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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 Reject H0 Type I error () Decision Fail to reject H0 No error 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 Tests Tails of a Test Two-tailed Test Left-tailed Test Right-tailed Test H0 = = or = or H1 < > Both tails Left tail Right tail Sum of areas beyond the test statistics Area to the left of the test statistic Area to the right of the test statistic Rejection region p-value 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 C2 C1 /2 1- Rejection Region =3.17 Non-rejection Region /2 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 Non-rejection Region Rejection Region 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 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 H0 with given data) 5. Check to see if the p-value is less than 6. Make decision Testing Hypothesis on the Mean with Variance Known (Z-Test) • Null Hypothesis: H0: = 0 • Test statistic: Z X 0 0 n Alt. Hypothesis P-value H1: 0 P(z>z0)+P(z<-z0) H1: > 0 P(z>z0) H1: < 0 P(z<-z0) Rejection Criterion z0 > z1-/2 or z0 < z/2 z0 > z1- z0 < z Testing Hypothesis on the Mean with Variance Known (Z-Test)– Example 16.1 • • • • • Claim: Burning time at least 3 hrs x 2.96 n=42, =0.23, =.10 Null Hypothesis: H0: 3 Alt. Hypothesis: H1: < 3 Test statistic: Z0 X 0 2.96 3 1.13 n 0.23 42 • Rejection region: z= z.10=-1.282 • P-value = P(z<-1.13) = .1299 • Fail to reject H0 C .9 .1 =3 Non-rejection Region -1.282 Rejection Region Testing Hypothesis on the Mean with Variance Known (Z-Test)– Example 16.2 • • • • • Claim: Width =38” x 38.02 n=80, =0.098, =.05 Null Hypothesis: H0: = 38 Alt. Hypothesis: H1: 38 Test statistic: X 0 38.02 38 Z0 1.825 n 0.098 80 C2 C1 .95 .025 .025 =38 Rejection Non-rejection Region Region 1.96 -1.96 Rejection Region • 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 Type II Error and Sample Size • Increasing sample size could reduce Type II error X 0 X ( 0 ) Z0 n n n C2 C1 1- n when H1 is true 2 = 0+ ,1) ( z1 = 0 Non-rejection Region Z0 N ( n n ) ( z 2 ( z1 2 z1 )2 2 2 when 0 n ) Testing Hypothesis on the Mean with Variance Known (Z-Test)– Example • • • • • Claim: Burning rate = 50 cm/s x 51.3 n=25, =2, =.05 Null Hypothesis: H0: = 50 Alt. Hypothesis: H1: 50 Test statistic: X 0 51.3 50 Z0 3.25 n 2 25 C2 C1 .95 .025 .025 =50 Rejection Non-rejection Region Region 1.96 -1.96 Rejection Region • 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 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 C2 C1 1- ( z1 2 n n ) ( z ) 2 1 25 1 25 ) ( 1.96 ) 2 2 ( 0.54) ( 4.46) 0.295 (1.96 = 0 Non-rejection Region n = 0+ ( z1 2 z1 )2 2 2 (1.96 1.28)2 22 12 42 Testing Hypothesis on the Mean with Variance Unknown (t-Test) • Null Hypothesis: H0: = 0 • Test statistic: T X 0 0 Alt. Hypothesis H1: 0 H1: > 0 H1: < 0 s n P-value 2*P(t>|t0|) P(t>t0) P(t<-t0) Rejection Criterion t0 > t/2,n-1 or t0 < -t/2,n-1 t0 > t, n-1 t0 <- t, n-1 Testing Hypothesis on the Mean with Variance Unknown (t-Test)– Example 16.3 • • • • • Claim: Length =2.5” x 2.49 n=49, s=0.021, =.05 Null Hypothesis: H0: = 2.5 Alt. Hypothesis: H1: 2.5 Test statistic: X 0 2.49 2.5 T0 3.33 s n 0.021 49 C2 C1 .95 .025 =2.5 Rejection Non-rejection Region Region 2.31 -2.31 Rejection Region • Rejection region: t.025,48= 2.3139 • P-value = 2*P(t>3.33)= .0033 • Reject H0 .025 Testing Hypothesis on the Mean with Variance Unknown (t-Test)– Example 16.4 • • • • • Claim: Battery life at least 65 mo. x 63 n=15, s=3, =.05 Null Hypothesis: H0: 65 Alt. Hypothesis: H1: < 65 Test statistic: T0 X 0 63 65 2.582 s n 3 15 • Rejection region: -t.05, 14=-2.14 • P-value = P(t<-2.582) = .0109 • Reject H0 C .95 .05 =65 Non-rejection Region -2.14 Rejection Region Testing Hypothesis on the Mean with Variance Unknown (t-Test)– Example 16.5 • • • • • Claim: Service rate = 22 customers/hr x 23.7 n=18, s=4.2, =.01 Null Hypothesis: H0: = 22 Alt. Hypothesis: H1: > 22 Test statistic: X 0 23.7 22 T0 1.717 s n 4.2 18 • Rejection region: t.01,17= 2.898 • P-value = P(t>1.717)= .0540 • Fail to reject H0 C .99 .01 =22 Rejection Non-rejection Region 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 H1: 0 S=Max(SH, SL) H1: > 0 H1: < 0 SH SL P-value (Binomial p=.5) 2*P(xS) P(xSH) P(xSL) 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 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: 2 (n 1)s 2 0 Alt. Hypothesis H1: 2 0 2 H1: 2 > 02 H1: 2 < 02 02 P-value Rejection Criterion 2*P(2 >02 ) or 2*P(2 <02 ) 02 > 2/2,n-1 or 02 < 21-/2,n-1 P(2 >02 ) P(2 <02 ) 02 > 2,n-1 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: 2 0 (n 1)s 2 2 0 (20 1)0.0153 29.07 0.01 • 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: z pˆ p0 0 p0 (1 p0 ) n Alt. Hypothesis P-value H1: p p0 P(z>z0)+P(z<-z0) H1: p > p0 P(z>z0) H1: p < p0 P(z<-z0) Rejection Criterion z0 > z1-/2 or z0 < z/2 z0 > z1- z0 < z Testing Hypothesis on the Population Proportion – Example 16.7 • • • • • Claim: Market share = 31.2% pˆ .29 n=400, =.01 Null Hypothesis: H0: p = .312 Alt. Hypothesis: H1: p .312 Test statistic: z0 pˆ p0 p0 (1 p0 ) n C2 C1 .005 .99 .005 =.312 .29 .312 Rejection .95 Rejection Non-rejection Region Region .312(1 .312) Region 400 2.576 -2.576 • 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 Testing Hypothesis on the Population Proportion – Example 16.8 • • • • • Claim: Defective rate 4% pˆ 7 / 300 .07 n=300, =.05 Null Hypothesis: H0: p = .04 Alt. Hypothesis: H1: p > .04 Test statistic: z0 pˆ p0 p0 (1 p0 ) n .07 .04 2.65 .04(1 .04) 300 • Rejection region: z1-= z.95= 1.645 • P-value = P(z>2.65) = .0040 • Reject H0 C .95 .05 =.04 Rejection Non-rejection Region Region 1.645

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