Chapter 10 Hypothesis Tests Regarding a Parameter
Ch. 10.1 The Language of Hypothesis Testing
Objective A: Set up a Hypothesis Testing
Hypothesis testing is a procedure, based on a sample evidence and probability, used to test
statements regarding a characteristic of one or more populations.
- The null hypothesis H 0 is a statement to be tested.
- The alternate hypothesis H1 is a statement that we are trying to find evidence to support.
Example 1:
Set up H 0 and H1 .
(a) In the past, student average income was $6000 per year. An administrator believes
the average income has increased.
𝐻0 :   6000 (No change)
𝐻1 :   6000 (A right-tailed test)
(b) The percentage of passing a Math course was 50%. A Math professor believes there
is a decrease in the passing rate.
𝐻0 : p  0.5 (No change)
𝐻1 : p  0.5 (A left-tailed test)
Objective B: Type I or Type II Error
Type I Error → Rejecting H 0 when H 0 is true.
Type II Error → Not rejecting H 0 when H1 is true.
We use  for the probability of making Type I error.
We use  for the probability of making Type II error.
For this statistics class, we only control the Type I error. (0.01    0.10)
 is also called the level of significance.
Objective C: State Conclusions to Hypothesis Tests
If H 0 is rejected, there is sufficient evidence to support the statement in H1 .
If H 0 is NOT rejected, there is NOT sufficient evidence to support the statement in H1 .
1
Example 1 :
In 2007, the mean SAT score on the reasoning test for all students was 710. A teacher
believes that, due to the heavy use of multiple choice test questions, the mean SAT
reasoning test has decreased.
(a) Determine H 0 and H1 .
H0: µ = 710
H1: µ< 710
(No change)
(A left-tailed test)
(b) Explain what it would mean to make a Type I error.
Type I Error  Rejecting the H0 when H0 is true.
A Type I error is made when the sample evidence leads the teacher to believe µ = 710
is not true when in fact µ = 710 is true.
(c) Explain what it would mean to make a Type II error.
Type II Error  Not rejecting the H0 when H1 is true.
A Type II error is made when the sample evidence leads the teacher to believe µ = 710
is true when in fact µ is less than 710.
(d) State the conclusion if the null hypothesis is rejected.
There is enough evidence to support the teacher's claim that the mean SAT reasoning test
score has decreased.
Example 2:
The mean score on the SAT Math Reasoning exam is 516. A test preparation company
states that the mean score of students who take its course is higher than 516.
(a) Determine the null and alternative hypotheses that would be used to test the effectiveness
of the marketing campaign.
H0: p = 516
H1: p > 516
(b) If sample data indicate that the null hypothesis should not be rejected, state the conclusion
of the company.
There is not sufficient evidence to support the mean score of the reasoning exam is higher
than 516.
2
(c) Suppose, in fact, that the mean score of students taking the preparatory course is 522.
Has a Type I or Type II error been made?
Type II error.
If we tested this hypothesis at α = 0.01 level, what is the probability of committing
a Type I error?
Since α is the probability of committing Type I error, there is a 1% chance of
committing a Type I error.
(d) If we wanted to decrease the probability of making a Type II error, wouldwe need
to increase or decrease the level of significance?
If we increase α, we decrease 𝛽 and vice versa.
Therefore, we need to increase the level of significance α in order to decrease the
probability of Type II error.
Example 3:
According to the Centers for Disease Control, 15.2% of adults experience migraine
headaches. Stress is a major contributor to the frequency and intensity of headaches.
A massage therapist feels that she has a technique that can reduce the frequency and
intensity of migraine headaches.
(a) Determine the null and alternative hypotheses that would be used to test the
effectiveness of the message therapist's techniques.
H0: p = 0.152
H1: p< 0.152
(b) A sample of 500 American adults who participated in the massage therapists program
results in data that indicate that the null hypothesis should be rejected. Provide a
statement that supports the massage therapists program.
There is sufficient evidence to support that the massage therapists’ program can reduce
the percentage of migraine headache from 15.2%.
(c) Suppose, in fact, that the percentage of patients in the program who experience
migraine headaches is 15.2%. Was a Type I or Type II error committed?
Type I error.
3
Ch10.2 Hypothesis Tests for a Population Proportion
Objective A: Classical Approach
Z  Test for a population proportion
A hypothesis test involving a population proportion can be considered as a binomial experiment.
x
The best point estimate of p , the population proportion, is a sample proportion, pˆ  .
n
There are two methods for testing hypothesis.
