Chapter 16 Single-Population Hypothesis Tests. Hypothesis Tests A statistical hypothesis is an...

Chapter 16

Single-Population Hypothesis Tests

Hypothesis Tests

• A statistical hypothesis is an assumption about a population parameter.

• There are two types of statistical hypotheses.– Null hypothesis -- The null hypothesis, H0, represents a

theory that has been put forward, either because it is believed to be true or because it is to be used as a basis for argument, but has not been proved.

– Alternative hypothesis (Research hypothesis) -- The alternative hypothesis, H1, is a statement of what a statistical hypothesis test is set up to establish.

Hypothesis Tests Examples

• Trials– H0: The person is innocent

– H1: The person is guilty

• Soda– H0: = 12 oz

– H1: < 12 oz

Hypothesis Tests

• Test Statistics -- the random variable X whose value is tested to arrive at a decision.

• Critical values-- the values of the test statistic that separate the rejection and non-rejection regions.

• Rejection Region -- the set of values for the test statistic that leads to rejection of H0

• Non-rejection region -- the set of values not in the rejection region that leads to non-rejection of H0

Errors in Hypothesis Tests

Actual Situation

H0 is true H0 is false

DecisionReject H0 Type I error () No error

Fail to reject H0 No error Type II error ()

• (Significance level): Probability of making Type I error • : Probability of making Type II error • 1-: Power of Test (Probability of rejecting H0 when H0 is

false)

Hypothesis TestsTails of a Test

Two-tailed Test

Left-tailed Test

Right-tailed Test

H0 = = or = or

H1 < >

Rejection region Both tails Left tail Right tail

p-value Sum of areas beyond the test

statistics

Area to the left of the test statistic

Area to the right of the test statistic

Hypothesis TestsExamples

Two-tailed test: According to the US Bureau of the Census, the mean family size was 3.17 in 1991. An economist wants to check whether or not this mean has changed since 1991.

1-/2 /2

C1C2H0: = 3.17

H1: 3.17

Non-rejectionRegion

RejectionRegion

RejectionRegion

=3.17


Left-tailed test: A soft-drink company claims that, on average, its cans contain 12 oz of soda. Suppose that a consumer agency wants to test whether the mean amount of soda per can is less than 12 oz.

1-

CH0: = 12

H1: < 12

Non-rejectionRegion

RejectionRegion

=12


Right-tailed test: According to the US Bureau of the Census, the mean income of all households was $37,922 in 1991. Suppose that we want to test whether the current mean income of all households is higher than $37,922.

1-

C

H0: = 37922

H1: > 37922

Non-rejectionRegion

RejectionRegion

=37922

Hypothesis TestsRejection Region Approach

1. Select the type of test and check the underlying conditions

2. State the null and alternative hypotheses

3. Determine the level of significance 4. Calculate the test statistics

5. Determine the critical values and rejection region

6. Check to see whether the test statistic falls in the rejection region

7. Make decision

Hypothesis TestsP-Value Approach

1. Select the type of test and check the underlying conditions

2. State the null and alternative hypotheses

3. Determine the level of significance 4. Calculate the p-value (the smallest level of

significance that would lead to rejection of the null hypothesis H0 with given data)

5. Check to see if the p-value is less than 6. Make decision

Testing Hypothesis on the Meanwith Variance Known (Z-Test)

Alt. Hypothesis P-value Rejection Criterion

H1: 0 P(z>z0)+P(z<-z0) z0 > z1-/2 or z0 < z/2

H1: > 0 P(z>z0) z0 > z1-

H1: < 0 P(z<-z0) z0 < z

• Null Hypothesis: H0: = 0

• Test statistic:n

XZ

0

0

Testing Hypothesis on the Meanwith Variance Known (Z-Test)– Example 16.1

• Claim: Burning time at least 3 hrs• n=42, =0.23, =.10• Null Hypothesis: H0: 3• Alt. Hypothesis: H1: < 3

• Test statistic:

• Rejection region: z= z.10=-1.282

• P-value = P(z<-1.13) = .1299• Fail to reject H0

13.14223.0

396.200

n

XZ

96.2x

.9.1

C

Non-rejectionRegion

RejectionRegion

=3

-1.282

Testing Hypothesis on the Meanwith Variance Known (Z-Test)– Example 16.2

• Claim: Width =38”• n=80, =0.098, =.05• Null Hypothesis: H0: = 38• Alt. Hypothesis: H1: 38• Test statistic:

• Rejection region: z/2= z.025=-1.96

z1-/2= z.975= 1.96• P-value = P(z>1.825)+P(z<-1.825) = .0679• Fail to reject H0

825.180098.0

3802.3800

n

XZ

02.38x.95.025 .025

C1C2

Non-rejectionRegion

RejectionRegion

RejectionRegion

=38

-1.96 1.96

Type II Error and Sample Size

• Increasing sample size could reduce Type II error

1-

C1C2

Non-rejectionRegion

= 0

nn

X

n

XZ

)( 00

0

= 0+

trueisHwhen

nNZ

1

0 )1,(

)()(221

nz

nz

0

2

2211 )(

2

when

zzn

Testing Hypothesis on the Meanwith Variance Known (Z-Test)– Example

• Claim: Burning rate = 50 cm/s• n=25, =2, =.05• Null Hypothesis: H0: = 50• Alt. Hypothesis: H1: 50• Test statistic:


z1-/2= z.975= 1.96• P-value = P(z>3.25)+P(z<-3.25) = .0012• Reject H0

25.3252

503.5100

n

XZ

3.51x.95.025 .025

C1C2

Non-rejectionRegion

RejectionRegion

RejectionRegion

=50

-1.96 1.96

Type II Error and Sample Size - Example

1-

C1C2

Non-rejectionRegion

= 0

= 0+

295.0

)46.4()54.0(

)2

25196.1()

