Answers - Millersville University

Millersville University
Department of Mathematics
MATH 130, Review of Hypothesis Testing
May 7, 2010
1. An engineer claims that a new fuel injection design increases the mean mileage on a
car above its current 31 mpg.
(a) Express the claim in symbolic form.
If µ represents the mean mileage of cars with the new fuel injection design then
the claim above can be expressed in the symbolic form: µ > 31.
(b) Identify the null hypothesis.
Since the null hypothesis always contains a condition of equality then the null
hypothesis can be expressed symbolically as, H0 : µ = 31.
(c) Identify the alternative hypothesis.
The alternative hypothesis does not contain a condition of equality, thus the alternative
hypothesis is the original claim in this case. It can be expressed symbolically
as, H1 : µ > 31.
(d) Identify the test as being left-tailed, right-tailed, or two-tailed.
The alternative hypothesis can be used to determine whether a test is left-tailed,
right-tailed, or two-tailed. The inequality symbol always “points” in the direction
of the tail of the test. In this example the inequality symbol is “greater than” (>)
and therefore this is a right-tailed test.
(e) Assuming the conclusion is to reject the null hypothesis, state the conclusion in
nontechnical terms.
Since the original claim is the alternative hypothesis and the null hypothesis is
rejected, the conclusion can be stated as,
“The sample data support the claim that the mean mileage of cars with
the new fuel injection design is greater than 31 mpg.”
(f) Assuming the conclusion is failure to reject the null hypothesis, state the conclusion
in nontechnical terms.
Since the original claim is the alternative hypothesis and the null hypothesis is not
rejected, the conclusion can be stated as,
“There is not sufficient sample data to support the claim that the mean
mileage of cars with the new fuel injection design is greater than 31 mpg.”
2. In a study of consumer habits, researchers designed a questionnaire to identify compulsive
buyers. For a sample of consumers who identified themselves as compulsive
buyers, questionnaire scores have a mean of 0.83 and a standard deviation of 0.24.
Assume that the subjects were randomly selected and that the sample size was 32.

At the 0.01 significance level test the claim that the self-identified compulsive buyer
population has a mean greater than 0.21, the mean for the entire population. Does
the questionnaire seem to be effective in identifying compulsive buyers?
In this example the important sample statistics are the sample size n = 32, the sample
mean x = 0.83, and the sample standard deviation s = 0.24.
The symbolic form of the claim made above is µ > 0.21 where µ represents the population
mean score of compulsive buyers on the questionnaire. Since this claim does not
contain a condition of equality, it must be the alternative hypothesis, H1 : µ > 0.21.
Therefore the null hypothesis is H0 : µ = 0.21. The test to be performed is a righttailed
test at the α = 0.01 significance level on a large sample (since n > 30). The
population standard deviation is unknown so we will use the Student’s t-distribution to
represent the distribution of the sample means. Thus from Table V, the critical value
of CV : tα = 2.326. The test statistic is
TS : t0 =
x − µ
s/ √ n
= 0.83 − 0.21
0.24/ √ 32
= 14.61.
The test statistic is much larger than the critical value and thus the test statistic lies
in the critical region of this right-tailed test. Since the test statistic falls into the
critical region, then the null hypothesis is rejected. Now since the original claim is the
alternative hypothesis and the null hypothesis is rejected, the conclusion can be stated
as,
“The sample data support the claim at the 0.01 significance level that the
score of compulsive buyers on the questionnaire is higher than the score of
the entire population.”
The questionnaire seems to be effective at identifying compulsive buyers.
3. According to the Insurance Information Institute, the mean expenditure for auto insurance
in the United States was $774 in 2002. An insurance salesman obtains a random
sample of 35 auto insurance policies and determines the mean expenditure to be $735
with a standard deviation of $48.31. Is there enough evidence to conclude that the
mean expenditure for auto insurance is different from the 2002 amount at the α = 0.01
level of significance?
Here the important sample statistics are the sample size n = 35, the sample mean
x = 735, and the sample standard deviation s = 48.31.
The symbolic form of the claim made by the insurance salesman is µ = 774 where
µ represents the population mean expenditure for auto insurance. This claim is the
alternative hypothesis, H1 : µ = 774. Therefore the null hypothesis is H0 : µ = 774.
The test to be performed is a two-tailed test at the α = 0.01 significance level on a
population whose standard deviation σ is unknown. Thus we will use the Student’s