Method 1: The Classical Approach
Method 2: The P  Value Approach
We willbriefly introduce the classical approach.
The preferred method for this class is the P  value approach.
Testing Hypotheses Regarding a Population Proportion, p .
Use the following steps to perform a Proportion Z  Test provided that
4
Objective A : Classical Approach
Z  Test for a population proportion
Example 1: Use the classical approach to test the following hypotheses.
H o : p  0.25 versus H1 : p  0.25
n  400 ; x  96 ;   0.1
(a) Setup H 0 and H 1
H0: p = 0.25
H1: p <0.25 (left-tailed test)
(b) Use the sample data of n  400 and x  96 to compute the test statistic
and round your answer to four decimal places.
5
𝑥
96
𝑝̂ = 𝑛 = 400 = 0.24
z0 
z0 
pˆ  p0

p0 (1  p0 )
n
0.24  0.25
(0.25)(1  0.25)
400
0.24  0.25
(0.25)(0.75)≈ -0.4619*(test statistic)
400
(c) Use  =0.1 level of significance and StatCrunch to determine the
critical value(s).
Use α to find 𝑧𝛼 (z critical value)
Open StatCrunch  Stat  Calculator  Normal  Input P(x ≤ ____) = 0.1 (alpha) 
Compute and record results
𝑧𝛼 = -1.28.
(d) Draw the Z  distribution that depicts the critical region and the Z  statistic.
α = 0.1
𝑍𝛼 = -1.28 𝑧0 ≈ -0.4619
*
(e) What conclusion can be drawn?
Since Z – statistic, 𝑧0 , is not within the critical region, fail to reject 𝐻0 .
There is not enough evidence to support 𝐻1 : p < 0.25.
6
Ch10.3 Hypothesis Testing about a Population Mean with unknown 
Objective A: Classical Approach
t  test for a population mean
The t  test procedure requires either that the sample was drawn from a normally distributed
population or the sample size was greater than 30. Minor departures from normality will not
adversely affect the results of the test. However, if the data include outliers, the t  test procedure
should not be used. A normality plot is used to test whether the sample was drawn from a normally
distributed population. A boxplot is used to detect outliers.
7
Objective A : Classical Approach
t  test for a population mean
Example 1:
Use StatCrunch to determine the critical value(s) for
(a) a right-tailed test for a population mean at the   0.05 level of significance based
on a sample size of n  18 .
Open StatCrunch  Stat  Calculator  T  DF: 17, P (x ≥ ____) = 0.05 (alpha),
change the direction of the inequality to ≥ for the right-tailed test 
Compute and record results
𝑡𝛼 = 1.7396067
(b) a left-tailed test for a population mean at the   0.01 level of significance base on
a sample size of n  15 .
Open StatCrunch  Stat  Calculator  T  DF: 14, P (x ≥ ____) = 0.01 (alpha), check the
direction of the inequality to ≤ for the left-tailed test  Compute and record results
𝑡𝛼 = −2.6244941
8
(c) a two-tailed test for a population mean at   0.05 level of significance based on a
sample size of n  25 .
Open StatCrunch  Stat  Calculator  T Between, DF: 24, P (___  x  ____) = 0.95 
Compute and record results
𝑡𝛼 = ± 2.0639
Example 2:
Determine the t  statistic and the tail(s) of testing.
(a) H 0 :   16 versus H1 :   16
A random sample of size n  10 was obtained from a population that was normally
distributed produced a mean of 16.5 with a standard deviation of 0.4.
(i) Standardize the sample data
t0 
x  0 16.5 − 16
=
= 0.3952847 ≈ 3.953
0.4/√10
s/ n
𝑡0 = 3.953
(ii) Based on 𝐻1 : µ> 16, this is a right-tailed test.
(b) H 0 :   72 versus H1 :   72
A random sample of size n  15 was obtained from a population that was normally
distributed produced a mean of 73.5 with a standard deviation of 17.1.
(i) Standardize the sample data
t0 
x  0 73.5−72
=
= 0.33973538 ≈ 0.340
s / n 17.1/√15
𝑡0 = 0.340
9
(ii) Based on 𝐻1 : µ ≠ 72, this is a two-tailed test.
Example 3:
Use the classical approach to test H 0 :   0.3 versus H1 :   0.3 , a random sample of
size n  16 is obtained from a population that is known to be normally distributed.
(a) Setup H 0 and H 1
𝐻0 : µ = 0.3𝜇0 = 0.3,
𝐻1 : µ ≠ 0.3
t0 
n = 16,
x  0 0.295  0.3

  0.119
s n
0.168 16
𝑥̅ = 0.295,
s = 0.168
*
(b) If x =0.295, s  0.168 , compute the test statistic and round the answer to three decimal
places.
