Lecture 15: Inference Based on Two Samples

Transcription

1 Lecture 15: Inference Based on Two Samples MSU-STT 351-Sum17B (P. Vellaisamy: STT 351-Sum17B) Probability & Statistics for Engineers 1 / 26

2 9.1 Z-tests and CI s for (µ 1 µ 2 ) The assumptions: (i) X = {X 1,..., X m } is a random sample from N(µ 1, σ 2 1 ) (ii) Y = {Y 1,..., Y n } is a random sample from N(µ 2, σ 2 2 ) (iii) The samples X and Y are independent. Case I: σ 2 1 and σ2 2 are known. Note E(X Y) = µ 1 µ 2 ; V(X Y) = σ2 1 m + σ2 2 n == σ2 x y. Hence, Z = (X Y) (µ 1 µ 2 ) N(0, 1), σ 2 1 m + σ2 2 n and is used to test hypothesis concerning (µ 1 µ 2 ). (P. Vellaisamy: STT 351-Sum17B) Probability & Statistics for Engineers 2 / 26

3 Case I: σ 2 1 and σ2 2 known. Suppose H 0 : µ 1 µ 2 = 0. Then the test statistic is Z = (X Y) 0 σ 2 1 m + σ2 2 n and the test can be carried out in the usual way. When µ 1 µ 2 = 1, the probability of type II error for H 1 : µ 1 µ 2 > 1 is ( β( 1 ) = Φ z α 1 ) 0. σ x y Also, the sample sizes m and n that satisfy specified α and β (when µ 1 µ 2 = 1 ) are given by σ 2 1 m + σ2 2 n = ( 1 0 ) 2 (z α + z β ) 2. (P. Vellaisamy: STT 351-Sum17B) Probability & Statistics for Engineers 3 / 26

4 Example 1 (Ex 6): An experiment to compare the tension bond strength of polymer latex modified mortar to that of unmodified mortar resulted in X = kgf/cm2 for the modified mortar (m = 40) and Y = kgf/cm2 for the unmodified mortar (n = 32). Let µ 1 and µ 2 be the true average tension bond strengths for the modified and unmodified mortars, respectively. Assume that the bond strength distributions are both normal. (a) Assume that σ 1 = 1.6 and σ 2 = 1.4, test H 0 : µ 1 µ 2 = 0 versus H 1 : µ 1 µ 2 > 0 at level α = (b) Compute the probability of a type II error for the test of Part(a) when µ 1 µ 2 = 1. (c) Suppose the investigator decided to use a level α = 0.05 test and wished β = 0.10 when µ 1 µ 2 = 1. If m = 40, what value of n is necessary? (d) How would the analysis and conclusion of Part (a) change if σ 1 and σ 2 were unknown but S 1 = 1.6 and S 2 = 1.4? (P. Vellaisamy: STT 351-Sum17B) Probability & Statistics for Engineers 4 / 26

5 Solution: (a) H 0 should be rejected if z 2.33 = z Since z = = Hence, H 0 should be rejected at level.01. ( (b) β(1) = Φ ) = Φ(.50) = (c) n = =.1169 ( ) 2 n =.0529 n = 37.06, So use n = 38. (d) Since n = 32 is a small sample, a small sample t-procedure should be used and the appropriate conclusion would follow. Note, however, that the test statistic value 3.53 would not change, and thus we would still reject H 0 at the.01 significance level. (P. Vellaisamy: STT 351-Sum17B) Probability & Statistics for Engineers 5 / 26

6 Case II: Large sample z-tests (unknown σ 2 1 σ2 2 variances; any population) Assume (i) Let X 1,..., X m is a random sample from any population with mean µ 1 and variance σ 2 1 ; (ii) Let Y 1,..., Y n is a random sample from any population with mean µ 2 and variance σ 2 2 ; (iii) The samples X = (X 1,..., X n ) and Y = (Y 1,..., Y n ) are independent. Our interest is on µ 1 µ 2, where both σ 2 1 and σ2 2 are unknown. Assume also both m and n are large. (P. Vellaisamy: STT 351-Sum17B) Probability & Statistics for Engineers 6 / 26

