sst5e_ppt_ch10.pptx

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STATISTICS

INFORMED DECISIONS USING DATA

Fifth Edition, Global Edition

Chapter 10

Hypothesis Tests

Regarding a

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10.1 The Language of Hypothesis Testing

Learning Objectives

1. Determine the null and alternative hypotheses

2. Explain Type I and Type II errors

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10.1 The Language of Hypothesis Testing

10.1.1 Determine the Null and Alternative Hypotheses

(1 of 17)

A

hypothesis

is a statement regarding a characteristic of

one or more populations.

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10.1 The Language of Hypothesis Testing

10.1.1 Determine the Null and Alternative Hypotheses

(2 of 17)

Examples of Claims Regarding a Characteristic of a

Single Population

• In 2008, 62% of American adults regularly volunteered their time for charity work. A researcher believes that this percentage is different today.

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10.1 The Language of Hypothesis Testing

10.1.1 Determine the Null and Alternative Hypotheses

(3 of 17)

Examples of Claims Regarding a Characteristic of a

Single Population

• According to a study published in March, 2006 the mean length of a phone call on a cellular telephone was 3.25 minutes. A

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10.1 The Language of Hypothesis Testing

10.1.1 Determine the Null and Alternative Hypotheses

(4 of 17)

Examples of Claims Regarding a Characteristic of a

Single Population

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10.1 The Language of Hypothesis Testing

10.1.1 Determine the Null and Alternative Hypotheses

(5 of 17)

CAUTION!

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10.1 The Language of Hypothesis Testing

10.1.1 Determine the Null and Alternative Hypotheses

(6 of 17)

Hypothesis

testing

is a procedure, based on sample

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10.1 The Language of Hypothesis Testing

10.1.1 Determine the Null and Alternative Hypotheses

(7 of 17)

Steps in Hypothesis Testing

1. Make a statement regarding the nature of the population.

2. Collect evidence (sample data) to test the statement.

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10.1 The Language of Hypothesis Testing

10.1.1 Determine the Null and Alternative Hypotheses

(8 of 17)

The

null hypothesis

, denoted

H

₀

, is a statement to be

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10.1 The Language of Hypothesis Testing

10.1.1 Determine the Null and Alternative Hypotheses

(9 of 17)

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10.1 The Language of Hypothesis Testing

10.1.1 Determine the Null and Alternative Hypotheses

(10 of 17)

In this chapter, there are three ways to set up the null and

alternative hypotheses:

1. Equal versus not equal hypothesis (two-tailed test)

H₀: parameter = some value

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10.1 The Language of Hypothesis Testing

10.1.1 Determine the Null and Alternative Hypotheses

(11 of 17)

In this chapter, there are three ways to set up the null and

alternative hypotheses:

2. Equal versus less than (left-tailed test)

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10.1 The Language of Hypothesis Testing

10.1.1 Determine the Null and Alternative Hypotheses

(12 of 17)

In this chapter, there are three ways to set up the null and

alternative hypotheses:

3. Equal versus greater than (right-tailed test)

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10.1 The Language of Hypothesis Testing

10.1.1 Determine the Null and Alternative Hypotheses

(13 of 17)

“In Other Words”

The null hypothesis is a statement of status quo or no difference and always contains a statement of equality. The null hypothesis is assumed to be true until we have evidence to the contrary. We seek evidence that supports the statement in the alternative

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10.1 The Language of Hypothesis Testing

10.1.1 Determine the Null and Alternative Hypotheses

(14 of 17)

Parallel Example 2: Forming Hypotheses

For each of the following claims, determine the null and alternative hypotheses. State whether the test is two-tailed, left-tailed or right-tailed.

a) In 2008, 62% of American adults regularly volunteered their time for charity work. A researcher believes that this percentage is different today.

b) According to a study published in March, 2006 the mean length of a phone call on a cellular telephone was 3.25 minutes. A researcher believes that the mean length of a call has increased since then.

