Section 7.2 Hypothesis Testing for the Mean (Large Samples) Larson/Farber 4th ed

Section 7.2

Hypothesis Testing for the Mean (Large Samples)

Larson/Farber 4th ed.

Section 7.2 Objectives

• Find P-values and use them to test a mean μ• Use P-values for a z-test• Find critical values and rejection regions in a normal

distribution• Use rejection regions for a z-test


Using P-values to Make a Decision

Decision Rule Based on P-value• To use a P-value to make a conclusion in a

hypothesis test, compare the P-value with α. If P ≤ α, then reject H0.

1. If P > α, then fail to reject H0.


Example: Interpreting a P-value

The P-value for a hypothesis test is P = 0.0132. What is your decision if the level of significance is

Solution:Because 0.0132 < 0.05, you should reject the null hypothesis.

Solution:Because 0.0132 > 0.01, you should fail to reject the null hypothesis.


2.0.01?

1.0.05?

Finding the P-value

After determining the hypothesis test’s standardized test statistic and the test statistic’s corresponding area, do one of the following to find the P-value.

a.For a left-tailed test, P = (Area in left tail).

b.For a right-tailed test, P = (Area in right tail).

c.For a two-tailed test, P = 2(Area in tail of test statistic).


Example: Finding the P-value

Find the P-value for a left-tailed hypothesis test with a test statistic of z = -1.99. Decide whether to reject H0 if the level of significance is α = 0.01.

z0-1.99

P = 0.0233

Solution:For a left-tailed test, P = (Area in left tail)

Because 0.0233 > 0.01, you should fail to reject H0


z0 1.82

Example: Finding the P-value

Find the P-value for a two-tailed hypothesis test with a test statistic of z = 1.82. Decide whether to reject H0 if the level of significance is α = 0.05.

Solution:For a two-tailed test, P = 2(Area in tail of test statistic)

Because 0.0688 > 0.05, you should fail to reject H0

0.9838

1 – 0.9656 = 0.0344

P = 2(0.0344) = 0.0688


Z-Test for a Mean μ

• Can be used when the population is normal and σ is known, or for any population when the sample size n is at least 30.

• The test statistic is the sample mean • The standardized test statistic is z

• When n ≥ 30, the sample standard deviation s can be substituted for σ.


Using P-values for a z-Test for Mean μ

1. State the claim mathematically and verbally. Identify the null and alternative hypotheses.

2. Specify the level of significance.

3. Determine the standardized test statistic.

4. Find the area that corresponds to z.

State H0 and Ha.

Identify α.

Use Table 4 in Appendix B.


In Words In Symbols

Using P-values for a z-Test for Mean μ

Reject H0 if P-value is less than or equal to α. Otherwise, fail to reject H0.

5. Find the P-value.a. For a left-tailed test, P = (Area in left tail).b. For a right-tailed test, P = (Area in right tail).c. For a two-tailed test, P = 2(Area in tail of test

statistic).6. Make a decision to reject or

fail to reject the null hypothesis.

7. Interpret the decision in the context of the original claim.


In Words In Symbols

Example: Hypothesis Testing Using P-values

A manufacturer of sprinkler systems designed for fire protection claims that the average activating temperature is at least 135˚F. To test the claim, you randomly select a sample of 32 systems and find the mean activation temperature to be 133˚F with a standard deviation of 3.3˚F. Is there enough evidence to support the claim at α = 0.01? Use a P-value.


Solution: Hypothesis Testing Using P-values

• H0:

• Ha:

• α = • Test Statistic:

μ ≥ 135 (Claim)

μ < 135

0.01

• Decision:

At the 10% level of significance, you have sufficient evidence to reject the manufacturer’s claim that the average activating temperature is at least 135 degrees.

0-3.43z

0.003

• P-value

0.0003 < 0.01Reject H0


Example: Hypothesis Testing Using P-values

An Alabama politician claims that the mean annual salary for engineering managers in Alabama is more than the national mean of $100,800. We take a random sample of 34 engineering managers salaries in Alabama. At α = 0.03, is there enough evidence to support the politician’s claim?


Solution: Hypothesis Testing Using P-values

• H0:

• Ha:

• α = • Test Statistic:

μ <= $100,800

μ > $100,800, (Claim)

0.03 (0.9700), z = 1.88

• Decision:

At the 3% level of significance, there is not enough evidence to support the politician’s claim!

