Lesson #23 Analysis of Variance

In Analysis of Variance (ANOVA), we have: H 0 :  1 =  2 =  3 = … =  k H 1 : at least one  i does not equal the others Assumptions - - the k populations are independent - the k populations are Normally distributed - the k populations all have equal variances

Model: Y is the called the “dependent variable”, or “response”, or “outcome”. Y ij =  + (  i –  ) + (Y ij –  i ) The variable (“factor”) whose levels define the k “treatments” is the “independent variable”. ii  ij Y ij =  +  i +  ij i = 1, 2, …, k j = 1, 2, …, n i

Let = n 1 + n 2 + … + n k = n. = n From each sample, we calculate – - the sample mean, - the sample variance, Also calculate the overall sample mean,

Sums of Squares SS TREATMENT = SS BETWEEN = SS AMONG = SS MODEL SS ERROR = SS RESIDUAL = SS WITHIN SS TOTAL = SS TREATMENT + SS ERROR

df TREATMENT = (k – 1) df ERROR = (n. – k) Degrees of freedom - Mean squares are sums of squares divided by df:

E ( MS ERROR ) =  2 Reject H 0 if F 0 = > F (k-1,n. -k),1- 

The F-distribution has two parameters, called the numerator df (df 1 ) and denominator df (df 2 ). The distribution is positively skewed, and defined only for positive values. In this case, df 1 = (k - 1), df 2 = (n. – k) F (5,20),.95 = 2.71

ANOVA Table Source df SS MS F Treatment Error Total k-1 n. –k n. –1 SS TRT SS ERROR SS TOTAL MS TRT MS ERROR F0F0

White Pink Red nini n. = 23

nini n. = 23 + (10)( ) 2 = (8)( ) 2 + (5)( ) 2 = SS TRT SS ERROR = (7)(0.1116) + (9)(0.2889) + (4)(0.1516) = 3.988

Source df SS MS F Treatment (color) Error Total Reject H 0 if F 0 > F (2,20),.90 = 2.59 Reject H 0, conclude the average diameter is not the same for all three colors of roses.