STAT 211 Handout 9 (Chapter 9): Inferences Based on Two Samples TESTS 1. One-sided (One tailed) test: A. Lower tailed: H0: population characteristics claimed constant value (Left-sided) Ha: population characteristics < claimed constant value B. Upper tailed: H0: population characteristics claimed constant value (Right-sided) Ha: population characteristics > claimed constant value 2. Two-sided (Two tailed) test: H0: population characteristics = claimed constant value Ha: population characteristics claimed constant value Hypothesis testing for: A. Independent Samples I. Population characteristics: Difference between two population means, 1-2. 0 is the claimed constant. 1 and 2 are the population means for X's and Y's, respectively. _ _ x and y are the sample means for X's and Y's, respectively. m and n are the sample sizes for X's and Y's, respectively. 12 and 22 are the population variances for X's and Y's, respectively. s12 and s 22 are the sample variances for X's and Y's, respectively. Test statistics: _ _ x y 0 z 2 1 m when both popn. distributions are normal and 12 , 22 are known n x y 0 z when there is large sample size (m>40 and n>40) and 12 , 22 are 2 2 s1 s 2 m n unknown _ _ x y 0 t when both popn. distributions are normal and at least one sample size is 2 s1 s 22 m n small with unknown 12 , 22 are assumed to be different ( 12 22 ) and degrees of freedom, _ 2 2 _ 2 s12 s 22 m n is used to look up the critical values. If v is not computed as integer, v 2 2 s12 / m s 22 / n m 1 n 1 it should be rounded down. Your textbook calls this two-sample t test. _ _ x y 0 (m 1) s12 (n 1) s 22 , s 2p when both popn. distributions are normal and at t mn2 1 1 sp m n least one sample size is small where unknown 12 , 22 are assumed to be the same ( 12 = 22 ) and degrees of freedom, v m n 2 is used to look up the critical values. Your textbook calls this pooled t test. Decision can be made in one of the two ways: a. Let z* or t* be the computed test statistic values. if test statistics is z Lower tailed test P-value = P(z<z*) Upper tailed test P-value = P(z>z*) Two-tailed test P-value = 2P(z>|z*|)= 2P(z<-|z*|) if test statistics is t P-value = P(t<t*) P-value = P(t>t*) P-value = 2P(t > |t*| )= 2P(t <- |t*| ) In each case, you can reject H0 if P-value and fail to reject H0 (accept H0) if P-value > b. Rejection region for level test: if test statistics is z Lower tailed test z -z Upper tailed test z z Two- tailed test z -z/2 or z z/2 if test statistics is t t -t;v t t;v t -t/2;v or t t/2;v 100(1-)% confidence Intervals with the same assumptions, 12 22 _ _ x y z /2 m n s2 s2 _ _ x y z / 2 1 2 m n s12 s 22 _ _ when you assume 12 22 x y t / 2 ; v m n 1 1 _ _ when you assume 12 = 22 x y t / 2 ;v s p m n Example 1: Two types of plastic are suitable for use by an electronics component manufacturer. The breaking strength of this plastic is very important. It is known that 1 = 2 =1 psi. From a _ _ random sample of size n1 =10 and n 2 =12, we obtain x 1 =162.5 and x 2 =155. The company will not adopt plastic 1 unless its mean breaking strength exceeds that of plastic 2 by at least 10 psi. Based on the sample information, would they use plastic 1? Use the significance level 0.05 in reaching a decision. H 0 : 1 2 10 (adopt plastic 1) versus H a : 1 2 10 (do not adopt plastic 1) _ _ x y 0 (162.5 155) 10 5.84 