t-distribution. The number of degrees of freedom is df = n − 1 = 34. Thus from Table
V, the critical value of CV : tα/2 = ±2.728. The test statistic is
TS : t0 =
x − µ
s/ √ n
= 735 − 774
48.31/ √ 35
= −4.776.
Since the test statistic falls into the critical region, the null hypothesis is rejected. Now
since the original claim is the alternative hypothesis and the null hypothesis is rejected,
the conclusion can be stated as,
“The sample data support the claim at the 0.01 significance level that the
mean expenditure for auto insurance is different from $774.”
4. An advertising agency claims that the mean length of television commercials is less
than 30 seconds.
(a) Express the claim in symbolic form.
If µ represents the mean length of commercials, then the claim above can be expressed
in the symbolic form: µ < 30.
(b) Identify the null hypothesis.
Since the null hypothesis always contains a condition of equality then the null
hypothesis is the logical opposite of the claim above. The null hypothesis can be
expressed symbolically as, H0 : µ = 30.
(c) Identify the alternative hypothesis.
The alternative hypothesis does not contain a condition of equality, thus the alternative
hypothesis is the original claim in this case. It can be expressed symbolically
as, H1 : µ < 30.
(d) Identify the test as being left-tailed, right-tailed, or two-tailed.
The alternative hypothesis can be used to determine whether a test is left-tailed,
right-tailed, or two-tailed. The inequality symbol always “points” in the direction
of the tail of the test. In this example the inequality symbol is “less than” (

“There is not sufficient sample data to support the claim that the mean
length of commercials is less than 30 seconds.”
5. A pharmaceutical company makes a pill intended for children susceptible to seizures.
The pill is supposed to contain 20.0 mg of phenobarbital. A random sample of 20 pills
yielded the amounts (in mg) listed below. Do the pills contain the proper amount of
phenobarbital at the α = 0.01 significance level? (Hint: The sample standard deviation
of the weights is s = 3.5.)
27.5 26.0 22.9 23.5 23.0
23.9 26.3 20.9 22.5 21.7
28.4 16.1 19.3 17.2 17.9
21.7 23.0 16.5 24.1 22.3
(a) State the original claim in symbolic form.
If µ represents the mean amount of phenobarbital measured in milligrams in the
pills, then the claim above can be expressed in the symbolic form: µ = 20.0.
(b) In non-technical language describe the type II error for this hypothesis test.
A type II error occurs when a we fail to reject a false null hypothesis. In this
case a type II error occurs when we fail to reject claim that the pills contain an
average of 20.0 milligrams of phenobarbital, when in fact they do not contain that
amount.
(c) Find the critical value of t for this hypothesis test.
Since the sample size is only n = 20, which is considered a small sample, we must
use the Student-t Distribution. The degrees of freedom becomes df = 19. The
original claim is the null hypothesis. The alternative hypothesis is the claim that
µ = 20.0 and thus we are conducting a two-tailed test. The critical value of t is
the value of t for a two-tailed test with α = 0.01 and df = 19. From the Student-t
table this value is tα/2 = ±2.86.
(d) Do the pills contain the proper amount of phenobarbital at the α = 0.01 significance
level?
We must calculate the sample mean before we can find the test statistic.
The test statistic is then
t0 =
x =
x
n
= 444.7
20
≈ 22.24
x − µ
s/ √ 22.24 − 20.0
=
n 3.5/ √ ≈ 2.862.
20
The test statistic is (barely) larger than the critical value. Thus we reject the null
hypothesis. We conclude that the sample data warrant rejection of the claim that
the mean amount of phenobarbital in the pills is 20.0 milligrams.

