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 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)

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

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. H0:  = 3.17 H1:   3.17 C1 C2 1- /2 /2 Rejection Region =3.17 Rejection Region Non-rejection Region

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

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. C H0:  = 37922 H1:  > 37922 1-  =37922 Rejection Region Non-rejection Region

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

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

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

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

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

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

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

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

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

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

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

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

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)

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

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

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

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

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

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