test statistics, z 2 2 1 2 12 12 10 12 m n Decision: (i) reject H0 if z -z=-1.645. z =-5.84 < -1.645 then reject H0. (ii) P-value = P(Z<-5.84)=P(Z>5.84)=0. Since the P-value =0.05, reject H0. Conclusion: Do not adopt plastic 1. Example 2 (Exercise 9.2): Sample sizes are large while population variances are unknown. Notice that this is a two-tailed test. _ _ x y 0 (42500 40400) 0 Test statistics: z =4.8462 2 2 s1 s 2 2200 2 1900 2 45 45 m n Decision: (i) Reject H0 if z -z/2=-1.96 or z z/2 =1.96. z =4.8462 > 1.96 then reject H0. (ii) P-value = 2P(Z>4.8462)=2(0)=0. Since the P-value =0.05, reject H0. Conclusion: true average tread lives for two competing brand tires are different. If you prefer to answer the question computing the confidence interval, 95% confidence interval s12 s 22 2200 2 1900 2 _ _ would become x y z / 2 (42500 40400) 1.96 m n 45 45 =(1250.67, 2949.33). You would see that zero does not fall into interval and you would reject H0. Example 3 (Exercise 9.8): (a) Sample sizes are large while population variances are unknown. . Notice that this is a uppertailed test. H 0 : 2 1 10 versus H a : 2 1 10 _ _ y x 0 (123.6 107.6) 10 Test statistics: z =28.57 2 2 s1 s 2 1.3 2 2 2 129 129 m n Decision: (i) Reject H0 if z z =1.645 if =0.05. z =28.57 > 1.645 and reject H0. (ii) P-value = P(Z>28.57)=0. Since the P-value =0.05, reject H0. Conclusion: the data provide compelling evidence that the true average strength for the 1078 grade exceeds that for the 1064 grade by more than 10. Or you can answer the same question using H 0 : 1 2 10 versus H a : 1 2 10 s2 s2 _ _ (b) 95% confidence interval for 1 2 is x y z / 2 1 2 m n = 107.6 123.6 1.96 1.3 2 2 2 =(-16.84 , -15.16) 129 129 s2 s2 _ _ 95% confidence interval for 2 1 is y x z / 2 1 2 =(15.16 , 16.84) m n Example 4 (Exercise 9.18): Sample sizes are small and population variances are unknown. . Notice that this is a two-tailed test. _ _ x y 0 (22.73 21.95) 0 Test statistics: t =6.166 s12 s 22 0.164 2 0.24 2 6 5 m n Decision: (i) Reject H0 if t -t/2;v= -2.447 or t t/2;v =2.447 where =0.05 and v=6.886. t =6.166 > 2.447 and reject H0. (ii) P-value = 2P(t>6.166)=2(0)=0. Since the P-value =0.05, reject H0. Conclusion: the true average densities for two different types of brick are different. II. Population characteristics: Difference between two population proportions, p1-p2. p0 is the claimed constant. ^ ^ p 1 and p 2 are the sample proportions for X's and Y's, respectively. p1 and p 2 are the population proportions for X's and Y's, respectively. m and n are the large sample sizes for X's and Y's, respectively. X Y m ^ n ^ p1 p2 mn mn mn ^ ^ p p 1 2 p0 Test statistics: z ^ ^ 1 1 p1 p m n Decision can be made in one of the two ways: ^ The estimator for p is p (a) Let z* be the computed test statistic values. Lower tailed test P-value = P(z<z*) Upper tailed test P-value = P(z>z*) Two-tailed test P-value = 2P(z > |z*| )=2P(z <- |z*| ) In each case, you can reject H0 if P-value and fail to reject H0 (accept H0) if P-value > (b) Rejection region for level test: Lower tailed test z -z Upper tailed test z z Two- tailed test z -z/2 or z z/2 100(1-)% large sample confidence Interval: ^ ^ p1 p 2 z / 2 ^ ^ ^ ^ p1 1 p1 p 2 1 p 2 m n Example 5: Two different types of injection-molding machines are used to form plastic parts. A part is considered defective