# ANalysis Of VAriance can be used to test for the equality of three or more population means. H 0 :  1  =  2  =  3  = ... =  k H a : Not all population.

## Presentation on theme: "ANalysis Of VAriance can be used to test for the equality of three or more population means. H 0 :  1  =  2  =  3  = ... =  k H a : Not all population."— Presentation transcript:

ANalysis Of VAriance can be used to test for the equality of three or more population means. H 0 :  1  =  2  =  3  = ... =  k H a : Not all population means are equal For each population, the response variable is The variances of the response variables are all equal to The observations must be normally distributed.  2 independent. ANOVA

Sample means are “close” together because there is only one sampling distribution when H 0 is true. Sampling Distribution of x given H 0 is true  ANOVA

There are k treatments: is computed from a random sample of size For j = 1 2 3 k The overall sample mean: yada, yada, yada ANOVA Dividing the following by k – 1 givens the MSTR

There are k treatments: is computed from a random sample of size For j = k The overall sample mean: ANOVA Dividing the following by k – 1 givens the MSTR

Sample means come from different sampling distributions, and so are not as “close” together when H 0 is false. Sampling Distribution of x when H 0 is false     ANOVA

There are k treatments: is computed from a random sample of size For j = 1 2 3 k yada, yada, yada The overall total number of observations in all samples: n T = n 1 + n 2 + n 3 + … + n k ANOVA Dividing the following by n T – k givens the MSE

There are k treatments: is computed from a random sample of size For j = The overall total number of observations in all samples: n T = n 1 + n 2 + n 3 + … + n k ANOVA k

The following is the SST which has n T – 1 degrees of freedom

MSE Treatment Error Total SSTR SSE k – 1 n T – k MSTR Source of Variation Sum of Squares Degrees of Freedom Mean Squares F -stat F …we reject H 0 If F-stat is “BIG” … … you cannot reject H 0 If F-stat is “small”… SST n T – 1 The above ANOVA procedure is an example of a completely randomized design, and is applicable when treatments are randomly assigned to the experimental units useful when the experimental units are homogenous ANOVA

Example 1 Janet Reed would like to know if there is any significant difference in the mean number of hours worked per week for the department managers at her three manufacturing plants (in Buffalo, Pittsburgh, and Detroit). A simple random sample of five managers from each of the three plants was taken and the number of hours worked by each manager for the previous week is shown on the next slide. Conduct an F test at the 5% level of significance. NOTE: k = 3 and n 1 = n 2 = n 3 = 5 Completely Randomized Design

1234512345 48 54 57 54 62 73 63 66 64 74 51 63 61 54 56 Plant 1 Buffalo Plant 2 Pittsburgh Plant 3 Detroit Observation nini 5 5 5 xixi si2si2 55 68 57 26.0 26.5 24.5 Average weekly hours worked by department managers Completely Randomized Design

H 0 :  1  =  2  =  3  H a : Not all the means are equal 1. Develop the hypotheses. Completely Randomized Design 490 = 490 2. Determine the critical value 3. Compute MSTR MSTR = SSTR = 5( 55 – 60 ) 2 + 5( 68 – 60 ) 2 + 5( 57 – 60 ) 2 = ( 55 + 68 + 57 )/3 = 60 = 245 F  = 3.89 Row: n T – k = 12  / 2 Column k – 1 = 2

308 4. Compute the MSE MSE = SSE = 4( 26.0 ) + 4( 26.5 ) + 4( 24.5 ) F = MSTR/MSE 5. Compute the F -stat nTnT k = 308 = = 9.55 245 25.667 /(15 – 3) = / Completely Randomized Design

25.667 Treatment Error Total 490 308 2 12 245 Source of Variation Sum of Squares Degrees of Freedom Mean Squares 9.55 F 798 14 At 5% significance, the mean hours worked by department managers is not the same.  Do Not Reject H 0 Reject H 0 3.89 9.55  1 Completely Randomized Design

Example: Crescent Oil Co. Crescent Oil has developed 3 new blends of gasoline and must decide which blend or blends to produce and distribute. A study of the MPG ratings of the 3 blends is being conducted to determine if the mean ratings are the same for the three blends at a 10% level of significance. Each of the 3 gasoline blends have been tested on 5 automobiles. The MPG ratings for the 15 automobiles are shown in the table on the next slide. If the experimental units are heterogeneous, use Randomized Block Design ANOVA

29.8 Treatment Means 1234512345 31 30 29 33 26 30 29 31 25 30 29 28 29 26 30.333 Type of Gasoline (Treatment) Block Means Blend XBlend YBlend Z Automobile (Block) 28.828.4 29.333 28.667 31.000 25.667 29 Randomized Block Design

62 SSE =   SSBL = (3) [ ] SSTR = (5) [ ] SST = (31 – 29) 2 = 62 = 5.2 = 51.33 Most of the variation is across blocks = 5.47 + (30 – 29) 2 +... (29.8 – 29 ) 2 + (28.8 – 29 ) 2 + (28.4 – 29 ) 2 ( 30.333 – 29 ) 2 + ( 29.333 – 29 ) 2 + ( 25.667 – 29 ) 2 + (30 – 29) 2 + (29 – 29) 2... + (26 – 29) 2 + ( 28.667 – 29 ) 2 + ( 31.000 – 29 ) 2 5.2 51.33 Randomized Block Design

Source of Variation Sum of Squares Degrees of Freedom Mean Squares F Treatments Error Total 2 14 5.20 5.47 62.00 8 2.60.68 3.82 ANOVA Table Blocks51.3312.834 NOTE: when the F-stat is “big” the estimated variances are “far apart”, and so the population means are probably different. Randomized Block Design

0 F  Do Not Reject H 0 Reject H 0 3.11 Row: denominator df = 8,  =.10 F-stat 3.82 At 10% significance, the MPG ratings differ for the three gasoline blends. Column: numerator df = 2  1 Randomized Block Design

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