Introduction to Statistical Hypothesis Testing

Transcription

1 Introduction to Statistical Hypothesis Testing Arun K. Tangirala Hypothesis Testing of Variance and Proportions Arun K. Tangirala, IIT Madras Intro to Statistical Hypothesis Testing 1

2 Learning objectives I One-sample test for variance I Two sample tests for ratio of variances I One-sample test for proportion I Two-sample test for difference in proportions with illustrations in R. Arun K. Tangirala, IIT Madras Intro to Statistical Hypothesis Testing

3 One sample test for variance Goal: Test for variance of a normal population Assumption: Random sample 1. Null: H 0 : = 0 Alternate: H a : <, >, 6= 0 (lower-, upper-, Two-tailed). Test statistic Q T : C (n 1)S = (n 1) 0 3. Critical region R c : c < 1 (n 1), c > (n 1), {c < 1 / (n 1) or c > / (n 1)}. R: Use vartest from the EnvStats package Arun K. Tangirala, IIT Madras Intro to Statistical Hypothesis Testing 3

4 Graphical understanding: One sample test for variance Cri.cal region Acceptance region Cri.cal region Cri$cal region Cri$cal region (a) Two-sided test (b) Lower-tailed test (c) Upper-tailed test Arun K. Tangirala, IIT Madras Intro to Statistical Hypothesis Testing 4

5 One sample test for variance: Example 1 Example: Automated filling machine To test: The variability of fill volume by an automated filling machine is greater than From data, n = 0, s = Solution: H 0 : =0.01, H a : > 0.01, Statistic: c (n 1)s = = Critical value: 0.05 (19) = Fail to reject H 0 at = R: Use qchisq with df=19 to compute the critical value Arun K. Tangirala, IIT Madras Intro to Statistical Hypothesis Testing 5

6 One sample test for variance: Example Example: Variance of Process A yield To test: The yield of process A is from N (75.5, 1.5 ). From data, n = 50, SA =.05 Solution: H 0 : A =1.5, H a : A 6=1.5, Statistic: c = Boundaries: 0.05 (49) = 31.6, (49) = 70.. Fail to reject H 0 at =0.05. R: Use qchisq with df=49 to compute critical values Arun K. Tangirala, IIT Madras Intro to Statistical Hypothesis Testing 6

7 Two sample test for variance Goal: Compare variances of two normal populations (test for ratio) Assumption: Mutually independent populations 1. Null: H 0 : 1 = Alternate: H a : 1 <, >, 6= (lower-, upper-, Two-tailed). Test statistic Q T : F = S 1 S F ( 1, ), 1 = n 1 1, = n Critical region R c : f<f 1 ( 1, ), f>f ( 1, ), {f <F 1 / ( 1, ) or f>f / ( 1, )}. R: Use var.test from the stats package Arun K. Tangirala, IIT Madras Intro to Statistical Hypothesis Testing 7

8 Two sample test for variance: Example 1 Example: Comparing oxide layer thickness in semiconductor wafers To test: Variability in oxide layer thickness is superior with one mixture gas than another. From data, n 1 = 16, s 1 =1.96 A ; n = 16,s =.13 A. Solution: H 0 : 1 =, H a : 1 6=, Statistic: f =0.85. Boundaries: f 0.05 (15, 15) =.86,f (15, 15) = 1/f 0.05 (15, 15) = Fail to reject H 0 at =0.05. Arun K. Tangirala, IIT Madras Intro to Statistical Hypothesis Testing 8

9 Two sample test for variance: Example Example: Comparing variances of yields To test: The variability in yields of two processes are identical. From data, n 1 = 50, SA =.05; n = 50,SB =7.64. Solution: H 0 : A = B, H a : A 6= B, Statistic: f =0.7. Boundaries: f 0.05 (49, 49) = 0.567,f (49, 49) = Reject H 0 at =0.05. Arun K. Tangirala, IIT Madras Intro to Statistical Hypothesis Testing 9

10 Tests for proportion Goal: Test for proportion of successes from a binomial population Assumption: Random and large sample Test for large sample assumption: The interval I 0 = p 0 ± 3 p p 0 (1 p 0 )/n should not include 0 or Hypothesis: H 0 : p = p 0, H a : p<,>,6= p 0 X np 0. Test statistic Q T : Z = p np0 (1 p 0 ) = p p 0 N(0, 1), =X/n. p0 (1 p 0 )/n 3. Critical region R c : z< z, z>z, {z < z / or z>z / }. Arun K. Tangirala, IIT Madras Intro to Statistical Hypothesis Testing 10

