8.4 - Triola textbook 10.4- Sullivan textbook

1. Elementary Statistics
Chapter 8:
Hypothesis Testing
8.4 Testing a Claim
about a Stndard
Deviation or Variance
1

2. 8.1 Basics of Hypothesis Testing
8.2 Testing a Claim about a Proportion
8.3 Testing a Claim About a Mean
8.4 Testing a Claim About a Standard Deviation or Variance
2
Objectives:
• Understand the definitions used in hypothesis testing.
• State the null and alternative hypotheses.
• State the steps used in hypothesis testing.
• Test proportions, using the z test.
• Test means when  is known, using the z test.
• Test means when  is unknown, using the t test.
• Test variances or standard deviations, using the chi-square test.
• Test hypotheses, using confidence intervals.
Chapter 8: Hypothesis Testing

3. 1. The traditional method (Critical Value Method) (CV)
2. The P-value method
P-Value Method: In a hypothesis test, the P-value is the probability of getting a value of the
test statistic that is at least as extreme as the test statistic obtained from the sample data,
assuming that the null hypothesis is true.
3. The confidence interval (CI)method
Because a confidence interval estimate of a population parameter contains the likely values of that parameter,
reject a claim that the population parameter has a value that is not included in the confidence interval.
Equivalent Methods: A confidence interval estimate of a proportion might lead to a conclusion different
from that of a hypothesis test.
Recall: 8.1 Basics of Hypothesis Testing: 3 methods used to test hypotheses:
3
Construct a confidence interval with
a confidence level selected:
Significance Level for
Hypothesis Test: α
Two-Tailed Test:
1 – α
One-Tailed
Test: 1 – 2α
0.01 99% 98%
0.05 95% 90%
0.10 90% 80%
A statistical hypothesis is a assumption about a population parameter. This conjecture may or may not be
true. The null hypothesis, symbolized by H0, and the alternative hypothesis, symbolized by H1

4. 4
Type I error: The mistake of rejecting the null hypothesis when it is
actually true. The symbol α (alpha) is used to represent the
probability of a type I error. (A type I error occurs if one rejects the
null hypothesis when it is true.)
The level of significance is the maximum probability of committing
a type I error: α = P(type I error) = P(rejecting H0 when H0 is
true) and Typical significance levels are: 0.10, 0.05, and 0.01
For example, when a = 0.10, there is a 10% chance of rejecting a
true null hypothesis.
Type II error: The mistake of failing to reject the null hypothesis
when it is actually false. The symbol β(beta) is used to represent the
probability of a type II error. (A type II error occurs if one does not
reject the null hypothesis when it is false.) β = P(type II error) =
P(failing to reject H0 when H0 is false)
Procedure for Hypothesis Tests
Step 1 State the null and alternative
hypotheses and identify the claim (H0 , H1).
Step 2 Test Statistic (TS): Compute
the test statistic value that is relevant to
the test and determine its sampling
distribution (such as normal, t, χ²).
Step 3 Critical Value (CV) :
Find the critical value(s) from the appropriate
table.
Step 4 Make the decision to
a. Reject or not reject the null
hypothesis.
b. The claim is true or false
c. Restate this decision: There is / is
not sufficient evidence to support
the claim that…

5. Objective: Conduct a formal hypothesis test of a claim about a population
proportion p.
Recall: 8.2 Testing a Claim about a Proportion
Notation
n = sample size or number of
trials
p = population proportion (used
in the the null hypothesis)
𝑝 =
𝑥
𝑛
= Sample proportion
Requirements
1. The sample observations are a simple random sample.
2. The conditions for a binomial distribution are
satisfied:
• There is a fixed number of trials.
• The trials are independent.
• Each trial has two categories of “success” and “failure.”
• The probability of a success remains the same in all
trials.
3. The conditions np ≥ 5 and nq ≥ 5 are both satisfied, so
the binomial distribution of sample proportions can be
approximated by a normal distribution with
𝜇 = 𝑛𝑝, 𝜎 = 𝑛𝑝𝑞
5
ˆ 

p p
z
pq n
TI Calculator:
1 - Proportion Z - test
1. Stat
2. Tests
3. 1 ‒ PropZTest
4. Enter Data or
Stats (p, x, n)
5. Choose RTT, LTT,
or 2TT
TI Calculator:
Confidence Interval:
proportion
1. Stat
2. Tests
3. 1-prop ZINT
4. Enter: x, n & CL

6. Key Concept: Testing a claim about a population mean
Objective: Use a formal hypothesis test to test a claim about a
population mean µ.
1. The population standard deviation σ is not known.
2. The population standard deviation σ is known.
Recall: 8.3 Testing a Claim About a Mean
The z test is a statistical test for the
mean of a population. It can be used
when n  30, or when the population is
normally distributed and  is known.
The formula for the z test is (Test
Statistic): 𝑍 =
𝑥−𝜇
𝜎/ 𝑛
where
𝑥 = sample mean
μ = hypothesized population mean
 = population standard deviation
n = sample size
The t test is a statistical test for the
mean of a population. It can be
used when n  30, or when the
population is normally distributed
and  is not known.
The formula for the t test is
(Test Statistic): 𝑡 =
𝑥−𝜇
𝑠/ 𝑛
where
𝑥 = sample mean
μ = hypothesized population mean
 = population standard deviation
n = sample size 6
TI Calculator:
Mean: T ‒ Test
1. Stat
2. Tests
3. T ‒ Test
4. Enter Data or Stats (p, x, n)
5. Choose RTT, LTT, or 2TT
6. Calculate
TI Calculator:
Mean: Z ‒ Test
1. Stat
2. Tests
3. Z ‒ Test
4. Enter Data or Stats (p, x, n)
5. Choose RTT, LTT, or 2TT
6. Calculate