2

25196.1(

)()(221

nz

nz

421

2)28.196.1(

)(

2

22

2

2211 2

zz

n

• Claim: Burning rate = 50 cm/s• n=25, =2, =.05, =.10• Hypothesis: H0: = 50; H1: 50• If = 51

Testing Hypothesis on the Meanwith Variance Unknown (t-Test)


H1: 0 2*P(t>|t0|) t0 > t/2,n-1 or t0 < -t/2,n-1

H1: > 0 P(t>t0) t0 > t, n-1

H1: < 0 P(t<-t0) t0 <- t, n-1


• Test statistic:ns

XT 0

0

Testing Hypothesis on the Meanwith Variance Unknown (t-Test)– Example 16.3

• Claim: Length =2.5”• n=49, s=0.021, =.05• Null Hypothesis: H0: = 2.5• Alt. Hypothesis: H1: 2.5• Test statistic:

• Rejection region: t.025,48= 2.3139• P-value = 2*P(t>3.33)= .0033• Reject H0

33.349021.0

5.249.200

ns

XT

49.2x.95.025 .025

C1C2

Non-rejectionRegion

RejectionRegion

RejectionRegion

=2.5

-2.31 2.31


• Claim: Battery life at least 65 mo.• n=15, s=3, =.05• Null Hypothesis: H0: 65

• Alt. Hypothesis: H1: < 65

• Test statistic:

• Rejection region: -t.05, 14=-2.14

• P-value = P(t<-2.582) = .0109• Reject H0

582.2153

656300

ns

XT

63x

.95.05

C

Non-rejectionRegion

RejectionRegion

=65

-2.14


• Claim: Service rate = 22 customers/hr• n=18, s=4.2, =.01• Null Hypothesis: H0: = 22• Alt. Hypothesis: H1: > 22• Test statistic:

• Rejection region: t.01,17= 2.898• P-value = P(t>1.717)= .0540• Fail to reject H0

717.1182.4

227.2300

ns

XT

7.23x.99 .01

C

Non-rejectionRegion

RejectionRegion

=22

2.31

Testing Hypothesis on the Median

Alt. Hypothesis Test Statistic P-value (Binomial p=.5)

H1: 0 S=Max(SH, SL) 2*P(xS)

H1: > 0 SH P(xSH)

H1: < 0 SL P(xSL)


• Test statistic: SH = No. of observations greater than 0

SL = No. of observations less than 0

Testing Hypothesis on the Median– Example 16.6

• Claim: Median spending = $67.53• n=12, =.10• Null Hypothesis: H0: = 67.53• Alt. Hypothesis: H1: > 67.53• Test statistic: SH = 9, SL = 3• P-value = P(x9)= P(x=9)+P(x=10)+P(x=11)+P(x=12)

=.0730• Reject H0

415365697478798397

119161203

Testing Hypothesis on the Variance of a Normal Distribution


H1: 2 02

2*P(2 >02 ) or

2*P(2 <02 )

02 > 2

/2,n-1 or

02 < 2

1-/2,n-1

H1: 2 > 02 P(2 >0

2 ) 02 > 2

,n-1

H1: 2 < 02 P(2 <0

2 ) 02 < 2

1-,n-1

• Null Hypothesis: H0: 2 = 02

• Test statistic:20

22 )1(0

sn

• Claim: Variance 0.01• s2=0.0153, n=20, =.05• Null Hypothesis: H0: 2 = .01• Alt. Hypothesis: H1: 2 > .01• Test statistic:

• Rejection region: 2.05,19= 30.14

• p-value = P(2 >29.07)=0.0649• Fail to reject H0

.95 .05

C

Non-rejectionRegion

RejectionRegion

30.14

Testing Hypothesis on the Variance– Example

07.2901.0

0153.0)120()1(20

22

0

sn

Testing Hypothesis on the Population Proportion


H1: p p0 P(z>z0)+P(z<-z0) z0 > z1-/2 or z0 < z/2

H1: p > p0 P(z>z0) z0 > z1-

H1: p < p0 P(z<-z0) z0 < z

• Null Hypothesis: H0: p = p0

• Test statistic:

npp

ppz

)1(

ˆ

00

00

• Claim: Market share = 31.2%• n=400, =.01• Null Hypothesis: H0: p = .312• Alt. Hypothesis: H1: p .312• Test statistic:


z1-/2= z.995= 2.576• P-value = P(z>.95)+P(z<-.95) = .3422• Fail to reject H0

29.ˆ p.99.005 .005

C1C2

Non-rejectionRegion

RejectionRegion

RejectionRegion

=.312

-2.576 2.576

Testing Hypothesis on the Population Proportion – Example 16.7

95.

400)312.1(312.

312.29.

)1(

ˆ

00

00

npp

ppz

• Claim: Defective rate 4%• n=300, =.05• Null Hypothesis: H0: p = .04• Alt. Hypothesis: H1: p > .04• Test statistic:

• Rejection region: z1-= z.95= 1.645• P-value = P(z>2.65) = .0040• Reject H0

07.300/7ˆ p.95 .05

C

Non-rejectionRegion

RejectionRegion

=.04

1.645

Testing Hypothesis on the Population Proportion – Example 16.8

65.2

300)04.1(04.

04.07.

)1(

ˆ

00

00

npp

ppz

Chapter 16 Single-Population Hypothesis Tests. Hypothesis Tests A statistical hypothesis is an...

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