• P-valueP ≈ 1


=> ~ 0

Fail to reject H0

z0 1.88

0.03

-6.58

1 > 0.03

Rejection Regions and Critical Values

Rejection region (or critical region) • The range of values for which the null hypothesis is

not probable. • If a test statistic falls in this region, the null

hypothesis is rejected.

• A critical value z0 separates the rejection region from the nonrejection region.


Rejection Regions and Critical Values

Finding Critical Values in a Normal Distribution1. Specify the level of significance α.2. Decide whether the test is left-, right-, or two-tailed.

3. Find the critical value(s) z0. If the hypothesis test isa. left-tailed, find the z-score that corresponds to an area of α,b. right-tailed, find the z-score that corresponds to an area of 1

– α,c. two-tailed, find the z-score that corresponds to ½α and 1 –

½α.d. Sketch the standard normal distribution. Draw a vertical

line at each critical value and shade the rejection region(s).


Example: Finding Critical Values

Find the critical value and rejection region for a two-tailed test with α = 0.05.

z0 z0z0

½α = 0.025 ½α = 0.025

1 – α = 0.95

The rejection regions are to the left of -z0 = -1.96 and to the right of z0 = 1.96.

z0 = 1.96-z0 = -1.96

Solution:


Decision Rule Based on Rejection Region

To use a rejection region to conduct a hypothesis test, calculate the standardized test statistic, z. If the standardized test statistic1. is in the rejection region, then reject H0.2. is not in the rejection region, then fail to reject H0.

z

0z0

Fail to reject H0.

Reject H0.

Left-Tailed Test

z < z0

z

0 z0

Reject Ho.

Fail to reject Ho.

z > z0

Right-Tailed Test

z0−z0

Two-Tailed Testz0z < -z0 z > z0

Reject H0

Fail to reject H0

Reject H0


Using Rejection Regions for a z-Test for a Mean μ

1. State the claim mathematically and verbally. Identify the null and alternative hypotheses.

2. Specify the level of significance.

3. Sketch the sampling distribution.

4. Determine the critical value(s).

5. Determine the rejection region(s).

State H0 and Ha.

Identify α.

Use Table 4 in Appendix B.


In Words In Symbols

Using Rejection Regions for a z-Test for a Mean μ

6. Find the standardized test statistic.

7. Make a decision to reject or fail to reject the null hypothesis.

7. Interpret the decision in the context of the original claim.

If z is in the rejection region, reject H0. Otherwise, fail to reject H0.


In Words In Symbols

Example: Testing with Rejection Regions

Employees in a large accounting firm claim that the mean salary of the firm’s accountants is less than that of its competitor’s, which is $45,000. A random sample of 30 of the firm’s accountants has a mean salary of $43,500 with a standard deviation of $5200. At α = 0.05, test the employees’ claim.


Solution: Testing with Rejection Regions

• H0:

• Ha:

• α = • Rejection Region:

μ ≥ $45,000

μ < $45,000

0.05

• Decision:At the 5% level of significance, there is not sufficient evidence to support the employees’ claim that the mean salary is less than $45,000.

• Test Statistic

z0-1.645

0.05

-1.58

-1.645

Fail to reject H0


Example: Testing with Rejection Regions

The U.S. Department of Agriculture reports that the mean cost of raising a child from birth to age 2 in a rural area is $10,460. You believe this value is incorrect, so you select a random sample of 900 children (age 2) and find that the mean cost is $10,345 with a standard deviation of $1540. At α = 0.05, is there enough evidence to conclude that the mean cost is different from $10,460? (Adapted from U.S. Department of Agriculture Center for Nutrition Policy and Promotion)


Solution: Testing with Rejection Regions

• H0:

• Ha:

• α = • Rejection Region:

μ = $10,460

μ ≠ $10,460

0.05

• Decision:At the 5% level of significance, you have enough evidence to conclude the mean cost of raising a child from birth to age 2 in a rural area is significantly different from $10,460.

• Test Statistic

z0-1.96

0.025

1.96

0.025

-1.96 1.96

-2.24

Reject H0


Section 7.2 Summary

• Found P-values and used them to test a mean μ• Used P-values for a z-test• Found critical values and rejection regions in a

normal distribution• Used rejection regions for a z-test• HW: 1 - 45 EO


Documents

Section 7.2 Hypothesis Testing for the Mean (Large Samples) Larson/Farber 4th ed