7 Test statistic The Z-test statistic Z = X Y (µ 1 µ 2 ) N(0, 1), S 2 1 m + S2 2 n under H 0. This statistics could be used for testiong about for µ 1 µ 2. Also, the (1 α) level confidence interval for µ 1 µ 2 is x y ± z α/2 S 2 1 m + S2 2 n. (P. Vellaisamy: STT 351-Sum17B) Probability & Statistics for Engineers 7 / 26

8 The P-Value Approach Null hypothesis H 0 : µ 1 µ 2 = 0. Alternative hypothesis H 1 : µ 1 µ 2 0 ; or H 1 : µ 1 µ 2 0 ; or H 1 : µ 1 µ 2 0. Test statistic value z = x y 0. S 2 1 m + S2 2 n The p-value is 2P(Z > z ) or P(Z < z) or P(Z > z), as per alternative hypotheses given above. (P. Vellaisamy: STT 351-Sum17B) Probability & Statistics for Engineers 8 / 26

9 Example 2 (Ex 8): Tensile strength tests were carried out on two different grades of wine rod, resulting in the following data. Grader Sample Size Sample Mean (kg/mm 2 ) Sample SD AISI 1064 m=129 X = s 1 = 1.3 AISI 1078 n=129 Y = s 2 = 2.0 (a) Does the data suggest that true average strength for the 1078 grade exceeds that for the 1064 grade by more than 10kg/mm 2? Test the appropriate hypotheses using the p-value approach. (b) Estimate the difference between true average strengths for the two grades so that it provides information about precision and reliability. (P. Vellaisamy: STT 351-Sum17B) Probability & Statistics for Engineers 9 / 26

10 Solution: (a) 1 Parameter of interest: µ 1 µ 2 = the true difference of mean tensile of the 1064 grade and the 1078 grade wire rod. Let µ 1 = 1064 grade average and µ 2 = 1078 grade average. 2 Test H 0 : µ 1 µ 2 = 10 vs H 1 : µ 1 µ 2 < 10 3 The test statistic is Z = x y 0 S 2 1 m + S2 2 n = 4 Reject H 0 if p-value < α = x y ( 10). S 2 1 m + S2 2 n. (P. Vellaisamy: STT 351-Sum17B) Probability & Statistics for Engineers 10 / 26

11 1 The observed value of Z is z = ( ) ( 10) = = For a lower-tailed test, the p-value =Φ( 28.57) 0 < α (for any value), so reject H 0. The data suggests that the mean tensile strength of the 1078 grade exceeds that of the 1064 grade by more than 10. (b) The requested information can be provided by a 95% confidence interval for µ 1 µ 2 : s 2 1 (x y) ± 1.96 m + s2 2 = ( 16) ± 1.96(.210) = ( , ) n (P. Vellaisamy: STT 351-Sum17B) Probability & Statistics for Engineers 11 / 26

12 Example 3 (Ex 12): The accompanying table gives summary data on cube compressive strength (N/mm2) for concrete specimens made with a pulverized fuel-ash mix. Age (days) Sample Size Sample Mean Sample SD Calculate and interpret a 99% CI for the difference between true average 7-day strength and true average 7-day strength and true average 28-day strength. (P. Vellaisamy: STT 351-Sum17B) Probability & Statistics for Engineers 12 / 26

13 Solution: The normal confidence interval is (note z = 2.58) x ȳ ± 2.58 s 2 1 m + s2 2 n = ( 8.77) ± = 8.77 ± 2.46 = ( 11.23, 6.31). With 99% confidence, we may say that the true difference between the average 7-day and 28-day strengths is between and 6.31 N/mm 2. (P. Vellaisamy: STT 351-Sum17B) Probability & Statistics for Engineers 13 / 26