if it has excessive shrinkage or is discolored. Two random samples, each of size 300 are selected and 15 defective parts are found from machine 1 while 8 defective parts are found in the sample from machine 2. Is it reasonable to conclude that both machines produce the same fraction of defective parts, using the significance 0.05? If this analysis done by hand Sample sizes are large enough to satisfy the assumptions and it is a two-tailed test. ^ ^ ^ X Y 15 8 15 8 p =0.0383 where p 1 =0.05 and p 2 =0.0267 m n 300 300 300 300 ^ ^ p1 p 2 p0 (0.05 0.0267) 0 Test statistics: z =1.49 ^ ^ 0.0383(0.9617)(1 / 300 1 / 300) 1 1 p1 p m n (i) Reject H0 if z -z/2=-1.96 or z z/2 =1.96. -1.96< z=1.49 < 1.96 and fail to reject H0. . (ii) P-value = 2P(Z>1.49)=2(0.0681)=0.1362. Since the P-value > =0.05, fail to reject H0. Conclusion: Yes it is reasonable to assume that both machines produce the same fraction of defective parts The MINITAB output analyzing such data is Test and CI for Two Proportions Sample X N Sample p 1 15 300 0.050000 2 8 300 0.026667 Estimate for p(1) - p(2): 0.0233333 95% CI for p(1) - p(2): (-0.00733568, 0.0540023) Test for p(1) - p(2) = 0 (vs not = 0): Z = 1.49 P-Value = 0.136 Test for single true proportions as an example Test of p = 0.04 vs p not = 0.04 Sample X N Sample p 95.0% CI 1 15 300 0.050000 (0.028251, 0.081127) 2 8 300 0.026667 (0.011582, 0.051866) Exact P-Value 0.461 0.247 Example 6 (Exercise 9.48(a)): Sample sizes are large enough to satisfy the assumptions and it is a two-tailed test. ^ ^ ^ X Y 63 75 63 75 p =0.2875 where p 1 =0.21 and p 2 =0.4167 m n 300 180 300 180 ^ ^ p1 p 2 p0 (0.21 0.4167) 0 Test statistics: z =-4.844 ^ ^ 0.2875(0.7125)(1 / 300 1 / 180) 1 1 p1 p m n Decision: (i) Reject H0 if z -z/2=-1.96 or z z/2 =1.96. |z| =4.844 > 1.96 and reject H0. (ii) P-value = 2P(Z>4.844)=2(0)=0. Since the P-value =0.05, reject H0. Conclusion: it is different for two groups of residents. The MINITAB output analyzing such data is Test and CI for Two Proportions Sample X N Sample p 1 63 300 0.210000 2 75 180 0.416667 Estimate for p(1) - p(2): -0.206667 95% CI for p(1) - p(2): (-0.292174, -0.121159) Test for p(1) - p(2) = 0 (vs not = 0): Z = -4.74 P-Value = 0.000 Population characteristics: Ratio of the two population variances, 12 / 22 or standard deviations, 1 / 2 . X and Y's are random sample from a normal distribution. 12 and 22 are the population variances for X's and Y's, respectively. s12 and s 22 are the sample variances for X's and Y's, respectively. m and n are the sample sizes for X's and Y's, respectively. s2 Test statistics: F 12 s2 III. Decision can be made in one of the two ways: (a) Let F* be the computed test statistic values. Lower tailed test P-value = P(F<F*) Upper tailed test P-value = P(F>F*) Two-tailed test P-value = 2P(F > F*) In each case, you can reject H0 if P-value and fail to reject H0 (accept H0) if P-value > (b) Rejection region for level test: Lower tailed test F F1-;m-1,n-1 Upper tailed test F F;m-1,n-1 Two- tailed test F F1-/2;m-1,n-1 or F F/2;m-1,n-1 Notice that F1-/2;m-1,n-1 = 1 / F/2;n-1,m-1 100(1-)% confidence Interval for 12 / 22 : s12 / s 22 F / 2;m 1,n 1 12 s12 / s 22 22 F1 / 2;m 1,n 1 Example 7 (Exercise 9.57): (a) On the F-table, column for 5 and row for 8 will give the area on the right 0.05 with F0.05:5,8 = 3.69 (d) F0.95:8,5 = 1/ F0.05:5,8 = 1/3.69 =0.271 (e) The percentile is the area on the left of the value and it means the area on the right of the value is 0.01. On the F-table, look at column for 10 and row for 12 with the area on the right 0.01. P( F F0.01:10,12 )= 0.99 then F0.01:10,12 = 4.30 (h) P(0.177 F 4.74) = P(F0.99:10,5 F F0.05:10,5) =1-(0.01+0.05)=0.94 where F0.99:10,5 = 1/F0.01:5,10 = 1/5.64=0.177 Example 8: A study was performed to determine whether men and women differ in their repeatability in assembling components on printed circuit boards. Two samples of 26 men and 21 women were selected and each subject assembled the units. The two sample standard deviations of assembly time were smen=0.98 min and swomen=1.02 min. Is there evidence to support the claim that men and women differ in repeatability for this assembly task? Use the significance level 0.02 and state any necessary assumptions about underlying distribution of the data. 2 2 H 0 : men women 2 2 H a : men women =0.10, m=26, n=21 s2 0.98 2 0.9231 Test statistics: F 2men s women 1.02 2 Decision: Reject H0 if F F1-/2;m-1,n-1 = F0.99;25,20 =1/2.70=0.37 or F F/2;m-1,n-1= F0.01;25,20 =2.84. Since 0.37 < F=0.9231 <2.84, fail to reject H0. Conclusion: men and women do not differ in repeatability for this assembly task Example 9 (Exercise 9.62): H 0 : 12 22 H a : 12 22 =0.10, m=48, n=45 s 2 21.45 2 Test statistics: F 12 1.2219 s 2 19.45 2 Decision: Reject H0 if F F1-/2;m-1,n-1 = F0.95;47,44 =1/1.6336=0.61214 or F F/2;m-1,n-1= F0.05;47,44 =1.6412 (The F-table on your book is not very detailed, you can use both degrees of freedom as 40 to look at the table. I gave you the real values using the computer programs.) Since 1.2219 is not larger than 1.6412 or smaller than 0.61214, fail to reject H0. Conclusion: the variances are the same. B. Dependent Samples- Paired Data Population characteristics: Difference between two population means, D =1-2. 0 is the claimed constant. Assumption: the difference distribution should be normal. _ Test statistics: t d 0 _ where D=X-Y and d and s D are the corresponding sample sD / n average and the standard deviation of D. Both X and Y must have n observations. Decision can be made in one of the two ways: (a) Let t* be the computed test statistic values. Lower tailed test P-value = P(t<t*) Upper tailed test P-value = P(t>t*) Two-tailed test P-value = 2P(t > |t*| )=2P(t <- |t*| ) In each case, you can reject H0 if P-value and fail to reject H0 (accept H0) if Pvalue > (b) Rejection region for level test: Lower tailed test t -t;n-1 Upper tailed test t t;n-1 Two- tailed test t -t/2;n-1 or t t/2;n-1 100(1-)% confidence Intervals with the same assumptions, _ d t / 2;n 1 sD n Example 10: The manager of a fleet of automobiles is testing two brands of radial tires. He assigns one tire of each brand at random to the two rear wheels of eight cars and runs the cars until the tires wear out. The descriptive statistics for the data are shown below (in kilometers). Find the 99% confidence interval on the difference in mean life. Which brand would you prefer based on this calculation? Is there an alternative method to answer this question instead of computing the confidence interval? Variable N Mean Median StDev SE Mean Minimum Maximum Q1 Q3 Brand1 8 38479 37067 5590 1976 32100 48360 34185 43525 Brand2 8 37611 36655 5244 1854 