1. Chapter 16 Single-Population Hypothesis Tests

2. 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.

3. Hypothesis Tests Examples
Trials
H0: The person is innocent
H1: The person is guilty
Soda
H0:  = 12 oz
H1:  < 12 oz

4. 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

5. Errors in Hypothesis Tests
Actual Situation
H0 is true
H0 is false
Decision
Reject H0
Type I error ()
No error
Fail to reject H0
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)

6. Hypothesis Tests Tails 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

7. Hypothesis Tests Examples
Two-tailed test: According to the US Bureau of the Census, the mean family size was 3.17 in An economist wants to check whether or not this mean has changed since 1991.
H0:  = 3.17
H1:   3.17 C1 C2
1-
/2
/2
Rejection
Region
=3.17
Rejection
Region
Non-rejection
Region

8. Hypothesis Tests Examples
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.
H0:  = 12
H1:  < 12 C
1- 
Rejection
Region
=12
Non-rejection
Region

9. Hypothesis Tests Examples
Right-tailed test: According to the US Bureau of the Census, the mean income of all households was $37,922 in Suppose that we want to test whether the current mean income of all households is higher than $37,922. C
H0:  = 37922
H1:  > 37922
1- 
=37922
Rejection
Region
Non-rejection
Region

10. Hypothesis Tests Rejection Region Approach
Select the type of test and check the underlying conditions
State the null and alternative hypotheses
Determine the level of significance 
Calculate the test statistics
Determine the critical values and rejection region
Check to see whether the test statistic falls in the rejection region
Make decision

11. Hypothesis Tests P-Value Approach
Select the type of test and check the underlying conditions
State the null and alternative hypotheses
Determine the level of significance 
Calculate the p-value (the smallest level of significance that would lead to rejection of the null hypothesis H0 with given data)
Check to see if the p-value is less than 
Make decision

12. Testing Hypothesis on the Mean with Variance Known (Z-Test)
Null Hypothesis: H0:  = 0
Test statistic:
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

13. Testing Hypothesis on the Mean with 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 C .9 .1
Rejection
Region
=3
Non-rejection
Region
-1.282

14. Testing Hypothesis on the Mean with 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 C1 C2
.95
.025
.025
Rejection
Region
=38
Rejection
Region
Non-rejection
Region
-1.96
1.96

15. Type II Error and Sample Size
Increasing sample size could reduce Type II error C1 C2 
1-
= 0
= 0+
Non-rejection
Region

16. Testing Hypothesis on the Mean with 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 C1 C2
.95
.025
.025
Rejection
Region
=50
Rejection
Region
Non-rejection
Region
-1.96
1.96

17. Type II Error and Sample Size - Example
Claim: Burning rate = 50 cm/s
n=25, =2, =.05, =.10
Hypothesis: H0:  = 50; H1:   50
If  = 51 C1 C2 
1-
= 0
= 0+
Non-rejection
Region

18. Testing Hypothesis on the Mean with Variance Unknown (t-Test)
Null Hypothesis: H0:  = 0
Test statistic:
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

19. Testing Hypothesis on the Mean with 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 C1 C2
.95
.025
.025
Rejection
Region
=2.5
Rejection
Region
Non-rejection
Region
-2.31
2.31

20. Testing Hypothesis on the Mean with 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 C
.95
.05
Rejection
Region
=65
Non-rejection
Region
-2.14

21. Testing Hypothesis on the Mean with 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 C
.99
.01
=22
Rejection
Region
Non-rejection
Region
2.31

22. Testing Hypothesis on the Median
Null Hypothesis: H0:  = 0
Test statistic: SH = No. of observations greater than 0
SL = No. of observations less than 0
Alt. Hypothesis
Test Statistic
P-value (Binomial p=.5)
H1:   0
S=Max(SH, SL)
2*P(xS)
H1:  > 0 SH
P(xSH)
H1:  < 0 SL
P(xSL)

23. 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(x9)= P(x=9)+P(x=10)+P(x=11)+P(x=12)
=.0730
Reject H0 41 53 65 69 74 78 79 83 97
119
161
203

24. Testing Hypothesis on the Variance of a Normal Distribution
Null Hypothesis: H0: 2 = 02
Test statistic:
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 < 21-/2,n-1
H1: 2 > 02
P(2 >02 )
02 > 2,n-1
H1: 2 < 02
P(2 <02 )
02 < 21-,n-1

25. Testing Hypothesis on the Variance – Example
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 C
.95
.05
Rejection
Region
Non-rejection
Region
30.14

26. Testing Hypothesis on the Population Proportion
Null Hypothesis: H0: p = p0
Test statistic:
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

27. Testing Hypothesis on the Population Proportion – Example 16.7
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 C1 C2
.99
.005
.005
Rejection
Region
=.312
Rejection
Region
Non-rejection
Region
-2.576
2.576

28. Testing Hypothesis on the Population Proportion – Example 16.8
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 C
.95
.05
=.04
Rejection
Region
Non-rejection
Region
1.645