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10.1 The Language of Hypothesis Testing

10.1.1 Determine the Null and Alternative Hypotheses

(15 of 17)

Solution

a) In 2008, 62% of American adults regularly volunteered their time for charity work. A researcher believes that this

percentage is different today.

The hypothesis deals with a population proportion, p. If the percentage participating in charity work is no different than in 2008, it will be 0.62 so the null hypothesis is H₀: p = 0.62.

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10.1 The Language of Hypothesis Testing

10.1.1 Determine the Null and Alternative Hypotheses

(16 of 17)

Solution

b) According to a study published in March, 2006 the mean length of a phone call on a cellular telephone was 3.25

minutes. A researcher believes that the mean length of a call has increased since then.

The hypothesis deals with a population mean, μ. If the mean call length on a cellular phone is no different than in

2006, it will be 3.25 minutes so the null hypothesis is H₀: μ = 3.25.

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10.1 The Language of Hypothesis Testing

10.1.1 Determine the Null and Alternative Hypotheses

(17 of 17)

Solution

c) Using an old manufacturing process, the standard deviation of the amount of wine put in a bottle was 0.23 ounces. With new equipment, the quality control manager believes the standard deviation has decreased.

The hypothesis deals with a population standard deviation, σ. If the standard deviation with the new equipment has not

changed, it will be 0.23 ounces so the null hypothesis is H₀: σ = 0.23.

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10.1 The Language of Hypothesis Testing

10.1.2 Explain Type I and Type II Errors

(1 of 10)

Four Outcomes from Hypothesis Testing

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10.1 The Language of Hypothesis Testing

10.1.2 Explain Type I and Type II Errors

(2 of 10)

Four Outcomes from Hypothesis Testing

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10.1 The Language of Hypothesis Testing

10.1.2 Explain Type I and Type II Errors

(3 of 10)

Four Outcomes from Hypothesis Testing

3. Reject the null hypothesis when the null hypothesis is true. This decision would be incorrect. This type of error is called a

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10.1 The Language of Hypothesis Testing

10.1.2 Explain Type I and Type II Errors

(4 of 10)

Four Outcomes from Hypothesis Testing

4. Do not reject the null hypothesis when the alternative

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10.1 The Language of Hypothesis Testing

10.1.2 Explain Type I and Type II Errors

(5 of 10)

Parallel Example 3: Type I and Type II Errors

For each of the following claims, explain what it would mean to make a Type I error. What would it mean to make a Type II error?

a) In 2008, 62% of American adults regularly volunteered their time for charity work. A researcher believes that this

percentage is different today.

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10.1 The Language of Hypothesis Testing

10.1.2 Explain Type I and Type II Errors

(6 of 10)

Solution

a) In 2008, 62% of American adults regularly volunteered their time for charity work. A researcher believes that this percentage is different today.

A Type I error is made if the researcher concludes that p ≠ 0.62 when the true proportion of Americans 18 years or older who participated in some form of charity work is currently 62%. A Type II error is made if the sample evidence leads the

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10.1 The Language of Hypothesis Testing

10.1.2 Explain Type I and Type II Errors

(7 of 10)

Solution

minutes. A researcher believes that the mean length of a call has increased since then.

A Type I error occurs if the sample evidence leads the

researcher to conclude that μ > 3.25 when, in fact, the actual mean call length on a cellular phone is still 3.25 minutes.

A Type II error occurs if the researcher fails to reject the

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10.1 The Language of Hypothesis Testing

10.1.2 Explain Type I and Type II Errors

(8 of 10)

α

=

P

(Type I Error)

=

P

(rejecting

H

₀

when

H

₀

is true)

β

=

P

(Type II Error)

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10.1 The Language of Hypothesis Testing

10.1.2 Explain Type I and Type II Errors

(9 of 10)

The probability of making a Type I error,

α

, is chosen by the

researcher before

the sample data is collected.

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10.1 The Language of Hypothesis Testing

10.1.2 Explain Type I and Type II Errors

(10 of 10)

“

In Other Words

”

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10.1 The Language of Hypothesis Testing

10.1.3 State Conclusions to Hypothesis Tests

(1 of 4)

CAUTION!