*
𝑡0 ≈ -0.119
(c) If the researcher decides to test this hypothesis at the  =0.05 level of significance,
Use StatCrunch to determine the critical value(s).
Open StatCrunch  Stat  Calculator  T Between, DF: 15, P (___  x  ____) = .95
Compute and record results
𝑡𝛼 = ± 2.1314
(d) Draw a t  distribution that depicts the critical region and the t  statistic.
10
t0  0.119
*
(e) What conclusion can be drawn?
Since the test statistic, 𝑡0 , does not fall within the critical region, we fail to reject 𝐻0 . There is
not enough evidence to support the 𝐻1 : µ ≠ 0.3.
P-Value Approach
Ch10.2 Hypothesis Tests for a Population Proportion
Objective B: P  Value Approach
Z  Test for a population proportion
A hypothesis test involving a population proportion can be considered as a binomial experiment.
x
The best point estimate of p , the population proportion, is a sample proportion, pˆ  , provided the sample is
n
obtained by 1) simple random sampling;
2) np0 (1  p0 )  10 to guarantee that a normal distribution can be used to test hypothesis
for H 0 : p  p0 ;
3) the sampled values are independent of each other (n  0.05 N ) .
Testing Hypotheses Regarding a Population Proportion, p .
Use the following steps to perform a Z  Test for a Proportion.
11
Example 1: The percentage of physicians who are women is 27.9%. In a survey of physicians
employed by a large university health system, 45 of 120 randomly selected physicians were women. Use the
P  value approach to determine whether there is sufficient evidence at the 0.05 level of significance to
conclude that the proportion of women physicians at the university health system exceeds 27.9%?
x = 45,
n = 120,
α = 0.05
(a) Setup
𝐻0 : p = 0.279
𝐻1 : p> 0.279 (right-tailed test)
(b) Standardize the sample data
𝑥
45
𝑝̂ = 𝑛 = 120= 0.375
Z=
𝑝̂−𝑝0
𝑝 (1−𝑝0 )
√ 0
𝜎
=
0.375−0.279
√
0.279(1−0.279)
120
≈ 2.345
(c) Use StatCrunch to find the P  value.
Open StatCrunch → Stat → Normal → Standard → P (x ≥ 2.345) = ____ → compute
12
P-value ≈ 0.0095
(d) Draw the Z  distribution and shade the area that represents  and the P  value
respectively.
0.0095
𝑧0 = 2.344
(e) What conclusion can be drawn?
Since P-value < α, we reject the null hypothesis. There is sufficient evidence to support that the
percentage of physicians who are women exceeds 27.9%.
Example 2:
Redo Example --> Use StatCrunch to calculate part (b) and part (c) in one command.
(a) Setup
𝐻0 : p = 0.279
𝐻0 : p > 0.279
(right-tailed test)
(b) Use StatCrunch to find Z 0 and its corresponding P  value.
Open StatCrunch → Stat → Proportion Stats → one sample → with summary → enter # of
successes 45, observations 120, value for p = 0.279, → select HA: p>→ compute
Hypothesis test results:
p : Proportion of successes
H0 : p = 0.279
HA : p > 0.279
Proportion Count Total Sample Prop. Std. Err.
Z-Stat P-value
p
45
120
0.375
0.040942948 2.3447261 0.0095
Z0 ≈ 2.345
p-value ≈ 0.0095
(c) Conclusion
13
Since P-value <α, we reject the null hypothesis. There is sufficient evidence to support that the
percentage of physicians who are women exceeds 27.9%.
Example 3:
In 2000, 58% of females aged 15 years of age and older lived alone, according to the
U.S. Census Bureau. A sociologist tests whether this percentage is different today by
conducting a random sample of 500 females aged 15 years of age and older and finds
that 285 are living alone.
Use the P  value approach and StatCrunch to determine whether there is sufficient evidence at
the   0.10 level of significance to conclude that the proportion has changed since 2000.
𝐻0 : p = 0.58,
n = 500,
𝐻1 : p ≠ 0.58
x = 285,
(two-tailed test)
285
𝑝̂ = 500 = 0.57
Open StatCrunch → Stat → Proportion Stats → One Sample → with summary → input # of
successes 285, observations 500, select Hypothesis test value for p = 0.58, → H1: p ≠ 0.58 →
compute and record results
Hypothesis test results:
p : Proportion of successes
H0 : p = 0.58
HA : p ≠ 0.58
Proportion Count Total Sample Prop. Std. Err.