6. A newspaper survey was the basis for a report that “55% of gun owners favor stricter
gun laws.” Assume that the survey group consisted of 504 randomly selected gun
owners. Test the claim that the majority (more than 50%) of gun owners favor stricter
gun laws. Use a 0.01 significance level.
(a) State the null and alternative hypotheses in symbolic form.
The original claim is that the proportion of gun owners favoring stricter gun laws
is greater than 0.50. The original claim lacks a condition of equality and is thus
the alternative hypothesis. Therefore the null and alternative hypotheses can be
expressed in symbolic form as
H0 : p = 0.50
H1 : p > 0.50
(b) Find the value of the test statistic for this hypothesis test.
Based on the survey data mentioned we see that the sample proportion of gun
owners favoring stricter gun laws is ˆp = 0.55. The value of test statistic is then
z0 =
ˆp − p
p(1−p)
n
= 0.55 − 0.50
(0.50)(1−0.50)
504
≈ 2.24.
(c) Find the critical value for this hypothesis test at the 0.01 significance level.
By examining the alternative hypothesis, we see that we are conducting a righttailed
test with α = 0.01. According to the normal probability table, the critical
value zα = 2.33.
(d) Test the claim that the majority (more than 50%) of gun owners favor stricter
gun laws.
Since the test statistic does not fall in the critical region, we fail to reject the null
hypothesis. Therefore we conclude that the sample data do not support the claim
that the majority of gun owners favor stricter gun laws.
z 2.24
0 zΑ2.33 zΑ2.33

7. A dental X-ray machine bears a label stating that the machine gives radiation dosages
with a mean of less than 5.00 milliroentgens. Sample data consist of 23 randomly
selected observations with a mean of 4.13 milliroentgens and a standard deviation 1.91
milliroentgens. Using a 0.01 level of significance, test the claim stated on the label.
(a) State the label’s claim in symbolic form.
If µ represents the mean X-ray dosage given by the machine, then the original
claim is that µ < 5.00.
(b) Is the original claim the null or alternative hypothesis?
Since the original claim does not contain a condition of equality, it is the alternative
hypothesis, H1 : µ < 5.00.
(c) Find the critical value for this hypothesis test.
Since the sample size is only n = 23, which is considered a small sample, we must
use the Student-t Distribution. The degrees of freedom becomes df = n − 1 = 22.
The alternative hypothesis is the claim that µ < 5.00 and thus we are conducting
a left-tailed test. The critical value of t is the value of t for a one-tailed test with
α = 0.01 and df = 22. From the Student-t table this value is tα = −2.51.
(d) Using a 0.01 level of significance, test the claim stated on the label.
The test statistic is
t0 =
x − µ
s/ √ n
= 4.13 − 5.00
1.9/ √ 23
≈ −2.20.
The test statistic does not fall into the critical region and thus we fail to reject the
null hypothesis. We conclude that the sample data do not support the claim that
the mean radiation dosage given by the machine is less than 5.00 milliroentgens.
t 2.20
tΑ2.51 tΑ2.51
8. The quality control manager of a company which produces telephone-answering machines
considers production to be “out of control” when the overall rate of defects
0

exceeds 4%. Testing of a random sample of 150 machines reveals that 9 are defective,
so the sample percentage of defective answering machines is 6%. The manager claims
that this is only a chance occurrence, so that production is not really out of control.
Use a 0.05 significance level to test the manager’s claim.
In this example the important sample statistics are the sample size n = 150 and the
sample proportion ˆp = x/n = 9/150 = 0.06.
The symbolic form of the claim made above is p ≤ 0.04 where p represents the fraction
of defective answering machines. Since the manager is claiming that production is not
out of control then he is claiming that the defect rate is not exceeding (and therefore
less than or equal to) 4%. Since this claim contains a condition of equality, it is the
null hypothesis, H0 : p = 0.04. Therefore the alternative hypothesis is H1 : p > 0.04.
The test to be performed is a right-tailed test at the α = 0.05 significance level on a
large sample (since n > 30). Thus from the normal probability distribution, the critical
value of zα = 1.645. The test statistic is
z0 =
ˆp − p
p(1 − p)/n =
0.06 − 0.04
0.04(1 − 0.04)/150
z 1.25
0 zΑ1.645 zΑ1.645
= 1.25.
Since the test statistic does not fall into the critical region, then the null hypothesis is
not rejected. Now since the original claim is the null hypothesis and the null hypothesis
is not rejected, the conclusion can be stated as,
“The sample data do not warrant rejection of the claim at the 0.05 significance
level that the defect rate of the answering machines is less than or
equal to 4%.”
Production does not seem to be out of control at the answering machine factory.
9. A newspaper ran a report about a University of Southern California poll of 1245
students from the western United States. It was reported that 8% of those surveyed

