Statistics

Chapter 8: Inferences Based on a Single Sample: Tests of Hypotheses

Statistics

McClave, Statistics, 11th ed. Chapter 8: Inferences Based on a Single Sample: Tests of Hypotheses

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Where We’ve Been

Calculated point estimators of population parameters

Used the sampling distribution of a statistic to assess the reliability of an estimate through a confidence interval


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Where We’re Going

Test a specific value of a population parameter

Measure the reliability of the test

8.1: The Elements of a Test of Hypotheses


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Confidence IntervalWhere on the number line do the data point us?(No prior idea about the value of the parameter.)

Hypothesis TestDo the data point us to this particular value?(We have a value in mind from the outset.)

µ? µ?

µ0?



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Null Hypothesis: H0

•This will be supported unless the data provide evidence that it is false• The status quo Alternative Hypothesis: Ha

•This will be supported if the data provide sufficient evidence that it is true• The research hypothesis


If the test statistic has a high probability when H0 is true, then H0 is not rejected.

If the test statistic has a (very) low probability when H0 is true, then H0 is rejected.


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Reality ↓ / Test Result → Do not reject H0 Reject H0

H0 is true Correct!Type I Error: rejecting a true null hypothesisP(Type I error) = α

H0 is falseType II Error: not rejecting a false null hypothesisP(Type II error) = β

Correct!


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Note: Null hypotheses are either rejected, or else there is insufficient evidence to reject them. (I.e., we don’t accept null hypotheses.)


• Null hypothesis (H0): A theory about the values of one or more parameters

• Ex.: H0: µ = µ0 (a specified value for µ)

• Alternative hypothesis (Ha): Contradicts the null hypothesis

• Ex.: H0: µ ≠ µ0

• Test Statistic: The sample statistic to be used to test the hypothesis• Rejection region: The values for the test statistic which lead to rejection of

the null hypothesis• Assumptions: Clear statements about any assumptions concerning the

target population• Experiment and calculation of test statistic: The appropriate calculation for

the test based on the sample data• Conclusion: Reject the null hypothesis (with possible Type I error) or do

not reject it (with possible Type II error)


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Suppose a new interpretation of the rules by soccer referees is expected to increase the number of yellow cards per game. The average number of yellow cards per game had been 4. A sample of 121 matches produced an average of 4.7 yellow cards per game, with a standard deviation of .5 cards. At the 5% significance level, has there been a change in infractions called?


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H0: µ = 4Ha: µ ≠ 4Sample statistic: = 4.7α= .05

Assume the sampling distribution is normal.

Test statistic:

Conclusion: z.025 = 1.96. Since z* > z.025 , reject H0.(That is, there do seem to be more yellow cards.)

94.10064.

47.4* 0

xs

xz

8.2: Large-Sample Test of a Hypothesis about a Population Mean


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The null hypothesis isusually stated as an equality …

… even though the alternative hypothesis can be either an equality or an inequality.



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Lower Tailed Upper Tailed Two tailed

α = .10 z < - 1.28 z > 1.28 | z | > 1.645

α = .05 z < - 1.645 z > 1.645 | z | > 1.96

α = .01 z < - 2.33 z > 2.33 | z | > 2.575


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Rejection Regions for Common Values of α


One-Tailed Test H0 : µ = µ0

Ha : µ < or > µ0

Test Statistic:

Rejection Region: | z | > zα

Two-Tailed Test


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x

xz

0

H0 : µ = µ0

Ha : µ ≠ µ0

Test Statistic:

Rejection Region: | z | > zα/2

x

xz

0

Conditions: 1) A random sample is selected from the target population.2) The sample size n is large.


The Economics of Education Review (Vol. 21, 2002) reported a mean salary for males with postgraduate degrees of $61,340, with an estimated standard error (s ) equal to $2,185. We wish to test, at the α = .05 level, H0: µ = $60,000.


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The Economics of Education Review (Vol. 21, 2002) reported a mean salary for males with postgraduate degrees of $61,340, with an estimated standard error (s) equal to $2,185. We wish to test, at the α = .05 level, H0: µ = $60,000.


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H0 : µ = 60,000

Ha : µ ≠ 60,000

Test Statistic:

Rejection Region: | z | > z.025 = 1.96

613.