7. Key Concept:
Conduct a formal hypothesis test of a claim made about a population standard
deviation σ or population variance σ².
The chi-square distribution is also used to test a claim about a single variance or standard deviation.
8.4 Testing a Claim About a Standard Deviation or Variance
Notation
n = sample size
d.f. = n – 1
s = sample standard deviation
σ = population standard deviation
s² = sample variance
σ² = population variance
When testing claims about σ or σ²,
the P-value method, the critical
value method, and the confidence
interval method are all equivalent
in the sense that they will always
lead to the same conclusion.
Requirements
1. The sample is a simple
random sample.
2. The population has a normal
distribution. (This is a fairly
strict requirement.)
Test Statistic
7
2
2
2
( 1)n s





8. Properties of the Chi-Square Distribution
1. All values of χ² are nonnegative, and the distribution is not symmetric.
2. There is a different χ² distribution for each number of degrees of freedom.
3. The critical values are found in Chi-Square Table using degrees of freedom = n – 1
An important note if using Chi-Square Table for finding critical values:
In Chi-Square Table, each critical value of χ² in the body of the table corresponds to an area given in the top
row of the table, and each area in that top row is a cumulative area to the right of the critical value.
8.4 Testing a Claim About a Standard Deviation or Variance
8
Assumptions for the χ² Test for a Variance or a Standard Deviation
1. The sample must be randomly selected from the population.
2. The population must be normally distributed for the variable under study.
3. The observations must be independent of one another.

9. 9
a. Find the critical chi-square value for 15 degrees of freedom when α = 0.05 and the test is right-
tailed.
b. Find the critical chi-square value for 10 degrees of freedom when α = 0.05 and the test is left-tailed.
Example 1
2
24.996 2
3.940
left-tailed: 1 – α = 1
– 0.05 = 0.95.
The chi-square table
gives the area to the
right of the CV.

10. 10
Find the critical chi-square value for 22 degrees of freedom when α = 0.05 and a two-tailed test is
conducted.
Example 2
df = 22
Areas 0.025 and 0.975 correspond
to chi-square values of :
𝝌 𝑳
𝟐
= 𝟏𝟎. 𝟗𝟖𝟐, 𝝌 𝑹
𝟐
= 𝟑𝟔. 𝟕𝟖𝟏

11. 11
An instructor wishes to see whether the variation in scores of the 23 students in her class is less than the
variance of the population. The variance of the class is 198. Is there enough evidence to support the
claim that the variation of the students is less than the population variance (2 =225) at α = 0.05?
Assume that the scores are normally distributed.
Example 3
CV: α = 0.05 & df = n − 1
= 22 →CV: χ² = 12.338
H0: 2 = 225 & H1: 2 < 225, claim, LTTSolution: ND,
n = 23, s2 = 198,
2 =225 , α = 0.05
Decision:
a. Do not Reject (Fail to reject) H0
b. The claim is False
c. There is not enough evidence to support the claim that the
variation in test scores of the instructor’s students is less than
the variation in scores of the population.
19.36
2
2
2
( 1)n s




2
2
2
( 1)
:
n s
TS 



(23 1)198
225


Step 1: H0 , H1, claim & Tails
Step 2: TS Calculate (TS)
Step 3: CV using α
Step 4: Make the decision to
a. Reject or not H0
b. The claim is true or false
c. Restate this decision: There is /
is not sufficient evidence to
support the claim that…

12. 12
A researcher wishes to test the claim that the variance of the nicotine content
of a cigarette manufacturer is 0.644. Nicotine content is measured in milligrams, and
assume that it is normally distributed. A sample of 20 cigarettes has a standard
deviation of 1.00 milligram. At α = 0.05, test the claim that that the variance of the
nicotine content of its cigarettes is 0.644.
Example 4
CV: α = 0.05 & df = n − 1 = 19 →
CV: The critical values are 32.852 and 8.907
H0: 2 = 0.644 (claim) and H1: 2  0.644, 2TT
Solution: ND, n = 20,
2 = 0.644, s = 1.00,
α = 0.05
Decision:
a. Do not reject H0
b. The claim is True
c. There is sufficient evidence to support the claim
that the variance of the nicotine content of its
cigarettes is 0.644.
2
19(1)
0.644
 29.5
2
2
2
( 1)n s




2
2
2
( 1)
:
n s
TS 



Step 1: H0 , H1, claim & Tails
Step 2: TS Calculate (TS)
Step 3: CV using α
Step 4: Make the decision to
a. Reject or not H0
b. The claim is true or false
c. Restate this decision: There is /
is not sufficient evidence to
support the claim that…