Z-Stat
P-value
p
285 500
0.57
0.022072607 -0.45305024 0.6505
Z0 ≈ -0.453
p-value ≈ 0.6505
Z0 = -0.453
Conclusion:
Since P-value (0.6505) is not less than α (0.1), we fail to reject the null hypothesis. There is
insufficient evidence to support the claim that the percentage of females aged 15 years and lied
alone is different today.
14
P-Value Approach
Ch10.3 Hypothesis Testing about a Population Mean with unknown 
Objective B: P  Value Approach
t  test for a population mean
15
The t  test procedure requires either that the sample was drawn from a normally distributed
population or the sample size was greater than 30. Minor departures from normality will not
adversely affect the results of the test. However, if the data include outliers, the t  test
procedure should not be used. A normality plot is used to test whether the sample was drawn
from a normallydistributed population. A boxplot is used to detect outliers.
16
Example 1:
(a) Draw a t  distribution with the area that represents the P  value shaded.
(b) Use StatCrunch to find the P  value for each test value, t o .
(c) Use   0.05 to determine whether to reject H 0 or not.
(i) t0  2.624, n  15,
(Right-tailed test)
(a) Draw a t  distribution with the area that represents the P  value shaded.
(b) Use StatCrunch to find the P  value for each test value, t o .
Open StatCrunch → Stat → Calculators → T →DF: 14 →P (x ≥ 2.624) → Compute.
P-value = (0.0100963) ≈ 0.0101
(c) Use   0.05 to determine whether to reject H 0 ornot.
Since P-value (0.0100963) is less than α (0.05), we reject the null hypothesis.
(ii) t0  2.321, n  23,
left-tailed
(a) Draw a t  distribution with the area that represents the P  value shaded.
17
(b) Use StatCrunch to find the P  value for each test value, t o .
Open StatCrunch → Stat → Calculators → T →DF: 22 →P (x ≤ -2.321) → Compute.
P-value = (0.01497327) ≈ 0.0150
(c) Use   0.05 to determine whether to reject H 0 or not.
Since P-value (0.01497327) is less than α (0.05), we reject the null hypothesis.
(iii) t0  1.562, n  17,
two-tailed
(a) Draw a t  distribution with the area that represents the P  value shaded.
(b) Use StatCrunch to find the P  value for each test value, t o .
Open StatCrunch → Stat → Calculators → T →Between, DF: 16→
P (-1.562≤ x ≤ 1.562) → Compute.
18
The P-value is 1 – 0.86215185 = 0.137848
P-value ≈ 0.1378
(c) Use   0.05 to determine whether to reject H 0 ornot.
Since P-value (0.1378) is not less than α (0.05), we fail to reject the null hypothesis.
There is insufficient evidence to support the claim.
Example 3:
A survey of 15 large U.S. cities finds that the average commute time one way is
25.4minutes. A chamber of commerce executive feels that the commute in his city
is lessand wants to publicize this. He randomly selects 25 commuters and finds the
average is 22.1 minutes with a standard deviation of 5.3 minutes. At   0.01 , is
there enough evidence to support the claim? Use the P  value approach.
(a) Setup
𝐻0 : µ = 25.4
𝐻1 : µ<25.4 (left-tailed test)
(b) Use StatCrunch to find t o and its corresponding P  value.
Hypothesis test results:
μ : Mean of population
H0 : μ = 25.4
HA : μ < 25.4
Mean Sample Mean Std. Err. DF T-Stat P-value
μ
22.1
1.06 24 -3.1132075 0.0024
(c) Conclusion.: 𝑡0 = -3.113; P-value ≈ 0.0024. Since P-value (0.0024) < α, reject the 𝐻0 .
There is sufficient evidence to support the executive’s claim that the commute time in his city is
less than 25.4 minutes.
19
Example 4:
Michael Sullivan, son of the author, decided to enroll in a reading course that allegedly
increases reading speed and comprehension. Prior to enrolling in the class, Michael read
198 words per minute (wpm). The following data represent the words per minute read
for 10 different passages read after the course.
(a) Because the sample size is small, we must verify that reading speed is normally
distributed and the sample does not contain any outliers. Perform a normal
probability plot (QQ plot) and a boxplot. Are the conditions for testing a t  hypothesis
satisfied?
The boxplot indicated there are no outliers. The QQ plot indicated the sample data may have
come from a population that is normally distributed because all observed data are close to the
straight line. The t-hypothesis testing conditions are satisfied.
