Chapter 17 Estimation and Hypothesis Tests: Two Populations

Two Independent Populations 2222 11 1212 22 n1n1 n2n2

Difference in Means: Variance Known 

Testing Hypothesis on the Difference in Means: Variance Known Alt. HypothesisP-valueRejection Criterion H 1 :    0 P(z>z 0 )+P(z<-z 0 )z 0 > z 1-  /2 or z 0 < z  /2 H 1 :  >  0 P(z>z 0 )z 0 > z 1-  H 1 :  <  0 P(z<-z 0 )z 0 < z  Null Hypothesis: H 0 :  1 -  2 =  0 Test statistic:

Type II Error and Sample Size

Type II Error (OC Curve) Two-sided,  =.05 Montgomery, D.C., Runger G.C., “Applied Statistics and Probability for Engineers”, 5 th Ed., 2010, John Wiley

Claim: Drying times of 2 different paints are same n 1 =n 2 =10,  1 =  2 =8,  =.05 Null Hypothesis: H 0 :  1 -  2 = 0 Alt. Hypothesis: H 1 :  1 >  2 Test statistic: Rejection region: z.95 = P-value = P(z>2.52)=.0059 Reject H C Non-rejection Region Rejection Region Testing Hypothesis on the Difference of Means with Variance Known – Example

Confidence Interval for Difference in Means: Variance Known Confidence Interval One-sided Upper Confidence Bound One-sided Lower Confidence Bound

Choice of Sample Size If n 1 = n 2 = n

Tensile Strength n 1 =10 n 2 =12,  1 =1  2 =1.5,  =.10 Confidence Interval on the Difference of Means with Variance Known – Example

Two Independent Populations Variance Unknown 11 22 n1n1 n2n2 Case 1:  1 2 =  2 2 =  2

Difference in Means: Variance Unknown (but equal)  Pooled Estimator of  2 Student – t

Testing Hypothesis on the Difference in Means: Variance Unknown (but equal) Null Hypothesis: H 0 :  1 -  2 =  0 Test statistic: Alt. HypothesisP-valueRejection Criterion H 1 :    0 2*P(t>|t 0 |) t 0 > t  /2,n1+n2-2 or t 0 < -t  /2, n1+n2-2 H 1 :  >  0 P(t>t 0 )t 0 > t , n1+n2-2 H 1 :  <  0 P(t<-t 0 )t 0 <- t , n1+n2-2

B 12 in Energy drink n 1 =14, n 2 =16, s 1 =4.7, s 2 =3.9,  =.01 Null Hypothesis: H 0 :  1 -  2 = 0 Alt. Hypothesis: H 1 :  1 ≠  2 Test statistic: Rejection region: ±t.005,28 = ±2.763 P-value = 2P(t>|3.822|)= Reject H 0 Errors in book! Testing Hypothesis on the Difference of Means with Variance Unknown (but equal) – Ex. 17.5

Claim: Yield from a catalyst n 1 =n 2 =8,  1 =  2,  =.05 Null Hypothesis: H 0 :  1 -  2 = 0 Alt. Hypothesis: H 1 :  1 ≠  2 Test statistic: Rejection region: ±t.025,14 = ±2.145 P-value = 2P(t>|-0.35|)=.7315 Fail to reject H 0 Testing Hypothesis on the Difference of Means with Variance Unknown (but equal) – Example Cat. 1Cat avg s

Confidence Interval for Difference in Means: Variance Unknown (but equal) Confidence Interval One-sided Upper Confidence Bound One-sided Lower Confidence Bound

Caffeine content n 1 =15 n 2 =12, s 1 =5.3 s 2 =6.7,  =.05 Confidence Interval on the Difference of Means with Variance Unknown (but equal) – Ex. 17.3

Two Independent Populations Variance Unknown 11 22 n1n1 n2n2 Case 2:  1 2 ≠  2 2

Difference in Means: Variance Unknown  Student – t

Testing Hypothesis on the Difference in Means: Variance Unknown Null Hypothesis: H 0 :  1 -  2 =  0 Test statistic: Alt. HypothesisP-valueRejection Criterion H 1 :    0 2*P(t>|t 0 |) t 0 > t  /2,  or t 0 < -t  /2, H 1 :  >  0 P(t>t 0 )t 0 > t , H 1 :  <  0 P(t<-t 0 )t 0 <- t ,

Promotional campaign effect n 1 =28, n 2 =24, s 1 =29, s 2 =13,  =.05 Null Hypothesis: H 0 :  1 -  2 = 0 Alt. Hypothesis: H 1 :  1 >  2 Test statistic: Rejection region: t.05,38 = P-value = P(t>5.5837)= 1.06E-6 Reject H 0 Testing Hypothesis on the Difference of Means with Variance Unknown – Example 17.6 Errors in book!

