Chapter 8 covers hypothesis testing, detailing methods for testing claims about population proportions, means, and variances. It introduces the critical value method, p-value method, and confidence intervals as tools for hypothesis testing, along with specific examples on how to apply these methods for various claims. The chapter emphasizes the necessity of using the correct statistical approach based on the given data and hypotheses.

1. Elementary Statistics
Chapter 8: Hypothesis
Testing
8.2 Testing a Claim about
a Proportion
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. Key Concept: A complete procedure for testing a claim made about a population proportion p.
1. The critical value (Traditional) method: In this section, the traditional method for solving
hypothesis-testing problems compares z-values:
 critical value
 test value
2. P-value: The P-value (or probability value) is the probability of getting a sample statistic
(such as the mean) or a more extreme sample statistic in the direction of the alternative
hypothesis when the null hypothesis is true.) The P-value method for solving hypothesis-
testing problems compares areas:
 alpha
 P-value
3. The Confidence intervals 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.
8.2 Testing a Claim about a Proportion
3
Test Value
P-Value

4. Objective:
Conduct a formal hypothesis test of a claim about a population proportion p.
8.2 Testing a Claim about a Proportion
Notation
n = sample size or number of
trials
p = population proportion (used
in 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
4

5. Critical Value (CV) : Procedure for Hypothesis Tests P-value
5
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…
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 P-value Method:
Find the P-value 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…
Both the P-value method and the critical value method use the same standard deviation based on the claimed
proportion p: 𝑝𝑞/𝑛, so they are equivalent to each other.The confidence interval method uses an estimated
standard deviation based on the sample proportion: 𝑝𝑞/𝑛. Therefore, it is not equivalent to the P-value and
critical value methods, so the confidence interval method could result in a different
conclusion.Recommendation: Use a confidence interval to estimate a population proportion, but use the P-
value method or critical value method for testing a claim about a proportion.
𝑧 =
𝑝 − 𝑝
𝑝𝑞
𝑛

6. 6
1009 consumers were asked if they are comfortable with having drones deliver their
purchases, and 54% (or 545) of them responded with “no.” Use these results to test the
claim that most consumers are uncomfortable with drone deliveries. We interpret
“most” to mean “more than half” or “greater than 0.5.” (α = 0.05)
Example 1
Step 1:
State H0 , H1, Identify the claim &
Tails
Step 2: TS
Calculate the test statistic (TS) that is
relevant to the test
Step 3: CV
Find the critical value /s 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…
Step 3: CV: α = 0.05 →CV: z = 1.645
Step 1: H0: p = 0.5, H1: p > 0.5, RTT, Claim
Given: BD, n = 1009, p = 0.5
→ q = 0.5, → np ≥ 5 and nq ≥ 5
→ Use ND, x = 545
𝑝 =
𝑥
𝑛
=
545
1009
= 0.540
Step 2: 𝑇𝑆: 𝑧 =
(545/1009)−0.5
0.5(0.5)
1009
𝑧 =
𝑝 − 𝑝
𝑝𝑞
𝑛
= 2.55
Step 4: Decision:
a. Reject H0
b. The claim is true
c. There is sufficient evidence to support the claim that the
majority of consumers are uncomfortable with drone deliveries.

7. The P-Value Method
7
Example 1 continued:
RTT: z = 2.55.
The P-value is the area to the right of
z = 2.55: P-value = 0.0054
Decision Criteria for the P-Value
Method:
P-value = 0.0054 ≤ α = 0.05
⇾ Same decision:
0.514 < p < 0.566.
The entire range of values in this CI > 0.5
We are 90% confident that the limits of 0.514 and 0.566 contain the
true value of p, the sample data appear to support the claim that most
(more than 0.5) consumers are uncomfortable with drone deliveries.
Confidence Interval Method: 90% CI
𝑝 ± 𝐸
𝑝 ± 𝐸, ,𝐸 = 𝑧𝛼
2
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
6. Calculate
TI Calculator:
Confidence Interval:
proportion
1. Stat
2. Tests
3. 1-prop ZINT
4. Enter: x, n & CL

8. 8
1009 consumers were asked if they are comfortable with having drones deliver their
purchases, and 54% (or 545) of them responded with “no.” Use these results to test the
claim that most consumers are uncomfortable with drone deliveries. We interpret
“most” to mean “more than half” or “greater than 0.5.” (α = 0.05)
Example 1
Step 3: CV: α = 0.05 →CV: z = 1.645
Step 1: H0: p = 0.5, H1: p > 0.5, RTT, Claim
Given: BD, n = 1009, p = 0.5
→ q = 0.5, → np ≥ 5 and nq ≥
5 → Use ND, x = 545
𝑝 =
𝑥
𝑛
=
545
1009
= 0.540
Step 2: 𝑇𝑆: 𝑧 =
(545/1009)−0.5
0.5(0.5)
1009
𝑧 =
𝑝 − 𝑝
𝑝𝑞
𝑛
= 2.55
Step 4: Decision:
a. Reject H0
b. The claim is true
c. There is sufficient evidence to support the
claim that the majority of consumers are
uncomfortable with drone deliveries.
The P-Value Method
Step 1:
H0: p = 0.5, H1: p > 0.5, RTT, Claim
Step 2:
TS: z = 2.55
Step 3: P-Value
P-value = Area to the right of TS = 0.0054
Step 4: Make the decision to
The same
Traditional & The P-Value Methods Side by Side

