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Essay by   •  November 2, 2016  •  Research Paper  •  1,756 Words (8 Pages)  •  1,223 Views

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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            

Variation source

Sum of squares

Degrees of freedom

 

Mean square

F

P-value

 

Treatment

    SSTR

    k - 1

MSTR= [pic 21]

[pic 22]

Error

    SSE

     - k[pic 23]

MSE= [pic 24]

Total

    SST

     - 1[pic 25]

        The ANOVA table for the given experiment is

Variation Source

Sum of Squares

Degrees

    of

Freedom

Mean Square

   F

 p - value

Treatment

1488

    2

    744

5.50

0.0162

Error

2030

   15

  135.3

Total

3518

   17

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

Area in the Upper Tail

0.1

0.05

0.025

0.01

F value (=2,=15)[pic 29][pic 30]

2.7

3.64

4.77

6.36

                

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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