Presentation is loading. Please wait. # 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.

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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

3.500 4.125 3.875 3.500 4.250 4.000 3.625 4.250 4.625 4.500 4.875 5.000 3.500 4.125 3.750 4.750 2.750 3.750 3.500 3.375 3.625 White Pink Red nini 8 10 5 3.875 4.3 3.4 0.1116 0.2889 0.1516 n. = 23

nini 8 10 5 3.875 4.3 3.4 0.1116 0.2889 0.1516 n. = 23 + (10)(4.3 - 3.9565) 2 = (8)(3.875 - 3.9565) 2 + (5)(3.4 - 3.9565) 2 = 2.782 SS TRT SS ERROR = (7)(0.1116) + (9)(0.2889) + (4)(0.1516) = 3.988

Source df SS MS F Treatment (color) Error Total 2 20 22 2.782 3.988 6.770 1.391 0.1994 6.98 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.

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