185,2

000,60340,61

0

z

z

xz

x

Do not reject H0

8.3:Observed Significance Levels: p - Values


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Suppose z = 2.12. P(z > 2.12) = .0170.

Reject H0 at the α= .05 level Do not reject H0 at the α= .01 level

But it’s pretty close, isn’t it?



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The observed significance level, or p-value, for a test is the probability of observing the results actually observed (z*) assuming the null hypothesis is true.

The lower this probability, the less likely H0 is true.

)|*( 0HzzP

Let’s go back to the Economics of Education Review report (= $61,340, s = $2,185). This time we’ll test H0: µ = $65,000.


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H0 : µ = 65,000

Ha : µ ≠ 65,000

Test Statistic:

p-value: P(|z| > 1.675) = .047 * 2 =0.094

675.1

185,2

000,65340,61

0

z

z

xz

x



Reporting test results Choose the maximum tolerable value of α If the p-value ≤ α, reject H0

If the p-value > α, do not reject H0


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8.4: Small-Sample Test of a Hypothesis about a Population Mean


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If the sample is small and is unknown, testing hypotheses about µ requires the t-distribution instead of the z-distribution.

ns

xt

/0

One-Tailed Test H0 : µ = µ0

Ha : µ < or > µ0

Test Statistic:

Rejection Region: | t | > tα

Two-Tailed Test


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ns

xt

/0

H0 : µ = µ0

Ha : µ ≠ µ0

Test Statistic:

Rejection Region: | t | > tα/2

ns

xt

/0

Conditions: 1) A random sample is selected from the target population.2) The population from which the sample is selected is

approximately normal.3) The value of tα is based on (n – 1) degrees of freedom



Suppose copiers average 100,000 between paper jams. A salesman claims his are better, and offers to leave 5 units for testing. The average number of copies between jams is 100,987, with a standard deviation of 157. Does his claim seem believable?


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H0 : µ = 100,000

Ha : µ > 100,000

Test Statistic:

p-value: P(tdf=4 > 14.06) < .0005




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06.145/157

000,100987,100/

0

t

t

ns

xt

H0 : µ = 100,000

Ha : µ > 100,000

Test Statistic:

p-value: P(|tdf=4| > 14.06) < .0005




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06.145/157

000,100987,100/

0

t

t

ns

xt

Reject the null hypothesis based on the very low probability of seeing the observed results if the null were true.So, the claim does seem plausible.

One-Tailed Test H0 : p = p0

Ha : p < or > p0

Test Statistic:

Rejection Region: | z | > zα

Two-Tailed Test


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p

ppz

ˆ

0ˆ

H0 : p = p0

Ha : p ≠ p0

Test Statistic:

Rejection Region: | z | > zα/2

p

ppz

ˆ

0ˆ

Conditions: 1) A random sample is selected from a binomial population. 2) The sample size n is large (i.e., np0 and nq0 are both ≥ 15).

8.5: Large-Sample Test of a Hypothesis about a Population Proportion

p0 = hypothesized value of p, , and q0 = 1 - p0 n

qpp

00ˆ


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Rope designed for use in the theatre must withstand unusual stresses. Assume a brand of 3” three-strand rope is expected to have a breaking strength of 1400 lbs. A vendor receives a shipment of rope and needs to (destructively) test it.

The vendor will reject any shipment which cannot pass a 1% defect test (that’s harsh, but so is falling scenery during an aria). 1500 sections of rope are tested, with 20 pieces failing the test. At the α = .01 level, should the shipment be rejected?


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The vendor will reject any shipment that cannot pass a 1% defects test . 1500 sections of rope are tested, with 20 pieces failing the test. At the α = .01 level, should the shipment be rejected?

H0: p = .01

Ha: p > .01

Rejection region: |z| > 2.33

Test statistic:

14.1

1500/)987)(.013(.

01.013.

ˆ

ˆ

0

z

z

ppz

p


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The vendor will reject any shipment that cannot pass a 1% defects test . 1500 sections of rope are tested, with 20 pieces failing the test. At the α = .01 level, should the shipment be rejected?

H0: p = .01

Ha: p > .01

Rejection region: |z| > 2.33

Test statistic:

14.1

1500/)987)(.013(.

01.013.