Confidence Interval for Difference in Means: Variance Unknown Confidence Interval One-sided Upper Confidence Bound One-sided Lower Confidence Bound

Construction workers’ wage vs. Manuf. workers’ wage n 1 =500 n 2 =700, s 1 =3.26 s 2 =1.42,  =.10 Confidence Interval on the Difference of Means with Variance Unknown – Ex. 17.4

Testing Hypothesis on the Difference of Medians – Mann Whitney Test Null Hypothesis: H 0 :  1 =  2 Test statistic: H 1 :  1 <  2 Critical value U c (from Table VI, with n 1, n 2,  ) Reject H 0 if U<U c H 1 :  1 >  2 Find U’ (from Table VI, with n 1, n 2,  ), U c = n 1 n 2 - U’ Reject H 0 if U>U c H 1 :  1 ≠  2 Find U lower (from Table VI, with n 1, n 2,  /2), U upper = n 1 n 2 - U lower Reject H 0 if U U upper Where S 1 is the sum of ranks of sample 1

Testing Hypothesis on the Difference of Medians – Example 17.9  =.05 Null Hypothesis: H 0 :  1 =  2 Test statistic: H 1 :  1 ≠  2 U lower = 4 (Table VI, with n 1 =5, n 2 =6,.025), U upper = n 1 n 2 – U lower = (5)(6)-4 = 26 Reject H 0 if U U upper Fail to reject H 0

Alt. HypothesisP-valueCritical Regions H 1 :  1   2 2P(z  |z 0 |) z 0 z 1-  /2 H 1 :  1 <  2 P(z  z 0 ) z 0 <z  H 1 :  1 >  2 P(z  z 0 ) z 0 >z 1-  Null Hypothesis: H 0 :  =  0 Test statistic: Testing Hypothesis on the Difference of Medians – Large Sample Size

Paired t-Test 11 22 n n Before After Paired

Paired t-Test dd  d = Mean of the paired difference for the population  d = Std dev. of the paired difference for the population s d = Sample std dev of the paired difference n = number of pairs Student – t

Testing Hypothesis on the Mean of the Paired Differences Null Hypothesis: H 0 :  d =  0 Test statistic: Alt. HypothesisP-valueRejection Criterion H 1 :    0 2*P(t>|t 0 |) t 0 > t  /2,n-1 or t 0 < -t  /2, n-1 H 1 :  >  0 P(t>t 0 )t 0 > t , n-1 H 1 :  <  0 P(t<-t 0 )t 0 <- t , n-1

Before and after a course  =.01 Null Hypothesis: H 0 :  d = 0 Alt. Hypothesis: H 1 :  d > 0 Test statistic: Rejection region: t.01,5 = P-value = P(t>3.869)= Reject H 0 Testing Hypothesis on the Mean of the Paired Differences – Ex BeforeAfterd Mean4.167 Std. Deviation2.639

Assembly Time  =.05 Null Hypothesis: H 0 :  d = 0 Alt. Hypothesis: H 1 :  d  0 Test statistic: Rejection region: t.025,6 =  P-value = 2P(t>2.7914)= Reject H 0 Testing Hypothesis on the Mean of the Paired Differences – Ex OldNewd Mean Std. Dev

Confidence Interval on the Mean of the Paired Differences Confidence Interval One-sided Upper Confidence Bound One-sided Lower Confidence Bound

Parallel Parking  =.10 Confidence Interval on the Mean of the Paired Differences – Example Auto 1Auto 2d Mean1.21 Std. Dev.12.68

Testing Hypothesis on the Median of the Paired Differences – Wilcoxon Signed Rank Test Assuming symmetric continuous distribution Null Hypothesis: H 0 :  d = 0 Test statistic: Rank the absolute values of d from smallest to largest (discard pairs with d=0) For each d, assign a plus (+) or minus (-) sign Calculate W + (sum of ranks with +) and W - (sum of ranks with -) For H 1 :  d ≠ 0, test statistics W = min (W +, W - ) For H 1 :  d > 0, test statistics W = W - For H 1 :  d < 0, test statistics W = W + Decision: Reject H 0 if W  Wc (Critical Value in Table VII)

