 0.88  0.12

104
96
0.12980769
0.002115384615


0.12980769
0.1056 0.1056

104
96

0.12980769
 2.822316  2.82
0.045993310
 Critical Value  For an α = 0.05 in a right-tailed test, the critical value is: 1.645
 Conclusion About Null  Since 2.822 is in the critical region, the conclusion is to
reject the null hypothesis.
 Final Conclusion  Based on the results of the hypothesis test, we can support the
claim that warmer patients recover better. Although the evidence does lead to the
potential of keeping patients warm, the test should be repeated for other surgeries
and for larger samples before it is assumed that the technique is effective in all
surgeries. Check: OK.
b) If a confidence interval is to be used for testing the claim in part (a), what confidence level
should be used?
For a significance level of 0.05 and a right-tailed test, use a confidence level of 90%.
This yields a zα/2 value of 1.645.Check: OK
c) Using the confidence level from part (b), construct a confidence interval estimate of the
difference between the two population proportions.
^
^
x
x
98
78
p1  1 
 0.9423
p2  2 
 0.8125
n1 104
n2 96
For a significance level of 0.05 and a right-tailed test, use a confidence level of 90%.
This yields a zα/2 value of 1.645.
^
E  z / 2 
^
^
^
p1 q1 p 2 q 2
0.9423  0.0766 0.8125  0.1875
0.07218 0.15234

 1.645 

 1.645 


n1
n2
104
96
104
96
1.645  0.000694  0.0015869  1.645  0.002280954  1.645  0.04775933  0.078564  0.07856
Construct the confidence interval:
^
^
 ^

 ^

 p1  p 2   E   p1  p 2    p1  p 2   E  0.1298  0.07856   p1  p 2   0.1298  0.07856 




0.05124   p1  p 2   0.20836
Due to the fact that zero is not in the confidence interval, we can conclude that there
is a difference between the proportions and we fail to reject the null hypothesis.
Check: OK
d) In general, if a confidence interval estimate of the difference between two population
proportions is used to test some claim about the proportions, will the conclusion based on the
confidence interval always be the same as the conclusion from a standard hypothesis test?
When dealing with population proportions, the conclusion based on a traditional
hypothesis test (or P-value) will not necessarily be the same as the conclusion based
on confidence intervals. Therefore, a hypothesis test should be done to address
claims dealing with population proportions.
Unit 9 – Cumulative Review Exercises
1) A section of Highway 405 in L.A. has a speed limit of 65 mph, and recorded speeds are listed
below for randomly selected cars traveling on northbound and southbound lanes.
NORTH:
SOUTH:
68 68 72 73 65 74 73 72 68 65 65 73 66 71 68 74 66 71 65 73
59 75 70 56 66 75 68 75 62 72 60 73 61 75 58 74 60 73 58 75
a) Using the speeds for the northbound lanes, find the mean, median, standard deviation,
variance, and range.
68  68  72  ...  71  65  73 1390

 69.5
Mean:
20
20
Median:
65 65 65 65 66 66 68 68 68 68 71 71 72 72 73 73 73 73 74 74
68  71 139

 69.5
The median is:
2
2
SD:
 x 2  652  652  652  ...  732  742  742  96826
 
 x 
2
s
 (65  65  65  ...  73  74  74) 2  1390  1932100
2
 
n x 2   x 
2
nn  1

20(96826)  1932100
1936520  1932100


2020  1
2019
4420

380
11.6315789  3.41051007144  3.411
Variance:
s 2  3.41051007144 2  11.6315789474  11.632
Range:
74  65  9 Check: OK
b) Using all of the speed combined, test the claim that the mean is greater than the posted
speed limit of 65 mph.
 Calculate the mean and sample standard deviation 
68  68  72  ...  73  58  75 2735
x

 68.375
40
40
 x 2  682  682  722  ...  732  582  752  188259
 
 x 
2
 (68  68  72  ...  73  58  75) 2  2735  7480225
2
 
n x 2   x 
s
2
nn  1

40(188259)  7480225
7530360  7480225
50135



4040  1
4039
1560
32.1378205128  5.66902288872  5.669
 Null hypothesis  H0:  = 65
 Alternative hypothesis  H1:  > 65
 Test statistic 
x  68.375
 x  65
s  5.669
n  40
df  39
t
x   x 68.375  65 3.375



s
5.669
5.669
n
40
40
3.375
3.375

 3.765265  3.765
5.669
0.89635122
6.324555
 Critical Value  For an α = 0.05 in a one-tailed test, the critical value is: 1.684
 Conclusion About Null  Since 3.765 is in the critical region, we can conclude to
reject the null hypothesis.
 Final Conclusion  Based on the results of the hypothesis test, we reject the null
hypothesis and mean speed of drivers is higher than the posted speed limit.
Check: OK.
c) Do the northbound speeds appear to come from a normally distributed population?
Explain.
Sketch a histogram of the data to visually check for normality:
Histogram of Speed
7
6
Frequency
5
4
3
2
1
0
57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76
Speed (mph)
Based on the histogram, the values for the northbound lane do not appear to be
normal. It has some characteristics, but it is skewed in one direction. Check: OK.
This is considered to be not too far from normal. Since there are no outliers, the t test
will be valid.
d) Assuming that the speeds are from normally distributed populations, test the claim that the
mean speed on the northbound lanes is equal to the mean speed on the southbound lanes.
Based on the results from part (c), does it appear that this hypothesis test is likely to be valid?
 Calculate the mean and sample standard deviation 
68  68  72  ...  71  65  73 1390
x1 

 69.5
20
20
 x   65  65  65  ...  73  74  74  96826
 x   (65  65  65  ...  73  74  74)  1390  1932100
2
2
2
2
2
2
2
2
2
2
 
n x 2   x 
s1 
2
nn  1

20(96826)  1932100
1936520  1932100


2020  1
2019
4420

380
11.6315789  3.41051007144  3.411
59  75  70  ...  73  58  75 1345
x2 

 67.25
20
20
 x 2  592  752  702  ...  732  582  752  91433
 
 x 
2
s1 
 (68  68  72  ...  73  58  75) 2  1345  1809025
2
 
n x 2   x 
2
nn  1

20(91433)  1809025
1828660  1809025
19635



2020  1
2019
380
51.6710526316  7.18825796919  7.188
 Null hypothesis  H0: 1 = 2
 Alternative hypothesis  H1: 1  2
 Test statistic 
x  x2   1   2   69.5  67.25  0 
2.25
t 1

11.6349 51.6673
s12 s 22
3.4112 7.188 2



20
20
20
20
n1 n2

2.25
3.16529868158

2.25
0.5817  2.5833
2.25
 1.2646640396  1.265
1.77912862
 Critical Value 
df  20  1  19
For an α = 0.05 in a right-tailed test, the critical value is: 1.729
 Conclusion About Null  Since 1.265 is NOT in the critical region, the conclusion is
to fail to reject the null hypothesis.
 Final Conclusion  Since we failed to reject the null hypothesis, it appears that there
is not a significance difference between the mean speeds in the north and southbound
lanes. Based on the results of the part (c), this is probably a valid hypothesis test.
Check: OK.
3) In an article from the Associated Press, it was reported that researchers “randomly selected
100 New York motorists who had been in an accident and 100 who had not. Of those in
accidents, 13.7 percent owned a cellular phone, while just 10.6 percent of accident-free drivers
had a phone in the car.” Analyze these results.
If the sample sizes are truly 100 people, then the percentages could not have been
decimal values. At some point, an error occurred. Check: OK. I looked at the answer
first to try and figure out what the author wanted analyzed. I understood the question
after I looked at the answer.