11 Tests for proportion: Example 1 Example: Defective controllers Manufacturer claims a maximum of 5% defective controllers. A random sample of n = 00 devices drawn reveals x =4(four) defective items. If the customer wishes to test that the proportion of defective items exceeds p 0 =0.05, H a : p>p 0 with H 0 : p = p 0. Solution: I 0 =(0.004, 0.096). Therefore,samplelargeenough. Statistic: z = Critical value: z c = Fail to reject H 0 at =0.05. Arun K. Tangirala, IIT Madras Intro to Statistical Hypothesis Testing 11

12 Tests for proportion: Example 1 Example: Defective controllers Manufacturer claims a maximum of 5% defective controllers. A random sample of n = 00 devices drawn reveals x =4(four) defective items. If the customer wishes to test that the proportion of defective items exceeds p 0 =0.05, H a : p>p 0 with H 0 : p = p 0. Solution: I 0 =(0.004, 0.096). Therefore,samplelargeenough. Statistic: z = Critical value: z c = Fail to reject H 0 at =0.05. Q.1: What if the manufacturer wishes to test his/her claim? Arun K. Tangirala, IIT Madras Intro to Statistical Hypothesis Testing 1

13 Tests for proportion: Example 1 Example: Defective controllers Manufacturer claims a maximum of 5% defective controllers. A random sample of n = 00 devices drawn reveals x =4(four) defective items. If the customer wishes to test that the proportion of defective items exceeds p 0 =0.05, H a : p>p 0 with H 0 : p = p 0. Solution: I 0 =(0.004, 0.096). Therefore,samplelargeenough. Statistic: z = Critical value: z c = Fail to reject H 0 at =0.05. Q.1: What if the manufacturer wishes to test his/her claim? Q.: What is the least value of x that would have resulted in a rejection of the H 0 by the customer? Arun K. Tangirala, IIT Madras Intro to Statistical Hypothesis Testing 13

14 Tests for proportion: Example Example: Exam type preference of students To test: Proportion of students preferring closed-book" exam format is 0.8. From data, n = 100, =0.75 Solution: I 0 =(0.68, 0.9). Therefore,samplelargeenough. H 0 : p =0.8, H a : p 6= 0.8 Statistic: z = 1.5. Critical region: z c = ±1.96. Fail to reject H 0 at =0.05. Arun K. Tangirala, IIT Madras Intro to Statistical Hypothesis Testing 14

15 Two-sample tests for proportion Goal: Test for difference in proportions from binomial populations Assumption: Random samples 1. Hypothesis: H 0 : p 1 p = 0, H a : p 1 p <, >, 6= 0. Test statistic: Z = ( 1 ) 0 q N(0, 1). p1 q 1 n 1 + p q n 1 = X 1 /n 1, = X /n.replacep 1,p with ˆp 1 = x 1 /n 1, ˆp = x /n. When 0 =0,usepooled proportion to estimate denominator: ˆp = x 1+x n 1 +n. 3. Critical region R c : z< z, z>z, {z < z / or z>z / }. Arun K. Tangirala, IIT Madras Intro to Statistical Hypothesis Testing 15

16 Two-sample tests for proportion: Example Example: Regional preference for PEPSI To test: Proportion of students preferring PEPSI on campus A is the same as that on campus B. From data, n 1 = 15,x 1 = 44, n = 15,x = 6 =) 1 =0.35, =0.08, ˆp =0.8. Solution: H 0 : p 1 = p, H a : p 1 p 6=0, Statistic: z =.54. Critical value: z c =1.96. Reject H 0 at =0.05. Arun K. Tangirala, IIT Madras Intro to Statistical Hypothesis Testing 16

17 Bibliography I Bendat, J. S. and A. G. Piersol (010). Random Data: Analysis and Measurement Procedures. 4 th edition. New York, USA: John Wiley & Sons, Inc. Johnson, R. A. (011). Miller and Freund s: Probability and Statistics for Engineers. Upper Saddle River, NJ, USA: Prentice Hall. Montgomery, D. C. and G. C. Runger (011). Applied Statistics and Probability for Engineers. 5 th edition. New York, USA: John Wiley & Sons, Inc. Ogunnaike, B. A. (010). Random Phenomena: Fundamentals of Probability and Statistics for Engineers. Raton, FL, USA: CRC Press, Taylor & Francis Group. Boca Tangirala, A. K. (014). Principles of System Identification: Theory and Practice. Group. CRC Press, Taylor & Francis Arun K. Tangirala, IIT Madras Intro to Statistical Hypothesis Testing 17