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10.1 The Language of Hypothesis Testing

10.1.3 State Conclusions to Hypothesis Tests

(2 of 4)

Parallel Example 4: Stating the Conclusion

According to a study published in March, 2006 the mean length of a phone call on a cellular telephone was 3.25 minutes. A

researcher believes that the mean length of a call has increased since then.

a) Suppose the sample evidence indicates that the null hypothesis should be rejected. State the wording of the conclusion.

b) Suppose the sample evidence indicates that the null

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10.1 The Language of Hypothesis Testing

10.1.3 State Conclusions to Hypothesis Tests

(3 of 4)

Solution

a) Suppose the sample evidence indicates that the null hypothesis should be rejected. State the wording of the conclusion.

The statement in the alternative hypothesis is that the mean call length is greater than 3.25 minutes. Since the null

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10.1 The Language of Hypothesis Testing

10.1.3 State Conclusions to Hypothesis Tests

(4 of 4)

Solution

b) Suppose the sample evidence indicates that the null

hypothesis should not be rejected. State the wording of the conclusion.

Since the null hypothesis (μ = 3.25) is not rejected, there is insufficient evidence to conclude that the mean length of a phone call on a cell phone is greater than 3.25 minutes. In

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10.3 Hypothesis Tests for a Population Mean

Learning Objectives

1. Test hypotheses about a mean

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10.3 Hypothesis Tests for a Population Mean

10.3.1 Test Hypotheses about a Mean

(1 of 39)

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10.3 Hypothesis Tests for a Population Mean

10.3.1 Test Hypotheses about a Mean

(2 of 39)

Properties of the

t-

Distribution

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10.3 Hypothesis Tests for a Population Mean

10.3.1 Test Hypotheses about a Mean

(3 of 39)

Properties of the

t-

Distribution

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10.3 Hypothesis Tests for a Population Mean

10.3.1 Test Hypotheses about a Mean

(4 of 39)

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10.3 Hypothesis Tests for a Population Mean

10.3.1 Test Hypotheses about a Mean

(5 of 39)

Properties of the

t-

Distribution

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10.3 Hypothesis Tests for a Population Mean

10.3.1 Test Hypotheses about a Mean

(6 of 39)

Properties of the

t-

Distribution

5. The area in the tails of the t-distribution is a little greater than the area in the tails of the standard normal distribution

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10.3 Hypothesis Tests for a Population Mean

10.3.1 Test Hypotheses about a Mean

(7 of 39)

Properties of the

t-

Distribution

6. As the sample size n increases, the density curve of t gets

closer to the standard normal density curve. This result occurs because as the sample size increases, the values of s get

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10.3 Hypothesis Tests for a Population Mean

10.3.1 Test Hypotheses about a Mean

(8 of 39)

Testing Hypotheses Regarding a Population Mean

To test hypotheses regarding the population mean, we use the following steps, provided that:

• The sample is obtained using simple random sampling.

• The sample has no outliers, and the population from which the sample is drawn is normally distributed or the sample size is large (n ≥ 30).

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10.3 Hypothesis Tests for a Population Mean

10.3.1 Test Hypotheses about a Mean

(9 of 39)

Step 1: Determine the null and alternative hypotheses. Again, the hypotheses can be structured in one of three ways:

Two-Tailed Left-Tailed Right-Tailed

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10.3 Hypothesis Tests for a Population Mean

10.3.1 Test Hypotheses about a Mean

(10 of 39)

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10.3 Hypothesis Tests for a Population Mean

10.3.1 Test Hypotheses about a Mean

(11 of 39)

Classical Approach

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10.3 Hypothesis Tests for a Population Mean

10.3.1 Test Hypotheses about a Mean

(12 of 39)

Classical Approach

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10.3 Hypothesis Tests for a Population Mean

10.3.1 Test Hypotheses about a Mean

(13 of 39)

Classical Approach

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10.3 Hypothesis Tests for a Population Mean