20
(b) Was the class effective? Use the   0.10 level of significance.
(i) Setup
𝐻0 : µ = 198
𝐻1 : µ> 198 (a right-tailed test)
(ii) Use StatCrunch to find t o and its corresponding P  value.
Open StatCrunch → Stats → T Stats → one sample → with data → select var1 → select
Hypothesis test forµ → 𝑯𝟎 :µ = 198, 𝐻𝐴 : µ> 198 → compute
Hypothesis test results:
μ : Mean of variable
H0 : μ = 198
HA : μ > 198
Variable Sample Mean Std. Err. DF T-Stat P-value
var1
208.4
2.9672284 9 3.5049543 0.0033
𝑡0 = 3.505
P-value ≈ 0.0033
(iii) Conclusion.
Since P-value (0.0033) < α (0.10), reject 𝐻0 . There is sufficient evidence to support the claim
that the reading course improves reading speed.
P-Value Approach
Ch10 Hypothesis Tests for a Population Standard Deviation (Supplemental Materials)
Objective B : P  Value Approach
 2  Test about a population variance or standard deviation
The concepts are similar to the P-value approach for Ch10.2 and Ch10.3 except aChi-Square distribution is
used. To test hypotheses about the population variance or standard deviation, two conditions must be met:
1)the sample is obtained using simple random sampling and 2) the population is normally distributed. Recall:
distribution is not symmetric and the values of are non-negative.
Example 1:
A machine fills bottles with 64 fluid ounces of liquid. The quality-control manager determines
that the fill levels are normally distributed with a mean of 64 ounces and a standard deviation
of 0.42 ounce. He has an engineer recalibrate the machine in an attempt to lower the standard
deviation. After the recalibration, the quality-control control manager randomly selects 19
bottles from the line and determines that the standard deviation is 0.38 ounce. Is there less
variability in the filling machine? Use the level of significance α (0.05).
(a) Setup
𝐻0 : 𝜎 = 0.42
𝐻1 : 𝜎< 0.42
(b) Use StatCrunch to find the P  value.
Open StatCrunch → Variance Stats → one sample → with summary → sample variance 0.382 =
0.1444 → n = 19 → select Hypothesis test for 𝐻0 : 𝜎 2 = 0.422 = 0.1764, 𝐻𝐴 : 𝜎 2 < 0.1764, →
compute
21
Hypothesis test results:
σ2 : Variance of population
H0 : σ2 = 0.1764
HA : σ2 < 0.1764
Variance Sample Var. DF Chi-Square Stat P-value
σ2
0.1444
18
14.734694
0.3199
(c) Draw the distribution and shade the area that represents and the value respectively.
(d) What conclusion can be drawn?
Since P-value (0.3199) is not less than α (0.05), fail to reject the 𝐻0 . There is insufficient
evidence to support the recalibration of the machine can lower the standard deviation.
22
Example 2:
Data obtained from the National Center for Health Statistics show that men between the ages of
20 to 29 have a mean height of 69.3 inches, with a standard deviation of 2.9 inches. A baseball
analyst wonders whether the standard deviation of heights of major-league baseball is less than
2.9 inches. The heights (in inches) of 20 randomly selected players are shown below.
µ = 69.3
s = 2.9
n = 20
72 74 71 72 76 70 77 75 72 72 77 72 75 70 73 73 75 73 74 74
(a) Verify the data are normally distributed by drawing a normal probability plot.
Open StatCrunch → Input raw data → Graph → QQ Plot → var1 → Compute
The sample data are close to the straight line. Thus, the data may have come from a population
that is normally distributed.
(b) Use StatCrunch to compute the sample standard deviation.
Open StatCrunch → Stat → Calculators → Summary Statistics → column → var1 → Statistics:
Variance → Compute
(c) Test the notion at the   0.10 level of significance.
µ = 69.3
𝜎 = 2.9
n = 20
𝑥̅ = 73.35
𝜎𝑥̅ = 2.0589982
Setup: 𝐻0 : 𝜎 = 2.9
𝐻1 : 𝜎< 2.9
23
Open StatCrunch → Variance Stats → one sample → with data → sample variance
2.0589982^2=4.2394736→ n = 20 → select Hypothesis test for 𝐻0 : 𝜎 2 = 2.92 = 8.41,
𝐻𝐴 : 𝜎 2 < 8.41 → compute
P-value: 0.0374
Conclusion:
Since P-value (0.0374) is less than α (0.05), reject the 𝐻0 . There is sufficient evidence to support
the claim that the standard deviation of heights of major-league baseball is less than 2.9 inches.
24