Pooled Variance t Test Tests means of independent populations having equal variances Parametric test procedure Assumptions Both populations are normally distributed If not normal, can be approximated by normal distribution (n 1 30 & n 30 ) Population variances are unknown but assumed equal
Two Independent Populations Examples An economist wishes to determine whether there is a difference in mean family income for households in socioeconomic groups. An admissions officer of a small liberal arts college wants to compare the mean SAT scores of applicants educated in rural high schools & in urban high schools.
Pooled Variance t Test Example You re a financial analyst for Charles Schwab. You want to see if there a difference in dividend yield between stocks listed on the NYSE & NASDAQ. NYSE NASDAQ Number 1 5 Mean 3.7.53 Std Dev 1.30 1.16 Assuming equal variances, is there a difference in average yield (a =.05)? 1984-1994 T/Maker Co.
Pooled Variance t Test Solution H0: m 1 - m = 0 (m 1 = m ) H1: m 1 - m 0 (m 1 m ) a =.05 df = 1 + 5 - = 44 Critical Value(s): Reject H 0 Reject H 0.05.05 -.0154 0.0154 t t Test Statistic: 3.7.53 = 1 1 1.510 1 5 Decision: Reject at a =.05 =.03 Conclusion: There is evidence of a difference in means
t = Test Statistic Solution c ha f X 1 X m 1 m F S P HG I a f af = 3. 7. 53 0 K J F H I = 1 1 1 1 1510. K n n 1 5 1. 03 S a f a f n 1 1 S 1 n 1 S P = a f a f n 1 1 n 1 = a f a f a f a f 1 1 130. 5 1 116. a f a f 1 1 5 1 = 1510.
Pooled Variance t Test Thinking Challenge You re a research analyst for General Motors. Assuming equal variances, is there a difference in the average miles per gallon (mpg) of two car models (a =.05)? You collect the following: Sedan Van Number 15 11 Mean.00 0.7 Std Dev 4.77 3.64 Alone Group Class
t = Test Statistic Solution* c ha f X 1 X m 1 m F S P HG I a f af =. 00 0. 7 0 K J F H I = 1 1 1 1 18. 793 K n n 15 11 1 100. S a f a f n 1 1 S 1 n 1 S P = a f a f n 1 1 n 1 = a f a f a f a f 15 1 4. 77 11 1 3. 64 a f a f 15 1 11 1 = 18. 793
One-Way ANOVA F-Test
& c-sample Tests with Numerical Data & C-Sample Tests Mean Variance Median # Samples C F Test ( Samples) # Samples C Pooled Variance t Test One-Way ANOVA Wilcoxon Rank Sum Test Kruskal- Wallis Rank Test
Experiment Investigator controls one or more independent variables Called treatment variables or factors Contain two or more levels (subcategories) Observes effect on dependent variable Response to levels of independent variable Experimental design: Plan used to test hypotheses
Completely Randomized Design Experimental units (subjects) are assigned randomly to treatments Subjects are assumed homogeneous One factor or independent variable or more treatment levels or classifications Analyzed by: One-Way ANOVA Kruskal-Wallis rank test
Randomized Design Example Factor levels (Treatments) Factor (Training Method) Level 1 Level Level 3 Experimental units K K K K K K K K K Dependent 1 hrs. 17 hrs. 31 hrs. variable 7 hrs. 5 hrs. 8 hrs. (Response) 9 hrs. 0 hrs. hrs.