10.3.1 Test Hypotheses about a Mean

(14 of 39)

Classical Approach

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10.3 Hypothesis Tests for a Population Mean

10.3.1 Test Hypotheses about a Mean

(15 of 39)

Classical Approach

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10.3 Hypothesis Tests for a Population Mean

10.3.1 Test Hypotheses about a Mean

(16 of 39)

Classical Approach

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10.3 Hypothesis Tests for a Population Mean

10.3.1 Test Hypotheses about a Mean

(17 of 39)

P

-Value Approach

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10.3 Hypothesis Tests for a Population Mean

10.3.1 Test Hypotheses about a Mean

(18 of 39)

P

-Value Approach

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10.3 Hypothesis Tests for a Population Mean

10.3.1 Test Hypotheses about a Mean

(19 of 39)

P

-Value Approach

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10.3 Hypothesis Tests for a Population Mean

10.3.1 Test Hypotheses about a Mean

(20 of 39)

P

-Value Approach

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10.3 Hypothesis Tests for a Population Mean

10.3.1 Test Hypotheses about a Mean

(21 of 39)

P

-Value Approach

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10.3 Hypothesis Tests for a Population Mean

10.3.1 Test Hypotheses about a Mean

(22 of 39)

P

-Value Approach

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10.3 Hypothesis Tests for a Population Mean

10.3.1 Test Hypotheses about a Mean

(23 of 39)

P

-Value Approach

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10.3 Hypothesis Tests for a Population Mean

10.3.1 Test Hypotheses about a Mean

(24 of 39)

P

-Value Approach

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10.3 Hypothesis Tests for a Population Mean

10.3.1 Test Hypotheses about a Mean

(25 of 39)

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10.3 Hypothesis Tests for a Population Mean

10.3.1 Test Hypotheses about a Mean

(26 of 39)

Parallel Example 1: Testing a Hypothesis about a

Population Mean, Large Sample

Assume the resting metabolic rate (RMR) of healthy males in

complete silence is 5710 kJ/day. Researchers measured the RMR of 45 healthy males who were listening to calm classical music

and found their mean RMR to be 5708.07 with a standard deviation of 992.05.

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10.3 Hypothesis Tests for a Population Mean

10.3.1 Test Hypotheses about a Mean

(27 of 39)

Solution

We assume that the RMR of healthy males is 5710 kJ/day. This is a two-tailed test since we are interested in determining whether the RMR differs from 5710 kJ/day.

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10.3 Hypothesis Tests for a Population Mean

10.3.1 Test Hypotheses about a Mean

(28 of 39)

Solution

Step 1: H₀: μ = 5710 versus H₁: μ ≠ 5710

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10.3 Hypothesis Tests for a Population Mean

10.3.1 Test Hypotheses about a Mean

(29 of 39)

Solution: Classical Approach

Since this is a two-tailed test, we determine the critical values at the α = 0.05 level of significance with n

− 1

= 45

−

1 = 44 degrees of freedom to be approximately

−

t_0.025 =

−

2.021 and t_0.025 = 2.021.

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10.3 Hypothesis Tests for a Population Mean

10.3.1 Test Hypotheses about a Mean

(30 of 39)

Solution:

P

-value Approach

Step 3: Since this is a two-tailed test, the P-value is the area under the t-distribution with n

−

1 = 44 degrees of

freedom to the left of

−

t_0.025=

−

0.013 and to the right of

t_0.025 = 0.013.

That is, P-value = P(t <

−

0.013) + P(t > 0.013)

= 2 P(t > 0.013). 0.50 < P-value.

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10.3 Hypothesis Tests for a Population Mean

10.3.1 Test Hypotheses about a Mean

(31 of 39)

Solution

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10.3 Hypothesis Tests for a Population Mean

10.3.1 Test Hypotheses about a Mean

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Parallel Example 3: Testing a Hypothesis about a

Population Mean, Small Sample

According to the United States Mint, quarters weigh 5.67 grams. A researcher is interested in determining whether the “state”

quarters have a weight that is different from 5.67 grams. He

randomly selects 18 “state” quarters, weighs them and obtains the following data.