31950 47800 33491 41214 Difference 8 868 475 1290 456.1 -805 3020 N/A N/A Brand1 Brand2 Difference 36925 45300 36240 32100 37210 48360 38200 33500 34318 42280 35500 31950 38015 47800 37810 33215 2607 3020 740 150 -805 560 390 285 _ 99% confidence interval for 1 2 is d t / 2;n 1 sD 868 3.499 1290 =(-727.84 , 2463.84) n 8 Confidence interval only tells us that there is no difference between those brands. Alternative method is the confidence interval H 0 : 1 2 versus H a : 1 2 _ test statistics, t d 0 868 0 s D / n 1290 / 8 Or H 0 : 1 2 versus H a : 1 2 _ test statistics, t d 0 =1.90 < t;n-1 = t0.01;7 =2.998 then fail to reject H0. 868 0 =1.90 > - t;n-1 = -t0.01;7 = -2.998 then fail to reject H0. s D / n 1290 / 8 So alternative method gives us the same conclusion. The MINITAB output analyzing such data is B1 B2 Difference N 8 8 8 Mean 38479 37611 868 StDev 5590 5244 1290 SE Mean 1976 1854 456 Test for equal true means in independent samples assuming equal true variances Difference = mu B1 - mu B2 Estimate for difference: 868 95% CI for difference: (-4944, 6681) T-Test of difference = 0 (vs not =): T-Value = 0.32 P-Value = 0.753 DF = 14 Both use Pooled StDev = 5420 Test for equal true means in independent samples assuming unequal true variances Difference = mu B1 - mu B2 Estimate for difference: 868 95% CI for difference: (-4986, 6723) T-Test of difference = 0 (vs not =): T-Value = 0.32 P-Value = 0.754 DF = 13 Test for equal true means in dependent samples 95% CI for mean difference: (-211, 1948) 99% CI for mean difference: (-728, 2465) T-Test of mean difference = 0 (vs not = 0): T-Value = 1.90 Test for true means in single samples Test of mu = 3800 vs mu not = 3800 Variable 95.0% CI T B1 ( 33803, 43156) 17.55 B2 ( 33224, 41998) 18.24 P-Value = 0.099 P 0.000 0.000 Test for equal variances: B1 versus B2 F-Test (normal distribution) Test Statistic: 1.136 P-Value : 0.870 Example 11 : An experiment to compare the yield (kg/ha) of Sundance winter wheat and Manitou spring wheat is considered. Data from nine different plots is given in the following table. Is there sufficient evidence to conclude that true average yield for the Sundance winter wheat is more than 500 kg/ha than the Manitou spring wheat? Check the plausibility of any assumptions needed to carry out an appropriate test of hypothesis. 1 2 3 4 5 6 7 8 9 S 3201 3095 3297 3644 3604 2860 3470 2042 3689 M 2386 2011 2616 3094 3069 2074 2308 1525 2779 D=S-M 815 1084 681 550 535 786 1162 517 910 H 0 : S M 500 or H 0 : S M 500 H a : S M 500 or H a : S M 500 The difference distribution should be normal. Normal Probability Plot for D ML Estimates - 95% CI 99 ML Estimates 95 Mean 782.222 StDev 223.197 90 Goodness of Fit Percent 80 AD* 70 60 50 40 30 20 10 5 1 0 500 1000 Data 1500 1.416 _ Use the differences and compute the sample mean, d =782.2222 s D =236.736. _ Test statistics: t d 0 sD / n 782.2222 500 236.736 / 9 and the standard deviation, =3.5764 Decision: (i) Since this is an upper tailed test, reject H0 if t t;n-1= t0.05;8 =1.86 or if the Pvalue=P(t>3.5764) is less than =0.05. Notice that test statistics, 3.5764 is large than 1.86. (ii) 0.001 <P-value < 0.005 by looking at the t-table. Reject H0 Conclusion: there is sufficient evidence to conclude that the true average yield in winter wheat is more than 500 higher than for spring wheat.