ˆ

ˆ

0

z

z

ppz

p

There is insufficient evidence to reject the null hypothesis based on the sample results.

There is insufficient evidence to reject the null hypothesis based on the sample results.

8.6: Calculating Type II Error Probabilities: More about β

To calculate P(Type II), or β, …

1. Calculate the value(s) of that divide the “do not reject” region from the “reject” region(s).

Upper-tailed test:

Lower-tailed test:

Two-tailed test:


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n

szzx

n

szzx

n

szzx

n

szzx

xU

xL

x

x

2/02/00

2/02/00

000

000


To calculate P(Type II), or β, …

1. Calculate the value(s) of that divide the “do not reject” region from the “reject” region(s).

2. Calculate the z-value of 0 assuming the alternative hypothesis mean is the true µ:

The probability of getting this z-value for the acceptance region is β.


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x

axz

0


The power of a test is the probability that the test will correctly lead to the rejection of the null hypothesis for a particular value of µ in the alternative hypothesis. The power of a test is calculated as (1 - β ).


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The Economics of Education Review (Vol. 21, 2002) reported a mean salary for males with postgraduate degrees of $61,340, with an estimated standard error (s) equal to $2,185. We wish to test, at the α = .05 level, H0: µ = $60,000.

H0 : µ = 60,000

Ha : µ ≠ 60,000

Test Statistic: z = .613;

zα/2=.025 = 1.96

We did not reject this null hypothesis earlier, but what if the true mean were $62,000?



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The Economics of Education Review (Vol. 21, 2002) reported a mean salary for males with postgraduate degrees of $61,340, with s equal to $2,185.

We did not reject this null hypothesis earlier, but what if the true mean were $62,000?

xO,L= 60,000 - 1.96*2,185 = 55,717.4xO,U= 60,000 +1.96*2,185 = 64,282.6ZL = (55,717.4 – 62,000)/2,185 = -2.875ZU = (64,282.6 - 62,000)/2,185 = 1.045β = P(-2.875≤Z≤1.045) = 0.8511The power of this test is 1 – β = 1 – 0.8511 = 0.1489


For fixed n and α, the value of β decreases and the power increases as the distance between µ0 and µa increases.

For fixed n, µ0 and µa, the value of β increases and the power decreases as the value of α is decreased.

For fixed α, µ0 and µa, the value of β decreases and the power increases as n is increased.


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8.7: Tests of Hypotheses about a Population Variance


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2

22 )1(

sn

The chi-square distribution is really a family of distributions, depending on the number of degrees of freedom.

But, the population must be normally distributed for the hypothesis tests on 2 (or ) to be reliable!


One-Tailed Test

Test statistic:

Rejection region:


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2221

2

20

22

20

220

2

20

20

)1(

:

:

sn

H

H

a

Two-Tailed Test

Test statistic:

Rejection region:22/

222/1

2

20

22

20

2

20

20

)1(

:

:

or

sn

H

H

a


Conditions Required for a Valid

Large- Sample Hypothesis Test for 2

1. A random sample is selected from the target population.

2. The population from which the sample is selected is approximately normal.


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Earlier, we considered the average number of copies between jams for a brand of copiers. The salesman also claims his copiers are more predictable, in that the standard deviation of jams is 125. In the sample of 5 copiers, that sample standard deviation was 157. Does his claim seem believable, at the α = .10 level?


Two-Tailed Test

Test statistic:

Rejection criterion:


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48773.931.6

31.6125

)157)(15(

125:

125:

205.

2

2

22

2

20

aH

H

Earlier, we considered the average number of copies between jams for a brand of copiers. The salesman also claims his copiers are more predictable, in that the standard deviation of jams is 125. In the sample of 5 copiers, that sample standard deviation was 157. Does his claim seem believable, at the α = .10 level?


Two-Tailed Test

Test statistic:

Rejection criterion:


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48773.931.6

31.6125

)157)(15(

125:

125:

205.

2

2

22

2

20

aH

H

Earlier, we considered the average number of copies between jams for a brand of copiers. The salesman also claims his copiers are more reliable, in that the standard deviation of jams is 125. In the sample of 5 copiers, that sample standard deviation was 157. Does his claim seem believable, at the α = .10 level?

Do not reject the null

hypothesis.

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