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Lecture Slides
Elementary Statistics
Twelfth Edition
and the Triola Statistics Series
by Mario F. Triola
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 8.4-1
Chapter 8
Hypothesis Testing
8-1 Review and Preview
8-2 Basics of Hypothesis Testing
8-3 Testing a Claim about a Proportion
8-4 Testing a Claim About a Mean
8-5 Testing a Claim About a Standard Deviation or
Variance
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 8.4-2
Key Concept
This section presents methods for testing a claim about a
population mean.
Part 1 deals with the very realistic and commonly used
case in which the population standard deviation σ is not
known.
Part 2 discusses the procedure when σ is known, which is
very rare.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 8.4-3
Part 1
When σ is not known, we use a “t test” that incorporates
the Student t distribution.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 8.4-4
Notation
n = sample size
x
= sample mean
 x = population mean
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 8.4-5
Requirements
1) The sample is a simple random sample.
2) Either or both of these conditions is satisfied:
The population is normally distributed or n > 30.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 8.4-6
Test Statistic
x  x
t
s
n
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 8.4-7
Running the Test
P-values: Use technology or use the Student t
distribution in Table A-3 with degrees of freedom
df = n – 1.
Critical values: Use the Student t distribution with
degrees of freedom df = n – 1.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 8.4-8
Important Properties of the
Student t Distribution
1. The Student t distribution is different for different sample
sizes (see Figure 7-5 in Section 7-3).
2. The Student t distribution has the same general bell shape
as the normal distribution; its wider shape reflects the
greater variability that is expected when s is used to
estimate σ.
3. The Student t distribution has a mean of t = 0.
4. The standard deviation of the Student t distribution varies
with the sample size and is greater than 1.
5. As the sample size n gets larger, the Student t distribution
gets closer to the standard normal distribution.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 8.4-9
Example
Listed below are the measured radiation emissions (in W/kg)
corresponding to a sample of cell phones.
Use a 0.05 level of significance to test the claim that cell
phones have a mean radiation level that is less than 1.00
W/kg.
0.38
0.55
1.54
1.55
0.50
The summary statistics are:
0.60
0.92
0.96
1.00
0.86
1.46
x  0.938 and s  0.423 .
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 8.4-10
Example - Continued
Requirement Check:
1. We assume the sample is a simple random sample.
2. The sample size is n = 11, which is not greater than 30, so
we must check a normal quantile plot for normality.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 8.4-11
Example - Continued
The points are reasonably close to a straight line and there is
no other pattern, so we conclude the data appear to be from a
normally distributed population.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 8.4-12
Example - Continued
Step 1: The claim that cell phones have a mean radiation
level less than 1.00 W/kg is expressed as μ < 1.00 W/kg.
Step 2: The alternative to the original claim is μ ≥ 1.00 W/kg.
Step 3: The hypotheses are written as:
H 0 :   1.00 W/kg
H1 :   1.00 W/kg
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 8.4-13
Example - Continued
Step 4: The stated level of significance is α = 0.05.
Step 5: Because the claim is about a population mean μ, the
statistic most relevant to this test is the sample mean: x .
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 8.4-14
Example - Continued
Step 6: Calculate the test statistic and then find the P-value or
the critical value from Table A-3:
x   x 0.938  1.00
t

 0.486
s
0.423
n
11
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 8.4-15
Example - Continued
Step 7: Critical Value Method: Because the test statistic of
t = –0.486 does not fall in the critical region bounded by the
critical value of t = –1.812, fail to reject the null hypothesis.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 8.4-16
Example - Continued
Step 7: P-value method: Technology, such as a TI-83/84 Plus
calculator can output the P-value of 0.3191. Since the P-value
exceeds α = 0.05, we fail to reject the null hypothesis.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 8.4-17
Example
Step 8: Because we fail to reject the null hypothesis, we
conclude that there is not sufficient evidence to support the
claim that cell phones have a mean radiation level that is less
than 1.00 W/kg.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 8.4-18
Finding P-Values
Assuming that neither software nor a TI-83 Plus calculator is
available, use Table A-3 to find a range of values for the Pvalue corresponding to the given results.
a) In a left-tailed hypothesis test, the sample size is n = 12,
and the test statistic is t = –2.007.
b) In a right-tailed hypothesis test, the sample size is n = 12,
and the test statistic is t = 1.222.
c) In a two-tailed hypothesis test, the sample size is n = 12,
and the test statistic is t = –3.456.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 8.4-19
Example – Confidence
Interval Method
We can use a confidence interval for testing a claim about μ.
For a two-tailed test with a 0.05 significance level, we
construct a 95% confidence interval.
For a one-tailed test with a 0.05 significance level, we
construct a 90% confidence interval.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 8.4-20
Example – Confidence
Interval Method
Using the cell phone example, construct a confidence interval
that can be used to test the claim that μ < 1.00 W/kg,
assuming a 0.05 significance level.
Note that a left-tailed hypothesis test with α = 0.05
corresponds to a 90% confidence interval.
Using methods described in Section 7.3, we find:
0.707 W/kg < μ < 1.169 W/kg
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 8.4-21
Example – Confidence
Interval Method
Because the value of μ = 1.00 W/kg is contained in the
interval, we cannot reject the null hypothesis that μ = 1.00
W/kg .
Based on the sample of 11 values, we do not have sufficient
evidence to support the claim that the mean radiation level is
less than 1.00 W/kg.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 8.4-22
Part 2
When σ is known, we use test that involves the standard
normal distribution.
In reality, it is very rare to test a claim about an unknown
population mean while the population standard deviation is
somehow known.
The procedure is essentially the same as a t test, with the
following exception:
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 8.4-23
Test Statistic for Testing a Claim
About a Mean (with σ Known)
The test statistic is:
z
x  x

n
The P-value can be provided by technology or the
standard normal distribution (Table A-2).
The critical values can be found using the standard normal
distribution (Table A-2).
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 8.4-24
Example
If we repeat the cell phone radiation example, with the
assumption that σ = 0.480 W/kg, the test statistic is:
z
x  x

n
0.938  1.00

 0.43
0.480
11
The example refers to a left-tailed test, so the P-value is the
area to the left of z = –0.43, which is 0.3336 (found in Table A2).
Since the P-value is large, we fail to reject the null and reach
the same conclusion as before.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 8.4-25