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  3. the following data are from a completely randomize

The following data are from a completely randomized design. Treatment A B C 163 142 126 142 158 121 168 129 138 145 142 143 147 133 153 189 148 123 Sample mean 159 142 134 Sample variance 325.2 108.4 162.4 Compute the sum of squares...

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Basic Math

The following data are from a completely randomized design.

Treatment
ABC
163142126
142158121
168129138
145142143
147133153
189148123
Sample mean159142134
Sample variance325.2108.4162.4
  1. Compute the sum of squares between treatments. Round theintermediate calculations to whole number.
      
  2. Compute the mean square between treatments.
      
  3. Compute the sum of squares due to error.
      
  4. Compute the mean square due to error (to 1 decimal).
      
  5. Set up the ANOVA table for this problem. Round all Sum ofSquares to the nearest whole number. Round all Mean Squares to onedecimal place. Round F to two decimal places. Roundp-value to four decimal places.
    Source of VariationSum of SquaresDegrees of FreedomMean SquareFp-value
    Treatments
    Error
    Total

Answer & Explanation Solved by verified expert
4.2 Ratings (898 Votes)

a)

A B C
count, ni = 6 6 6
mean , x? i = 159.000 142.00 134.00
std. dev., si = 18.033 10.412 12.744
sample variances, si^2 = 325.200 108.400 162.400
total sum 954 852 804 2610 (grand sum)
grand mean , x?? = ?ni*x?i/?ni =   145.00
square of deviation of sample mean from grand mean,( x? - x??)² 196.000 9.000 121.000
TOTAL
SS(between)= SSB = ?n( x? - x??)² = 1176.000 54.000 726.000 1956
SS(within ) = SSW = ?(n-1)s² = 1626.000 542.000 812.000 2980.0000

sum of squares between treatments = 1956

b)

no. of treatment , k =   3
df between = k-1 =    2
N = ?n =   18
df within = N-k =   15

mean square between treatments , = SSB/k-1 = 1956/2 = 978

c)

sum of squares due to error=2980

d)

mean square due to error MSE = SSE/N-k = 2980/15 = 198.7

e)

SS df MS F p-value
Treatments 1956 2 978.0 4.92 0.0227
Error 2980 15 198.7
Total: 4936 17

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