14 Example 4: Use the accompanying data to estimate with a 95% confidence interval for the difference between true average compressive strength (N/mm 2 ) for 7-day-old concrete specimens and true average strength for 28-day-old specimens. 7-day old : n 1 = 68, x 1 = 26.99, s 1 = day old:n 2 = 74, x 2 = 35.76, s 2 = Solution: A 95% confidence interval for the difference between the true average compressive strength for 7-day-old concrete specimens and the true average strength for 28-day-old concrete specimens is: ( s 2 1 x 1 x 2 ± s2 ) 2 (4.89) 2 = ( ) ± (6.43)2 n 1 n = 8.77 ± 1.87 = ( 10.64, 6.9). (P. Vellaisamy: STT 351-Sum17B) Probability & Statistics for Engineers 14 / 26

15 9.2 Two Sample t-test and Confidence Intervals (Small Sample Situation) Assumptions: (i) The two samples are independent. (ii) Both samples are simple random samples from normal populations. (iii) The variances are unknown and unequal. The test statistic is T = X Y (µ 1 µ 2 ) t ν, S 2 1 m + S2 2 n t-distribution with df ν which is estimated from the data as ν = (s2 1 /m + s2 2 /n)2 (s 2 1 /m)2 m 1 + (s2 2 /n)2 n 1. (P. Vellaisamy: STT 351-Sum17B) Probability & Statistics for Engineers 15 / 26

16 The two sample t-test : Null hypothesis: H 0 : µ 1 µ 2 = 0. Alternative hypotheses: H 1 : µ 1 µ 2 0 ; H 1 : µ 1 µ 2 0 ; or H 1 : µ 1 µ 2 0. The Test Statistic is: T ν = x y 0 s 2 1m + s2 2n. The p-value: 2P(T ν > t ); P(T ν < t); or P(T ν > t), as per H 1 defined above. The (1 α) CI based on two sample t-test: x ȳ ± t ν,α/2 s 2 1 m + s2 2 n. (P. Vellaisamy: STT 351-Sum17B) Probability & Statistics for Engineers 16 / 26

17 Note: (i) The two-sample T-statistic does not have a t-distribution and its exact distribution involves σ 1 and σ 2. The approximation used here is quite accurate when both n 1, n 2 5, and is used in most statistical softwares. (ii) Sometimes, the t-distribution with ν = min{n 1 1, n 2 1} is also used, for simplicity. Example 4 (Ex 18): Let µ 1 and µ 2 devote true average densities for two different types of brick. Assuming normality of the two density distributions, test H 0 : µ 1 µ 2 = 0 versus H 1 : µ 1 µ 2 0 using the following data: m = 6, X = 22.73, s 1 = 0.164; n = 5, Y = and s 2 = (P. Vellaisamy: STT 351-Sum17B) Probability & Statistics for Engineers 17 / 26

18 Solution: With H 0 : µ 1 µ 2 = 0 vs H 1 : µ 1 µ 2 0, we will reject H 0 if p-value < α. Now ( (.164) 2 ) + (.240) ν = = ((.164) 2 /6) 2 ((.240) 2 /5) , The test statistic value is t = (.164) 2 + (.240)2 6 5 = = 6.17 which leads to the p-value of 2[P(T 6 > 6.17)] = 2(.0005) =.001 < α. We reject H 0 and conclude that there is a difference in the densities of the two brick types. (P. Vellaisamy: STT 351-Sum17B) Probability & Statistics for Engineers 18 / 26

19 Example 5 (Ex 32): An article gave the following summary data on provisional stress limits for specimens constructed using two different types of wood: Type of Wood Sample Size Sample Mean Sample SD Red oak Douglas fir Assuming that both samples were selected from normal distributions, carry out a test of hypotheses to decide whether the true average proportional stress limit for red oak joints exceeds that for Douglas fir joints by more than 1 MPa. (P. Vellaisamy: STT 351-Sum17B) Probability & Statistics for Engineers 19 / 26