One-Way ANOVA F-Test Tests the equality of or more (c) population means Variables One nominal scaled independent variable or more (c) treatment levels or classifications One interval or ratio scaled dependent variable Used to analyze completely randomized experimental designs
One-Way ANOVA F-Test Assumptions Randomness & independence of errors Independent random samples are drawn Normality Populations are normally distributed Homogeneity of variance Populations have equal variances
One-Way ANOVA F-Test Hypotheses H 0 : m 1 = m = m 3 =... = m c All population means are equal No treatment effect H 1 : Not all m j are equal At least 1 population mean is different Treatment effect m 1 m... m c is wrong f(x) f(x) m 1 = m = m 3 m 1 = m m 3 X X
One-Way ANOVA Basic Idea Compares types of variation to test equality of means Ratio of variances is comparison basis If treatment variation is significantly greater than random variation then means are not equal Variation measures are obtained by partitioning total variation
ANOVA Partitions Total Variation Total variation Variation due to treatment Sum of squares among Sum of squares between Sum of squares model Among groups variation Variation due to random sampling Sum of squares within Sum of squares error Within groups variation
Response, X Total Variation e j e j e j e j e j e j 11 1 n c c SST = X X X X X X c `X Group 1 Group Group 3
Among-Groups Variation e j e j e j e j e j e j 1 1 c c SSA = n X X n X X n X X Response, X `X 3 `X `X 1 `X Group 1 Group Group 3
Within-Groups Variation c h c h c h c h c h c h 11 1 1 1 n c c SSW = X X X X X X Response, X c `X 3 `X 1 `X Group 1 Group Group 3
One-Way ANOVA Test Statistic Test statistic F = MSA / MSW MSA is Mean Square Among MSW is Mean Square Within Degrees of freedom df 1 = c -1 df = n - c c = # Columns (populations, groups, or levels) n = Total sample size
One-Way ANOVA Summary Table Source of Variation Degrees of Freedom Sum of Squares Mean Square (Variance) Among c - 1 SSA MSA = (Factor) SSA/(c - 1) Within n - c SSW MSW = (Error) SSW/(n - c) Total n - 1 SST = SSA+SSW F MSA MSW
One-Way ANOVA Critical Value If means are equal, F = MSA / MSW 1. Only reject large F! Reject H 0 Do Not Reject H 0 a 0 F U ( a ; c 1, n c ) F 1984-1994 T/Maker Co. Always One-Tail!
One-Way ANOVA F-Test Example As production manager, you want to see if 3 filling machines have different mean filling times. You assign 15 similarly trained & experienced workers, 5 per machine, to the machines. At the.05 level, is there a difference in mean filling times? Mach1MachMach3 5.40 3.40 0.00 6.31 1.80.0 4.10 3.50 19.75 3.74.75 0.60 5.10 1.60 0.40
One-Way ANOVA F-Test Solution H0: m 1 = m = m 3 H1: Not all equal a =.05 df1 = df = 1 Critical Value(s): a =.05 0 3.89 F F Test Statistic: = MSA MSW = 3. 580. 911 = Decision: Reject at a =.05 5. 6 Conclusion: There is evidence pop. means are different
Summary Table Solution Source of Degrees of Sum of Mean F Variation Freedom Squares Square (Variance) Among 3-1 = 47.1640 3.580 5.60 (Machines) Within 15-3 = 1 11.053.911 (Error) Total 15-1 = 14 58.17
Summary Table Excel Output
One-Way ANOVA Thinking Challenge You re a trainer for Microsoft Corp. Is there a difference in mean learning times of 1 people using 4 different training methods (a =.05)? M1 M M3 M4 10 11 13 18 9 16 8 3 5 9 9 5 1984-1994 T/Maker Co. Alone Group Class
One-Way ANOVA Solution* H0: m 1 = m = m 3 = m 4 H1: Not all equal a =.05 df1 = 3 df = 8 Critical Value(s): a =.05 0 4.07 F Test Statistic: F = MSA MSW = 116 10 = Decision: Reject at a =.05 116. Conclusion: There is evidence pop. means are different
Summary Table Solution* Source of Degrees of Sum of Mean F Variation Freedom Squares Square (Variance) Among 4-1 = 3 348 116 11.6 (Methods) Within 1-4 = 8 80 10 (Error) Total 1-1 = 11 48