elieve that Jim Morrison (of The Doors) is still alive. The newspaper article began
with the claim that “almost 1 out of 10” students thinks that Jim Morrison is alive.
At the α = 0.025 significance level, test the claim that the true percentage is less than
10%.
(a) Identify the type I error for this test.
Type I error is made when we reject a true null hypothesis. The original claim
is that the proportion of students from the western US who believe Jim Morrison
is still alive is less than 10%. Expressed in symbolic form this is p < 0.10. Since
that statement does not contain a condition of equality, the original claim is the
alternative hypothesis. The null hypothesis is H0 : p = 0.10. Thus the Type I
error in this situation would be the mistake of rejecting the claim that at least
10% of students in the western US believe that Jim Morrison is still alive, when
in fact it is true that at least 10% of students in the western US believe that Jim
Morrison is still alive.
(b) Is the original claim the null or alternative hypothesis?
Based on the discussion in the answer to the previous point, the original claim is
the alternative hypothesis.
(c) Find the critical value for this hypothesis test.
Examining the alternative hypothesis H1 : p < 0.10 we determine that we are
conducting a left-tailed test. The significance level is α = 0.025. From the normal
probability table then the critical value of z is zα = −1.96.
(d) At the 0.025 significance level, test the claim that the true percentage is less than
10%.
Based on the survey data mentioned we see that the sample proportion of gun
owners favoring stricter gun laws is ˆp = 0.08. The value of test statistic is then
z0 =
ˆp − p
p(1−p)
n
= 0.08 − 0.10
(0.10)(1−0.10)
1245
≈ −2.35.
Since the test statistic falls in the critical region we reject the null hypothesis.
Therefore we conclude that the sample data support the claim that the percentage
of students in the western United States who believe that Jim Morrison is alive is
less than 10%.

z 2.35
zΑ1.96 zΑ1.96
10. When the contents of 1200 boxes of cereal were weighed, 432 were found to be underweight.
Construct the 96% confidence interval estimate for the proportion of cereal
boxes which are underweight.
A cereal box is either underweight or not, so this is yet another example of a binomial
experiment. The sample size is large (n = 1200 > 30). The random variable describing
the number of underweight boxes in the sample is x = 432. The sample proportion of
underweight cereal boxes is
ˆp = x 432
= = 0.360
n 1200
The sample of proportion of non-underweight boxes is 1 − ˆp = 1 − 0.360 = 0.640.
We will use the normal distribution to approximate the binomial distribution. At the
96% confidence level the value of α = 0.04 and thus α/2 = 0.02. The critical value
corresponding to probability 0.5 − 0.02 = 0.48 is zα/2 = 2.054 according to Table V.
Thus the margin of error in this estimate of the population proportion is
E = zα/2
ˆp(1 − ˆp)
n
= 2.054
0
(0.360)(0.640)
1200
≈ 0.028
Note that we are rounding the margin of error to three decimal places just like the
sample proportions. Therefore the 96% confidence interval estimate of the sample proportion
of underweight cereal boxes is
(ˆp − E, ˆp + E) = (0.360 − 0.028, 0.360 + 0.028) = (0.332, 0.388).
11. An earlier survey showed that 57% of adults had seen the movie Titanic.
(a) How many people must be surveyed to produce a sample proportion in error by
no more than 3% at the 95% confidence level?
We are being asked to estimate a sample size when we know a previous estimate
of the proportion of adults who have seen the movie is 57% or in other words