Testing Hypothesis on the Median of the Paired Differences – Example  =.10 Null Hypothesis: H 0 :  d = 0 H 1 :  d < 0 Test statistics W = W + = 4 Critical value: W c = 10 Reject H 0 Curr.NewdRank

Alt. HypothesisP-valueCritical Regions H 1 :  d  02P(z  |z 0 |) z 0 z 1-  /2 H 1 :  d < 0P(z  z 0 ) z 0 <z  H 1 :  d > 0P(z  z 0 ) z 0 >z 1-  Null Hypothesis: H 0 :  d = 0 Test statistic: Testing Hypothesis on the Median of the Paired Differences – Large Sample Size

Let x 11, x 12,.. x 1n be a random sample from a normal distribution with  1 and  1 2, and x 21, x 22,.. x 2n be a random sample from a normal distribution with  2 and  2 2, and let s 1 2 and s 2 2 be the sample variances. Assume both normal population are independent, then the ratio has F distribution with n 1 -1 numerator degrees of freedom and n 2 -1 denominator degrees of freedom. Testing Hypothesis on the Ratio of Two Variances

F Distribution Let W, and Y be independent  2 random variables with u and degrees of freedom, the ratio F = (W/u)/(Y/ ) has F distribution. Probability Density Function, with u, degrees of freedom, Mean Variance

F Distribution Table V in Appendix A.

Testing Hypothesis on the Ratio of Two Variances Alt. Hypothesis P-valueRejection Criterion H 1 :  1 2   2 2 P(f >f 0 ) + P(f <1/f 0 ) f 0 > f  /2,n1-1, n2-1 or f 0 < f 1-  /2,n1-1, n2-1 H 1 :  1 2 >  2 2 P(f >f 0 )f 0 > f ,n1-1, n2-1 H 1 :  1 2 <  2 2 P(f <1/f 0 )f 0 < f 1- ,n1-1, n2-1 Null Hypothesis: H 0 :  1 2 =  2 2 Test statistic:

Semiconductor Etch Variability s 1 =1.96, s 2 =2.13, n=16,  =.05 Null Hypothesis: H 0 :  1 2 =  2 2 Alt. Hypothesis: H 1 :  1 2   2 2 Test statistic: Rejection region: F.025,15,15 = F.975,15,15 = p-value = P(F >0.85)= Fail to reject H C Non-rejection Region Rejection Region Testing Hypothesis on the Ratio of Two Variances – Example Rejection Region

Confidence Interval on the Ratio of Two Variances Confidence Interval One-sided Upper Confidence Bound One-sided Lower Confidence Bound

Surface finish for Titanium Alloy s 1 =5.1, n 1 =11, s 2 =4.7, n 2 =16,  =.10 Confidence Interval on the Ratio of Two Variances – Example

Inferences about Differences between Two Population Proportions Population x USL LSL p Population 22 x USL LSL p n 1, x 1 n 2, x 2 11

Testing Hypothesis on the Difference in Two Population Proportions Alt. HypothesisP-valueRejection Criterion H 1 : p  p 0 2P(z>|z 0| )z 0 > z 1-  /2 or z 0 < z  /2 H 1 : p > p 0 P(z>z 0 )z 0 > z 1-  H 1 : p < p 0 P(z<-z 0 )z 0 < z  Null Hypothesis: H 0 : p 1 = p 2 Test statistic:

Brands of toothpaste n 1 =500, n 2 =400,  =.01 Null Hypothesis: H 0 : p 1 = p 2 Alt. Hypothesis: H 1 : p 1 > p 2 Test statistic: Rejection region: z 1-  = z.99 =2.326 P-value = P(z>2.7852)= Reject H 0 Testing Hypothesis on the Difference in Two Population Proportions – Example 17.15

Types of machine x 1 =48, n 1 =800, x 1 =45, n 2 =900,  =.01 Null Hypothesis: H 0 : p 1 = p 2 Alt. Hypothesis: H 1 : p 1  p 2 Test statistic: Rejection region: z  /2 = z.005 = or z 1-  /2 = z.995 =2.576 P-value = P(z>.9050)=.3655 Fail to reject H 0 Testing Hypothesis on the Difference in Two Population Proportions – Example 17.16

Confidence Interval on the Difference in Two Population Proportions Confidence Interval One-sided Upper Confidence Bound One-sided Lower Confidence Bound

Confidence Interval on the Difference in Two Population Proportions – Ex Brands of toothpaste n 1 =500, n 2 =400,  =.05 Confidence Interval