Statistics

Chapter 13

1. a) From the problem, the given data is

The sample mean treatments for A,B and C are

 = 156[pic 1]

 = 142[pic 2]

 = 134[pic 3]

Number of treatments, k = 3

== = 6[pic 4][pic 5][pic 6]

 = [pic 7][pic 8]

      = 6 + 6 + 6

      = 18

The formula for the overall sample mean is

 = [pic 9][pic 10]

    =  = 14[pic 11]

The sum of squares between treatments is

SSTR = [pic 12]

          = 6(156 – 144)2 + 6(142 – 144)2 + 6(134 – 144)2 

          = 1488

b) Mean square treatment is

MSTR =  [pic 13]

           =  = 744[pic 14]

c) The given sample variances are

 = 164.4[pic 15]

 = 131.2        [pic 16]

 = 110.4[pic 17]

Sum of the squares after the error is followed by

SSE =[pic 18]

       = (6 – 1)164.4 + (6 – 1)131.2 + (6 – 1)110.4

       = 2030

d) Mean square due to the error is given by: MSE = [pic 19]

         = [pic 20]

         = 135.3

        e) ANOVA table for the entire randomized design is            

        The ANOVA table for the given experiment is

f) In order to test  the equality of three means for the test statistic is given by F = [pic 26]

   =    = 5.50[pic 27]

degree of freedom for the numerator is k – 1 = 2

 degree of freedom for the denominator is  18 – 3[pic 28]

 = 15.

Here we reject null hypothesis, because of the greater values of the test statistics.

Here the p-value is the higher tail area of F distribution to the right side of the test statistic.

F=5.50

F=5.50  which is between 4.77 and 6.36

The area in the upper tail at F=5.50 is between 0.025 and 0.01.

 So  the p-value is between 0.01 and 0.025.

 p-value = 0.0162.

  will be rejected With p-value ≤ α = 0.05[pic 31]

It is proved from the test that the three treatments are not equal

11) Assume the population 1 as drying times for paint 1, population 2 for the paint 2 and population 3 for the paint 3 and population 4 for the paint 4.

Here

  =  population 1 mean drying time [pic 32]

  = population 2 mean drying time [pic 33]

 = population 3 mean drying time [pic 34]

 = population 4 mean drying time[pic 35]

Total treatments, k = 4

and = ====5[pic 36][pic 37][pic 38][pic 39][pic 40]

 = + +++[pic 41][pic 42][pic 43][pic 44][pic 45][pic 46]

      = 5+5+5+5+5

      = 25

Treatment j sample mean =  = [pic 47][pic 48]

given data for the sample mean yields at the three different  temperature levels are

 = 133[pic 49]

 = 139[pic 50]

 = 136[pic 51]

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