Solution file for additional exercise 13.5
------------------------------------------
(continuation of additional exercise 10.4)

As suggested in the question, all analysis will be based on the litter means.

a)
MTB > WOpen "H:\VHM\VHM802\Data_csv\hs10_4.csv";
SUBC>   FType;
SUBC>     CSV;
SUBC>   DecSep;
SUBC>     Period;
SUBC>   Field;
SUBC>     Comma;
SUBC>   TDelimiter;
SUBC>     DoubleQuote.
Retrieving worksheet from file: ‘H:\VHM\VHM802\Data_csv\hs10_4.csv’
Worksheet was saved on 27/02/2012

MTB > Name c5 "ByVar1" c6 "ByVar2" c7 "Mean1"
MTB > Statistics 'pH';
SUBC>   By 'strain' 'litter';
SUBC>   GValues 'ByVar1'-'ByVar2';
SUBC>   Mean 'Mean1'.
MTB > name c5 'strain_1'
MTB > name c7 'mean_pH'

MTB > TwoT 'mean_pH' 'strain_1';
SUBC>   Confidence 95.0;
SUBC>   Test 0.0;
SUBC>   Alternative 0;
SUBC>   Pooled.
Two-Sample T-Test and CI: mean_pH, strain_1 

Two-sample T for mean_pH

strain_1  N    Mean   StDev  SE Mean
pHH       7  7.4771  0.0331    0.013
pHL       7  7.4554  0.0392    0.015

Difference = mu (pHH) - mu (pHL)
Estimate for difference:  0.0218
95% CI for difference:  (-0.0204, 0.0640)
T-Test of difference = 0 (vs not =): T-Value = 1.12  P-Value = 0.283  DF = 12
Both use Pooled StDev = 0.0363

MTB > TOST 'mean_pH' 'strain_1';
SUBC>   LLimit -.1;
SUBC>   ULimit .1;
SUBC>   Stacked;
SUBC>      Reference "pHL";
SUBC>   UEVariance;
SUBC>   Alpha 0.05;
SUBC>   GInterval;
SUBC>   TMethod;
SUBC>   TDStatistics;
SUBC>   TDMean;
SUBC>   TTest.
Two-Sample Equivalence Test: mean_pH, strain_1 

Method

Test mean = mean of pHH
Reference mean = mean of pHL
Equal variances were not assumed for the analysis.

Descriptive Statistics

strain_1  N    Mean     StDev   SE Mean
pHH       7  7.4771  0.033085  0.012505
pHL       7  7.4554  0.039169  0.014805

Difference: Mean(pHH) - Mean(pHL)

Difference        SE  95% CI for Equivalence  Equivalence Interval
  0.021786  0.019379   (-0.013017, 0.056588)       (-0.1, 0.1)

CI is within the equivalence interval. Can claim equivalence.

Test

Null hypothesis:         Difference = -0.1 or Difference = 0.1
Alternative hypothesis:  -0.1 < Difference < 0.1
a level:                 0.05

Null Hypothesis    DF  T-Value  P-Value
Difference = -0.1  11   6.2844    0.000
Difference = 0.1   11  -4.0360    0.001

The greater of the two P-Values is 0.001. Can claim equivalence.
 
Equivalence Test: Mean(pHH) - Mean(pHL) 

Comments:
---------
The t-value is related to the F-statistic from the ANOVA in Additional
Exercise 10.4 by: t^2 = 1.12^2 = 1.26 = F, and the two tests have the
same P-value (0.283). Note that we used the version of the two-sample 
t-test with equal variances in the two samples to get an exact agreement
between the two tests. Because the standard deviations are very similar
in the two samples, the inference without assuming equal variance is
almost identical (t=1.12, P=0.285; not shown). 

The output from the equivalence analysis shows that we could easily 
claim equivalence between the two strains up to 0.1 pH units. The
confidence computed as part of the equivalence analysis equals 
(-0.013,0.057), so we could claim equivalence up to any value 
greater than 0.057.


b)
MTB > Power;
SUBC>  TOST 2; 
SUBC>   Sample 7; 
SUBC>   Difference 0.0218; 
SUBC>   LLIMIT  -.1; 
SUBC>   ULIMIT  .1; 
SUBC>   Sigma 0.0363; 
SUBC>   Alpha 0.05.
Power and Sample Size 

2-Sample Equivalence Test

Power for difference:       Test mean - reference mean
Null hypothesis:            Difference = -0.1 or Difference = 0.1
Alternative hypothesis:     -0.1 < Difference < 0.1
a level:                    0.05
Assumed standard deviation: 0.0363

            Sample
Difference    Size     Power
    0.0218       7  0.984002

The sample size is for each group.

Comments:
---------
The computed power for demonstrating equivalence up to 0.1 is 0.98. We
were in fact pretty sure to succeed, assuming the estimates obtained. In
practice we would have needed to use guessed values for the standard
deviation and the observed difference between the two groups, and thus
any power calculation prior to the study would most likely have produced
a different value.


c)
MTB > Power;
SUBC>  TOST 2; 
SUBC>   Difference 0.0218; 
SUBC>   Power .8; 
SUBC>   LLIMIT  -.05; 
SUBC>   ULIMIT  .05; 
SUBC>   Sigma 0.0363; 
SUBC>   Alpha 0.05.
Power and Sample Size 

2-Sample Equivalence Test

Power for difference:       Test mean - reference mean
Null hypothesis:            Difference = -0.05 or Difference = 0.05
Alternative hypothesis:     -0.05 < Difference < 0.05
a level:                    0.05
Assumed standard deviation: 0.0363

            Sample  Target
Difference    Size   Power  Actual Power
    0.0218      22     0.8      0.813221

The sample size is for each group.

Comments:
---------
The calculation shows that with the observed values (and the same caveat
as above for their use), we would need 22 litters in each group to
demonstrate equivalence between the two strains up to 0.05 pH units.