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, H 0, 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, H 1, is a statement of what a statistical hypothesis test is set up to establish.

3. Hypothesis Tests Examples Trials –H 0 : The person is innocent –H 1 : The person is guilty Soda –H 0 :  = 12 oz –H 1 :  < 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 H 0 Non-rejection region -- the set of values not in the rejection region that leads to non-rejection of H 0

5. Errors in Hypothesis Tests Actual Situation H 0 is trueH 0 is false Decision Reject H 0 Type I error (  ) No error Fail to reject H 0 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 H 0 when H 0 is false)

6. Hypothesis Tests Tails of a Test Two-tailed Test Left-tailed Test Right-tailed Test H0H0 = = or  = or  H1H1  <> Rejection regionBoth tailsLeft tailRight tail p-valueSum 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 1991. An economist wants to check whether or not this mean has changed since 1991. 1-   /2 C1C1 C2C2 H 0 :  = 3.17 H 1 :   3.17 Non-rejection Region Rejection Region Rejection Region  =3.17

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. 1-   C H 0 :  = 12 H 1 :  < 12 Non-rejection Region Rejection Region  =12

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 1991. Suppose that we want to test whether the current mean income of all households is higher than $37,922. 1-   C H 0 :  = 37922 H 1 :  > 37922 Non-rejection Region Rejection Region  =37922

10. Hypothesis Tests Rejection 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

11. Hypothesis Tests P-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 H 0 with given data) 5.Check to see if the p-value is less than  6.Make decision

12. Testing Hypothesis on the Mean with Variance Known (Z-Test) Alt. HypothesisP-valueRejection Criterion H 1 :    0 P(z>z 0 )+P(z<-z 0 )z 0 > z 1-  /2 or z 0 < z  /2 H 1 :  >  0 P(z>z 0 )z 0 > z 1-  H 1 :  <  0 P(z<-z 0 )z 0 < z  Null Hypothesis: H 0 :  =  0 Test statistic:

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: H 0 :   3 Alt. Hypothesis: H 1 :  < 3 Test statistic: Rejection region: z  = z.10 =-1.282 P-value = P(z<-1.13) =.1299 Fail to reject H 0.9.1 C Non-rejection Region Rejection Region  =3 -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: H 0 :  = 38 Alt. Hypothesis: H 1 :   38 Test statistic: Rejection region: z  /2 = z.025 =-1.96 z 1-  /2 = z.975 = 1.96 P-value = P(z>1.825)+P(z<-1.825) =.0679 Fail to reject H 0.95.025 C1C1 C2C2 Non-rejection Region Rejection Region Rejection Region  =38 -1.96 1.96

15. Type II Error and Sample Size Increasing sample size could reduce Type II error 1-  C1C1 C2C2 Non-rejection Region  =  0   =  0 + 

16. Testing Hypothesis on the Mean with Variance Known (Z-Test)– Example Claim: Burning rate = 50 cm/s n=25,  =2,  =.05 Null Hypothesis: H 0 :  = 50 Alt. Hypothesis: H 1 :   50 Test statistic: Rejection region: z  /2 = z.025 =-1.96 z 1-  /2 = z.975 = 1.96 P-value = P(z>3.25)+P(z<-3.25) =.0012 Reject H 0.95.025 C1C1 C2C2 Non-rejection Region Rejection Region Rejection Region  =50 -1.96 1.96

17. Type II Error and Sample Size - Example 1-  C1C1 C2C2 Non-rejection Region  =  0   =  0 +  Claim: Burning rate = 50 cm/s n=25,  =2,  =.05,  =.10 Hypothesis: H 0 :  = 50; H 1 :   50 If  = 51

18. Testing Hypothesis on the Mean with Variance Unknown (t-Test) Alt. HypothesisP-valueRejection Criterion H 1 :    0 2*P(t>|t 0 |)t 0 > t  /2,n-1 or t 0 < -t  /2,n-1 H 1 :  >  0 P(t>t 0 )t 0 > t , n-1 H 1 :  <  0 P(t<-t 0 )t 0 <- t , n-1 Null Hypothesis: H 0 :  =  0 Test statistic:

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: H 0 :  = 2.5 Alt. Hypothesis: H 1 :   2.5 Test statistic: Rejection region:  t.025,48 =  2.3139 P-value = 2*P(t>3.33)=.0033 Reject H 0.95.025 C1C1 C2C2 Non-rejection Region Rejection Region Rejection Region  =2.5 -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: H 0 :   65 Alt. Hypothesis: H 1 :  < 65 Test statistic: Rejection region: -t.05, 14 =-2.14 P-value = P(t<-2.582) =.0109 Reject H 0.95.05 C Non-rejection Region Rejection Region  =65 -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: H 0 :  = 22 Alt. Hypothesis: H 1 :  > 22 Test statistic: Rejection region: t.01,17 = 2.898 P-value = P(t>1.717)=.0540 Fail to reject H 0.99.01 C Non-rejection Region Rejection Region  =22 2.31

22. Testing Hypothesis on the Median Alt. HypothesisTest StatisticP-value (Binomial p=.5) H 1 :    0 S=Max(S H, S L ) 2*P(x  S) H 1 :  >  0 SHSH P(x  S H ) H 1 :  <  0 SLSL P(x  S L ) Null Hypothesis: H 0 :  =  0 Test statistic: S H = No. of observations greater than  0 S L = No. of observations less than  0

23. Testing Hypothesis on the Median – Example 16.6 Claim: Median spending = $67.53 n=12,  =.10 Null Hypothesis: H 0 :  = 67.53 Alt. Hypothesis: H 1 :  > 67.53 Test statistic: S H = 9, S L = 3 P-value = P(x  9)= P(x=9)+P(x=10)+P(x=11)+P(x=12) =.0730 Reject H 0 41 53 65 69 74 78 79 83 97 119 161 203

24. Testing Hypothesis on the Variance of a Normal Distribution Alt. Hypothesis P-valueRejection Criterion H 1 :  2   0 2 2*P(  2 >  0 2 ) or 2*P(  2 <  0 2 )  0 2 >  2  /2,n-1 or  0 2 <  2 1-  /2,n-1 H 1 :  2 >  0 2 P(  2 >  0 2 )  0 2 >  2 ,n-1 H 1 :  2 <  0 2 P(  2 <  0 2 )  0 2 <  2 1- ,n-1 Null Hypothesis: H 0 :  2 =  0 2 Test statistic:

25. Claim: Variance  0.01 s 2 =0.0153, n=20,  =.05 Null Hypothesis: H 0 :  2 =.01 Alt. Hypothesis: H 1 :  2 >.01 Test statistic: Rejection region:  2.05,19 = 30.14 p-value = P(  2 >29.07)=0.0649 Fail to reject H 0.95.05 C Non-rejection Region Rejection Region 30.14 Testing Hypothesis on the Variance – Example

26. Testing Hypothesis on the Population Proportion Alt. HypothesisP-valueRejection Criterion H 1 : p  p 0 P(z>z 0 )+P(z<-z 0 )z 0 > z 1-  /2 or z 0 < z  /2 H 1 : p > p 0 P(z>z 0 )z 0 > z 1-  H 1 : p < p 0 P(z<-z 0 )z 0 < z  Null Hypothesis: H 0 : p = p 0 Test statistic:

27. Claim: Market share = 31.2% n=400,  =.01 Null Hypothesis: H 0 : p =.312 Alt. Hypothesis: H 1 : p .312 Test statistic: Rejection region: z  /2 = z.005 =-2.576 z 1-  /2 = z.995 = 2.576 P-value = P(z>.95)+P(z<-.95) =.3422 Fail to reject H 0.99.005 C1C1 C2C2 Non-rejection Region Rejection Region Rejection Region  =.312 -2.576 2.576 Testing Hypothesis on the Population Proportion – Example 16.7

28. Claim: Defective rate  4% n=300,  =.05 Null Hypothesis: H 0 : p =.04 Alt. Hypothesis: H 1 : p >.04 Test statistic: Rejection region: z 1-  = z.95 = 1.645 P-value = P(z>2.65) =.0040 Reject H 0.95.05 C Non-rejection Region Rejection Region  =.04 1.645 Testing Hypothesis on the Population Proportion – Example 16.8