Post on 26-Dec-2015
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
Hypothesis TestsExamples
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
Hypothesis TestsExamples
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:
• Rejection region: z/2= z.025=-1.96
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)
Alt. Hypothesis P-value Rejection Criterion
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
• Null Hypothesis: H0: = 0
• 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
Testing Hypothesis on the Meanwith Variance Unknown (t-Test)– Example 16.4
• 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
Testing Hypothesis on the Meanwith Variance Unknown (t-Test)– Example 16.5
• 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)
• Null Hypothesis: H0: = 0
• 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
Alt. Hypothesis P-value Rejection Criterion
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
Alt. Hypothesis P-value Rejection Criterion
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:
• Rejection region: z/2= z.005=-2.576
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