9. 9
There is a claim that 60% of people are trying to avoid trans fats in their diets. A
researcher randomly selected 200 people and found that 128 people stated that they
were trying to avoid trans fats in their diets. At α = 0.05, is there enough evidence to
reject this claim?
Example 2
Step 3: CV: α = 0.05 →CV: z = ±1.96
Step 1: H0: p = 0.60 (claim), H1: p  0.60 2TT
Given: BD, n = 200, p = 0.6 →
q = 0.4, α = 0.05, x = 128, →
np ≥ 5 and nq ≥ 5 → Use ND
𝑝 =
𝑥
𝑛
=
128
200
= 0.64
Step 2: 𝑇𝑆: 𝑧 =
0.64−0.60
0.60 0.40
200
Step 4: Decision:
a. Fail to Reject H0
b. The claim is true
c. There is sufficient evidence to support the claim that 60% of
people are trying to avoid trans fats in their diets.
= 1.15
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…
𝑧 =
𝑝 − 𝑝
𝑝𝑞
𝑛

10. 10
A study of sleepwalking or “nocturnal wandering” was described in Neurology magazine,
and it included information that 29.2% of 19,136 American adults have sleepwalked. What is
the actual number of adults who have sleepwalked? Let’s use a 0.05 significance level to test
the claim that for the adult population, the proportion of those who have sleepwalked is less
than 0.30.
Example 3
Step 3: CV: α = 0.05 →CV: z = ‒1.645
Step 1: H0: p = 0.30, H1: p < 0.30 (claim), LTT
Given: BD, n = 19,136, p = 0.3
→ q = 0.7 𝑝 = 0.292, α = 0.05,
np ≥ 5 and nq ≥ 5 → Use ND
Step 2: 𝑇𝑆: 𝑧 =
0.292−0.30
0.3 0.7
19136
Step 4: Decision:
a. Reject H0
b. The claim is true
c. There is sufficient evidence to support the claim that fewer than
30% of adults have sleepwalked.
= −2.41
𝑝 =
𝑥
𝑛
→ 𝑥 = 𝑛𝑝 = 19136(0.292) = 5587.7 → 5588
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…
−2.41 − 1.645
𝑧 =
𝑝 − 𝑝
𝑝𝑞
𝑛

11. The P-Values Method
11
Example 3 continued:
LTT: z = −2.41
The P-value = The area to the left of the test
statistic = 0.0080
Decision Criteria for the P-Value
Method:
P-value = 0.0080 ≤ α = 0.05
⇾ Same decision:
0.2866 < p < 0.2974
The entire range of values in this CI < 0.3
We are 90% confident that the limits of 0.2866 and 0.2974 contain the
true value of p, the sample data appear to support the claim that fewer
than 30% of adults have sleepwalked.
Confidence Interval Method: 90% CI
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
6. Calculate
TI Calculator:
Confidence Interval:
proportion
1. Stat
2. Tests
3. 1-prop ZINT
4. Enter: x, n & CL

12. 12
The Family and Medical Leave Act provides job protection and unpaid time off from work for a serious
illness or birth of a child. In 2000, 60% of the respondents of a survey stated that it was very easy to get
time off for these circumstances. A researcher wishes to see if the percentage who said that it was very
easy to get time off has changed. A sample of 100 people who used the leave said that 53% found it easy
to use the leave. At α = 0.01, has the percentage changed?
Example 4 (Time)
Step 3: CV: α = 0.01 →CV: z = ±2.58
Step 1: H0: p = 0.60, H1: p  0.60 2TT, (claim)
Given: BD, n = 100, p = 0.6
→ q = 0.4 𝑝 = 0.53, np ≥ 5
and nq ≥ 5 → Use ND 𝑝 =
𝑥
𝑛
=
128
200
= 0.64
Step 2: 𝑇𝑆: 𝑧 =
0.53−0.60
0.60 0.40
100
Step 4: Decision:
a. Don’t Reject H0
b. The claim is False
c. There is not enough evidence to support the claim that
the percentage of those using the medical leave said
that it was easy to get time off has changed.
= −1.43
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…
𝑧 =
𝑝 − 𝑝
𝑝𝑞
𝑛