20 Solution: Let µ 1 = the true average proportional stress limit for red oak and let µ 2 = the true average proportional stress limit for Douglas fir. We test H 0 : µ 1 µ 2 = 1 vs H 1 : µ 1 µ 2 > 1. The test statistic s value is t = ( ) = = and has degrees of freedom ν = (79 2 /14) 2 13 (.2084) 2 + (1.282 /10) 2 9 = The p-value=p(t 13 > 1.8) = We would reject H 0 at significance levels greater than.046 (e.g., the standard 5% significance level). At α =.05, the data suggests that true average proportional stress limit for red oak exceeds that of Douglas fir by more than 1 MPa. (P. Vellaisamy: STT 351-Sum17B) Probability & Statistics for Engineers 20 / 26

21 Pooled t-test (unknown but equal variances σ 2 1 = σ2 2 = σ2 ) When the normal population variances are the equal (i.e., σ 2 1 = σ2 2 = σ2 ), the pooled estimator of common σ 2 is S 2 p = (m 1) (n 1) m + n 2 S2 1 + m + n 2 S2 2. and the pooled t-statistic for H 0 : µ 1 = µ 2 is T = (X Y) (µ 1 µ 2 ), 1 s p m + 1 n which follows t distribution with ν = (m + n 2) df. (P. Vellaisamy: STT 351-Sum17B) Probability & Statistics for Engineers 21 / 26

22 Example 6 (Ex.34): Consider the pooled T-variable T = (X Y) (µ 1 µ 2 ) 1 s p m + 1 n t m+n 2 when both population distributions are normal with σ 1 = σ 2. (a) Using T variable, get a pooled t confidence interval for µ 1 µ 2. (b) A sample on maximum output of moisture (oz) in a controlled chamber of an ultrasonic humidifier (Brand 1) were 14.0, 14.3, 12.2, and A sample of the second brand (Brand 2) gave output values 12.1, 13.6, 11.9, and Use the pooled t formula from Part (a) to estimate the difference between true average outputs for the two brands with 95% confidence interval. (c) Estimate the difference between the two µ s using the two-sample t interval, and compare it to the interval of Part (b). (P. Vellaisamy: STT 351-Sum17B) Probability & Statistics for Engineers 22 / 26

23 Solution: (a) Following the usual format for most confidence intervals, a pooled variance confidence interval for the difference between two means is (x y) ± t α/2,m+n 2.S p 1 m + 1 n. (b) The sample means and standard deviations of the two samples are x = 13.90, s 1 = 1.225, y = 12.20, s 2 = The pooled variance estimate is ( Sp 2 m ( 1 = )S 21 m + n 2 + n 1 m + n 2 ( 4 ) ( 1 = (1.225) = ) S ) (1.010) 2 (P. Vellaisamy: STT 351-Sum17B) Probability & Statistics for Engineers 23 / 26

24 Hence, S p = With df = m + n 2 = 6, t.025,6 = Therefore, the desired interval is 1 ( ) ± (2.447)(1.1227) = 1.7 ± = (.24, 3.64). 4 This interval contains 0, so it does not support the conclusion that the two population means are different. (P. Vellaisamy: STT 351-Sum17B) Probability & Statistics for Engineers 24 / 26

25 (c) Using the two-sample t interval discussed earlier, we find the CI as follows: First, we need to calculate the degrees of freedom. ν = ( ( ( ) So, t.025,5 = Then the interval is ( ) ± ) 2 ) =.3971 = = 1.7 ±2.571(0.7938) = ( 0.34, 3.74). This interval is slightly wider, but it still supports the same conclusion. (P. Vellaisamy: STT 351-Sum17B) Probability & Statistics for Engineers 25 / 26

26 Home work Sec 9.1: 5, 7, 11 Sec 9.2: 19, 22, 30 (P. Vellaisamy: STT 351-Sum17B) Probability & Statistics for Engineers 26 / 26