8.2 - Triola textbook 10.2 - Sullivan textbook

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. 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
1. The traditional method (Critical Value Method) (CV)
The critical value-Method, separates the critical region from the noncritical region.
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.

4. 4
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)
Example: a = 0.10, there is a 10% chance of rejecting a true null
hypothesis.
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)
Recall: 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…
The critical value, C.V., separates the critical region from the noncritical region.
The critical or rejection region is the range of values of the test value that
indicates that there is a significant difference and that the null hypothesis should
be rejected.
The noncritical or nonrejection region is the range of values of the test value that
indicates that the difference was probably due to chance and that the null
hypothesis should not be rejected.
Two-tailed test: The critical region is in the two extreme regions (tails) under the curve.
Left-tailed test: The critical region is in the extreme left region (tail) under the curve.
Right-tailed test: The critical region is in the extreme right region (tail) under the
curve.

5. 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
5
Test Value
P-Value

6. 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
6

7. Procedure for Hypothesis Tests
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: 𝑝 𝑞/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.
7
ˆ 

p p
z
pq n
Step 1 State the null and alternative
hypotheses and identify the
claim (H0 , H1).
Step 2 Test Statistic (TS): Compute
the test statistic value.
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…

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 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, ClaimSolution: BD, n = 1009, p = 0.5 → q = 0.5,
→ np ≥ 5 and nq ≥ 5 → Use ND, x = 545,
α = 0.05, 𝑝 =0.54
𝑝 =
𝑥
𝑛
=
545
1009
= 0.540
Step 2:
(545/1009) 0.5
:
0.5(0.5) 1009
TS z


ˆ 

p p
z
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.

9. The P-Value Method
9
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
ˆp E
2
ˆ ˆpq
E z
n
a
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

10. 10
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.”
Example 1: Traditional (CV) Method & The P-Value Method side by side
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
Step 3: CV: α = 0.05 →CV: z = 1.645
Step 1: H0: p = 0.5, H1: p > 0.5, RTT, Claim
Solution: BD, n = 1009, p = 0.5 →
q = 0.5, → np ≥ 5 and nq ≥ 5 → Use
ND, x = 545, α = 0.05, 𝑝 =0.54
Step 2: 𝑝 =
𝑥
𝑛
=
545
1009
= 0.540
(545/1009) 0.5
:
0.5(0.5) 1009
S zT


ˆ 

p p
z
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.

11. 11
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.”
Example 1: Traditional (CV) Method & The P-Value Method side by side
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
Step 3: CV: α = 0.05 →CV: z = 1.645
Step 1: H0: p = 0.5, H1: p > 0.5, RTT, ClaimSolution: BD, n = 1009, p = 0.5 → q = 0.5,
→ np ≥ 5 and nq ≥ 5 → Use ND, x = 545, α
= 0.05, 𝑝 =0.54
Step 2: 𝑝 =
𝑥
𝑛
=
545
1009
=
0.540
Step 2:
(545/1009) 0.5
:
0.5(0.5) 1009
TS z


ˆ 

p p
z
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.

12. 12
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
CV: α = 0.05 →CV: z = ±1.96
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
  
0.64 0.60
:
0.60 0.40 200
TS z


ˆ 

p p
z
pq n
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…

13. 13
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
CV: α = 0.05 →CV: z = ‒1.645
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
  
0.292 0.30
:
0.3 0.7 19136
TS z


ˆ 

p p
z
pq n
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 = 𝑛 𝑝 = 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

14. The P-Values Method
14
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
TI Calculator:
Confidence Interval:
proportion
1. Stat
2. Tests
3. 1-prop ZINT
4. Enter: x, n & CL