ˆp = 0.570. Therefore 1 − ˆp = 1 − 0.570 = 0.430. The margin of error in our
estimate should be 3% or E = 0.03. At the 95% confidence level the value of
α = 0.05 and thus α/2 = 0.025. The critical value corresponding to probability
0.5 − 0.025 = 0.475 is zα/2 = 1.960 according to Table V. Thus the sample size
needed for this estimate is
n =
zα/2
2
ˆp(1 − ˆp) =
E
2 1.96
(0.570)(0.430) = 1046.2 ≈ 1047
0.03
Notice that we must always round sample sizes up to the next whole number.
(b) How many people must be surveyed if no prior estimate of the sample proportion
is available?
We are being asked to estimate a sample size when no previous estimate of the
proportion of adults who have seen the movie is available, in other words we do
not know ˆp. The margin of error in our estimate should be 3% or E = 0.03. At
the 95% confidence level the value of α = 0.05 and thus α/2 = 0.025. The critical
value corresponding to probability 0.5 −0.025 = 0.475 is zα/2 = 1.960 according to
Table V. Thus the sample size needed for this estimate is
n =
zα/2
2
0.25 =
E
2 1.96
0.25 = 1067.1 ≈ 1068
0.03
Notice that we must always round sample sizes up to the next whole number. Also
notice that when no prior estimate of sample proportion is available we must take
a larger sample.
12. A statistics student believes that an individual’s arm span is equal to the individual’s
height. The student used a random sample of 10 students and obtained the following
data.
Student: 1 2 3 4 5
Height (inches) 59.5 69.0 77.0 59.5 74.5
Arm span (inches) 62.0 65.5 76.0 63.0 74.0
Student: 6 7 8 9 10
Height (inches) 63.0 61.5 67.5 73.0 69.0
Arm span (inches) 66.0 61.0 69.0 70.0 71.0
(a) Is the sampling method dependent or independent? Why?
The samples are dependent since the students whose heights are measured determine
the students whose arm spans are measured. The same students are in both
samples.
(b) Does the evidence suggest that an individual’s height and arm span are different
at the α = 0.05 level of significance? Note: A normal probability plot indicates

that the data and the differenced data are normally distributed and that the data
and the differenced data have no outliers.
Let quantities with a subscript of “d” refer to the difference of height minus arm
span. The null and alternative hypotheses are respectively,
H0 : µd = 0
H1 : µd = 0 (two-tailed test).
From the table of data above we see can calculate the differences.
Student: 1 2 3 4 5 6 7 8 9 10
d −2.5 3.5 1.0 −3.5 0.5 −3.0 0.5 −1.5 3.0 −2.0
Thus the sample statistics of the difference are d = −0.40 and sd = 2.47.
We are performing a two-tailed test with α = 0.05 which implies α/2 = 0.025.
There are df = n − 1 = 10 − 1 = 9 degrees of freedom. Thus the critical values
are t0.025 = ±2.262.
The test statistic is
t0 = d
sd/ √ n
−0.40
=
2.47/ √ = −0.512.
10
Since the test statistic falls between the critical values (not in either of the tails
of the distribution), we fail to reject the null hypothesis.
Conclusion: the is insufficient evidence at the α = 0.05 level of significance to
reject the claim that a student’s height and arm span are the same.
(c) Construct and interpret a 95% confidence interval about the population mean
difference between height and arm span.
The margin of error is
sd sd
E = tα/2 √ = t0.025 √ = 2.262
n n 2.47
√ = 1.77.
10
Thus the 95% confidence interval estimate for µd is
d ± E = −0.40 ± 1.77 ⇐⇒ (−2.17, 1.37).
13. Zoloft is a drug that is used to treat obsessive-compulsive disorder (OCD). In randomized,
double-blind clinical trials, 926 patients diagnosed with OCD were randomly
divided into two groups. Subjects in Group 1 (experimental group) received 200 mg
per day of Zoloft, while the subjects in Group 2 (control group) received a placebo.
Of the 553 subjects in the experimental group, 77 experienced dray mouth as a side
effect. Of the 373 subjects in the control group, 34 experienced dry mouth as a side
effect.

