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Lesson #23 Analysis of Variance

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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

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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”. ii ij Y ij = + i + ij i = 1, 2, …, k j = 1, 2, …, n i

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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,

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

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

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

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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

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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

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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

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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

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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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