1 1 slide ©2009. econ-2030-applied statistics (dr. tadesse) chapter 9 learning objectives...

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©2009. Econ-2030-Applied Statistics (Dr. Tadesse)

Chapter 9 Learning Objectives

Population Mean: s Unknown

Population Proportion

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Tests About a Population Mean:s Unknown

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Test Statistic


nS

Xt 0

This test statistic has a t distribution with n - 1 degrees of freedom.

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Rejection Rule: p -Value Approach

H1: Reject H0 if t < -t

Reject H0 if t > t

Reject H0 if t < - t or t > t

H1:

H1:


Rejection Rule: Critical Value Approach

Reject H0 if p –value < a

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p -Values and the t Distribution

The format of the t distribution table provided in most statistics textbooks does not have sufficient detail to determine the exact p-value for a hypothesis test.

However, we can still use the t distribution table to identify a range for the p-value.

An advantage of computer software packages is that the computer output will provide the p-value for the t distribution.

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A State Highway Patrol periodically samplesvehicle speeds at various locationson a particular roadway. The sample speed is used to test the null hypothesis

Example: Highway Patrol

One-Tailed Test About a Population Mean: s Unknown

The locations where H0 is rejected are deemed

the best locations for radar traps.

H0: m < 65

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Example: Highway Patrol

At Location F, a sample of 64 vehicles shows a mean speed of 66.2 mph with a standard deviation of 4.2 mph. Use a = .05 to test the above null hypothesis.

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One-Tailed Test About a Population Mean:s Unknown

1. Determine the hypotheses.

2. Specify the level of significance.

3. Compute the value of the test statistic.

a = .05

H0: < 65

Ha: m > 65

0 66.2 65

2.286/ 4.2/ 64

xt

s n

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Using the Critical Value Approach

5. Compare Test Statistic with the Critical Value

We are at least 95% confident that the mean speed of vehicles at Location F is greater than 65 mph. Location F is a good candidate for a radar trap.

Because 2.286 > 1.669, we reject H0.


For a = .05 and d.f. = 64 – 1 = 63, t.05 = 1.669

4. Determine the Critical value and rejection rule.

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Using the p –Value Approach

5. Determine whether to reject H0.

4. Compute the p –value.

For t = 2.286, the p–value lies between .025(where t = 1.998) and .01 (where t = 2.387).

.01 < p–value < .025

Since any p–value between 0.01 and 0.025 is less than a = .05. Thus, we reject H0. That is, we are at least 95% confident that the mean speed of vehicles at Location F is greater than 65 mph.

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In general, a hypothesis test about the value of a population proportion p takes one of the following three forms (where p0

is the hypothesized value of the population proportion).

Hypotheses Testing of the Population Proportion

One-tailed(lower tail)

One-tailed(upper tail)

Two-tailed

0 0: H p p

0: aH p p0: aH p p0 0: H p p 0 0: H p p

0: aH p p

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The Test Statistic

zp p

p

0

pp p

n 0 01( )

Tests About a Population Proportion

where:

assuming np > 5 and n(1 – p) > 5

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When Using the p –Value Approach

H1 : pp Reject H0 if z > z

Reject H0 if z < -z

Reject H0 if z < -z or z > z

H1: p p

H1 pp

Tests About a Population Proportion

Reject H0 if p –value < a

When using the Critical Value Approach

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Example:

During a Christmas and New Year’s week, theNational Safety Council estimates that500 people would be killed and 25,000Would be injured on the nation’s roads. TheNSC claims that 50% of the accidents would be caused by drunk driving.

Two-Tailed Test About aPopulation Proportion

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A sample of 120 accidents showed that67 were caused by drunk driving. Usethese data to test the NSC’s claim witha = .05.


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1. Determine the hypotheses.

2. Specify the level of significance.

3. Compute the value of the test statistic.

a = .05

0: .5H p

: .5aH p

0 (67/ 120) .5 1.28

.045644p

p pz

0 0(1 ) .5(1 .5).045644

120p

p pn

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Using the Critical Value Approach

5. Compare the Test Statistic with the Critical Value

For a/2 = .05/2 = .025, z.025 = 1.96. Since the alternative is non-directional, the rejection regionwould be the area below -1.96 and above 1.96

4. Determine the critical value

As 1.278 falls between -1.96 and 1.96 (the acceptance region), we cannot reject H0.

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Using the p-Value Approach

4. Compute the p -value.

5. Compare the p-value with the significance level

As p–value = .2006 > a = .05, we fail to reject H0.


For z = 1.28, cumulative probability = .8997

p–value = 2(1 - .8997) = .2006

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