(a) Do a higher proportion of the subjects in the experimental group experience dry
mouth than the subjects in the control group at the α = 0.05 level of significance?
We are being asked to perform a hypothesis test regarding two population proportions.
The samples are independent. There are some conditions we must
check before we can conduct the hypothesis test using the z-distribution. Note that
n1 = 553 and x1 = 77. Thus the sample proportion for Group 1 is
ˆp1 = 77
553 = 0.1392 and n1ˆp1(1 − ˆp1) = 553(0.1392)(1 − 0.1392) = 66.3 ≥ 10.
Similarly we note that n2 = 373 and x2 = 34. Thus the sample proportion for
Group 2 is
ˆp2 = 34
373 = 0.0912 and n2ˆp2(1 − ˆp2) = 373(0.0912)(1 − 0.0912) = 30.9 ≥ 10.
Assuming that each sample is less than 5% of the population, we can conduct the
hypothesis test.
The hypotheses are
The pooled estimate is
H0 : p1 = p2
H1 : p1 > p2 (right-tailed test).
ˆp = x1 + x2
n1 + n2
= 77 + 34
111
= = 0.1199.
553 + 373 926
Since α = 0.05 the critical value of z is zα = z0.05 = 1.645. The test statistic is
ˆp1 − ˆp2
z0 =
ˆp(1 − ˆp) 0.1392 − 0.0912
=
1 1 + 0.1199(1 − 0.1199) n1 n2
= 2.21.
1 1 + 553 373
Since the test statistics lies to the right of the critical value we reject the null
hypothesis.
Conclusion: the sample data support the claim that a higher proportion of the
experimental group experience dry mouth than in the control group at the α = 0.05
level of significance.
(b) Construct a 90% confidence interval for the difference between the two population
proportions, p1 − p2.
When constructing a 90% confidence interval α = 0.10 which implies α/2 = 0.05
and zα/2 = 1.645. The margin of error is
E = zα/2
= 1.645
ˆp1(1 − ˆp1)
n1
+ ˆp2(1 − ˆp2)
n2
0.1392(1 − 0.1392)
553
+ 0.0912(1 − 0.0912)
373
= 0.0345.

Thus the 90% confidence interval estimate of p1 − p2 is
ˆp1 − ˆp2 ± E = 0.1392 − 0.0912 ± 0.0345 ⇐⇒ (0.0135, 0.0825).
14. A student wanted to determine whether the wait time in the drive-through at Mc-
Donald’s differed from that at Wendy’s. She used a random sample of 30 cars at
McDonald’s and a 27 cars at Wendy’s and found that
xM = 133.60 sM = 39.61
xW = 219.14 sW = 102.85
where these times are measured in seconds.
(a) Is the sampling method dependent or independent?
The samples are independent since the cars going through the drive-through at Mc-
Donald’s do not influence the cars passing through the drive-through at Wendy’s.
(b) Is there a difference in wait times at each restaurant’s drive-through at the α =
0.10 level of significance?
The hypotheses to be tested are
H0 : µM = µW
H1 : µM = µW (two-tailed test).
Since the population standard deviation of the difference in waiting times is unknown
we must use the t-distribution in this test with
df = min{30, 27} − 1 = 27 − 1 = 26
degrees of freedom and tα/2 = t0.05 = ±1.706.
The test statistic is
t0 = (xM − xW) − (µM − µW)
s 2 M
nM + s2 W
nW
= (133.60 − 219.14) − 0
(39.61) 2 (102.85)2
+ 30 27
= −4.059.
Since the test statistic is smaller than tα/2 = −1.706 we reject the null hypothesis.
Conclusion: there is sufficient sample evidence at the α = 0.10 level of significance
to conclude that the population mean wait times at the drive-throughs of
McDonald’s and Wendy’s are different.
(c) Construct and interpret a 95% confidence interval about µM − µW. How might
marketing executives with McDonald’s use this information?
When constructing a 95% confidence interval α = 0.05 which implies α/2 = 0.025
and tα/2 = 2.056 when df = 26. The margin of error is
E = tα/2
s 2 M
nM
+ s2 W
nW
(39.61)
= 2.056
2
30
+ (102.852
27
= 43.33

Thus the 95% confidence interval estimate of µM − µW is
µM − µW ± E = 133.60 − 219.14 ± 43.33 ⇐⇒ (−128.87, −42, 21).
The McDonald’s marketing executives may state that customers get served at
the McDonald’s drive-throughs on average 1-2 minutes faster than customers at
Wendy’s drive-throughs.