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10.3 Hypothesis Tests for a Population Mean

10.3.1 Test Hypotheses about a Mean

(33 of 39)

Solution

We assume that the weight of the state quarters is 5.67 grams. This is a two-tailed test since we are interested in determining whether the weight differs from 5.67 grams.

Since the sample size is small, we must verify that the data come from a population that is normally distributed with no outliers

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10.3 Hypothesis Tests for a Population Mean

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10.3 Hypothesis Tests for a Population Mean

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10.3 Hypothesis Tests for a Population Mean

10.3.1 Test Hypotheses about a Mean

(36 of 39)

Solution

Step 1: H₀: μ = 5.67 versus H₁: μ ≠ 5.67

Step 2: The level of significance is α = 0.05.

Step 3: From the data, the sample mean is calculated to be

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10.3 Hypothesis Tests for a Population Mean

10.3.1 Test Hypotheses about a Mean

(37 of 39)

Solution: Classical Approach

Step 3: Since this is a two-tailed test, we determine the critical values at the α = 0.05 level of significance with n

−

1 = 18

−

1 = 17 degrees of freedom to be

−

t_0.025 =

−

2.11 and

t_0.025 = 2.11.

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10.3 Hypothesis Tests for a Population Mean

10.3.1 Test Hypotheses about a Mean

(38 of 39)

Solution:

P

-Value Approach

Step 3: Since this is a two-tailed test, the P-value is the area under the t-distribution with n

−

1 = 17 degrees of

freedom to the left of

−

t_0.025 =

−

2.75 and to the right of t_0.025

= 2.75. That is,

P-value = P(t <

−

2.75) + P(t > 2.75)

= 2 P(t > 2.75). 0.01 < P-value < 0.02.

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10.3 Hypothesis Tests for a Population Mean

10.3.1 Test Hypotheses about a Mean

(39 of 39)

Solution

Step 5: There is sufficient evidence at the α = 0.05 level of

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10.3 Hypothesis Tests for a Population Mean

10.3.2 Understand the difference between Statistical

Significance and Practical Significance

(1 of 7)

When a large sample size is used in a hypothesis test, the

results could be statistically significant even though the

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10.3 Hypothesis Tests for a Population Mean

10.3.2 Understand the difference between Statistical

Significance and Practical Significance

(2 of 7)

Practical significance

refers to the idea that, while small

differences between the statistic and parameter stated in the

null hypothesis are statistically significant, the difference may

not be large enough to cause concern or be considered

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10.3 Hypothesis Tests for a Population Mean

10.3.2 Understand the difference between Statistical

Significance and Practical Significance

(3 of 7)

Parallel Example 7: Statistical versus Practical Significance

In 2003, the average age of a mother at the time of her first childbirth was 25.2. To determine if the average age has

increased, a random sample of 1200 mothers is taken and is found to have a sample mean age of 25.5 with a standard

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10.3 Hypothesis Tests for a Population Mean

10.3.2 Understand the difference between Statistical

Significance and Practical Significance

(4 of 7)

Solution

Step 1: To determine whether the mean age has increased, this is a right-tailed test with

H₀: μ = 25.2 versus H₁: μ > 25.2.

Step 2: The level of significance is α = 0.05.

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10.3 Hypothesis Tests for a Population Mean

10.3.2 Understand the difference between Statistical

Significance and Practical Significance

(5 of 7)

Solution

Step 3: Since this is a right-tailed test, the critical value with

α = 0.05 and 1200

−

1 = 1199 degrees of freedom is

P-value = P(t > 2.17).

From Table VI: .01 < P-value < .025

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10.3 Hypothesis Tests for a Population Mean

10.3.2 Understand the difference between Statistical

Significance and Practical Significance

(6 of 7)

Solution

Step 5: There is sufficient evidence at the 0.05 significance level to conclude that the mean age of a mother at the time of her first childbirth is greater than 25.2.