Elementary Statistics (Text Only)
Elementary Statistics (Text Only)
2nd Edition
ISBN: 9780077836351
Author: Author
Publisher: McGraw Hill
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Chapter 14, Problem 11RE
To determine

To find: The pairs of means which are different by Tukey-Kramer test.

Expert Solution & Answer
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Answer to Problem 11RE

The pairs of means which are different by Tukey-Kramer test are 1,2 and 1,3 .

Explanation of Solution

Given information:

The value of α is 0.05 and the given data is,

    PlantConcentration
    A438619732638  
    B857101411538831053 
    C9257861179786  
    D893891917695675595

Calculation:

The sample size are 4 and n1=4,n2=5,n3=4,n4=6 .

The total number in all the samples combined is,

  N=4+5+4+6=19

From the given data the sample means are,

  x1¯=438+619+732+6384=606.75x2¯=857+1014+1153+883+10535=992

Further solve,

  x3¯=925+786+1179+7864=919x4¯=893+891+917+695+675+5956=777.6667

The grand mean is,

  x¯¯=606.75+992+919+777.66674=823.8542

The value of i=14x1i2 is,

  i=14x 1i2=4382+6192+7322+6382=1517873

The standard of deviation of sample 1 is,

  s12=1n11( i=1 4 x 1i 2 n1 x 1 2 ¯)=141(15178734 ( 606.75 )2)s1=368145.5625

The value of i=15x2i2 is,

  i=15x 2i2=8572+10142+11532+8832+10532=4980552

The standard of deviation of sample 2 is,

  s22=1n21( i=1 5 x 2i 2 n2 x 2 2 ¯)=151(49805525 ( 992 )2)s2=122.711

The value of i=14x3i2 is,

  i=14x 3i2=9252+7862+11792+7862=3481258

The standard of deviation of sample 3 is,

  s32=1n31( i=1 4 x 3i 2 n3 x 3 2 ¯)=141(34812584 ( 919 )2)s3=185.3052

The value of i=16x4i2 is,

  i=16x 4i2=8932+8912+9172+6952+6752+5952=3724894

The standard of deviation of sample 4 is,

  s42=1n41( i=1 6 x 4i 2 n4 x 4 2 ¯)=161(37248946 ( 777.6667 )2)s4=138.7811

The treatment sum of squares is,

  SSTr=n1( x 1 ¯ x ¯ ¯)2+n2( x 2 ¯ x ¯¯)2+n3( x 3¯x¯¯)2+n4(x4¯x¯¯)2=4(606.75823.8542)2+5(992823.8542)2+4(919823.8542)2+6(777.6667823.8542)2=378912.5897

The error sum square is,

  SSE=(n11)s12+(n21)s22+(n31)s32+(n41)s42=(41)(15096.9167)+(51)(15058)+(41)(34338)+(61)(19260.2044)=304837.7721

The degree of freedom for treatment sum of square is,

  I1=41=3

The degree of freedom for error sum of square is,

  NI=194=15

The treatment mean sum of square is,

  MSTr=SSTrI1=378912.58973=126304.1966

The error mean sum of square is,

  MSE=SSENI=304837.772115=20322.5181

The mean of 1 and 2 sample is,

  q1,2=| x 1 ¯ x 2 ¯| MSE 2 ( 1 n 1 + 1 n 2 )=|606.75992| 20322.5181 2 ( 1 4 + 1 5 )=5.6972

The mean of 1 and 3 sample is,

  q1,3=| x 1 ¯ x 3 ¯| MSE 2 ( 1 n 1 + 1 n 3 )=|606.75919| 20322.5181 2 ( 1 4 + 1 4 )=4.3807

The mean of 1 and 4 sample is,

  q1,4=| x 1 ¯ x 4 ¯| MSE 2 ( 1 n 1 + 1 n 4 )=|606.75777.6667| 20322.5181 2 ( 1 4 + 1 6 )=2.6266

The mean of 2 and 3 sample is,

  q2,3=| x 2 ¯ x 3 ¯| MSE 2 ( 1 n 2 + 1 n 3 )=|992919| 20322.5181 2 ( 1 5 + 1 4 )=1.07955

The mean of 2 and 4 sample is,

  q2,4=| x 2 ¯ x 4 ¯| MSE 2 ( 1 n 2 + 1 n 4 )=|992777.6667| 20322.5181 2 ( 1 5 + 1 6 )=3.5112

The mean of 3 and 4 sample is,

  q3,4=| x 3 ¯ x 4 ¯| MSE 2 ( 1 n 3 + 1 n 4 )=|919777.6667| 20322.5181 2 ( 1 4 + 1 6 )=2.1720

The critical value q for the student zed range distribution at α=0.05 level of significance is 4.08 .

The comparison of pairwise test statistic values with critical values is shown in table below.

    MeansTest statisticCritical valueDecision
    1,25.69724.08reject null hypothesis.
    1,34.38074.08reject null hypothesis.
    1,42.62664.08Do not reject null hypothesis.
    2,31.079554.08Do not reject null hypothesis.
    2,43.51124.08Do not reject null hypothesis.
    3,42.17204.08Do not reject null hypothesis.

Therefore, the pairs of means which are different by Tukey-Kramer test are 1,2 and 1,3 .

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Chapter 14 Solutions

Elementary Statistics (Text Only)

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