Solution file for Additional exercise 12.1
------------------------------------------

Planning of a study on turtle shell lengths.
Standard deviation sigma=2.1 cm.

Question a):
------------
For a 95% CI with a length of 2 cm (and a margin of error of 1cm), we must solve:
1 >= 2*sigma/sqrt(n) => n >= (2*sigma/1)^2 = (2*2.1)^2 = 17.64; take n=18.

Because the resulting n is fairly small, we redo the calculation while
replacing 2 by t(17,0.975)=2.11:
1 >= 2.11*sigma/sqrt(n) => n >= (2.11*sigma/1)^2 = (2.11*2.1)^2 = 19.63; take n=20.

Minitab command and output:
MTB > SSCI; 
SUBC>   NMean; 
SUBC>     Sigma 2.1; 
SUBC>   Confidence 95.0; 
SUBC>   IType 0; 
SUBC>   MError 1.
Sample Size for Estimation 

Method

Parameter            Mean
Distribution         Normal
Standard deviation   2.1 (estimate)
Confidence level     95%
Confidence interval  Two-sided

Results

  Margin  Sample
of Error    Size
       1      20


Question b):
------------
Interest is in an effect size of 1cm, so according to Cox's rule, we take
1 = 3*sigma/sqrt(n) => n = (3*sigma/1)^2 = 39.69; take n=40.


Question c):
------------
We use Minitab's power menu (see appendices for Stata and R listings) to compute 
necessary sample sizes to obtain powers 0.7, 0.8 or 0.9 with a significance level 
of 0.05:

MTB > Power;
SUBC>   TOne;
SUBC>     Difference 1;
SUBC>     Power 0.7 0.8 0.9;
SUBC>     Sigma 2.1.
Power and Sample Size 

1-Sample t Test

Testing mean = null (versus not = null)
Calculating power for mean = null + difference
Alpha = 0.05  Assumed standard deviation = 2.1

            Sample  Target
Difference    Size   Power  Actual Power
         1      30     0.7      0.712635
         1      37     0.8      0.804705
         1      49     0.9      0.904219

Comments:
---------
It is seen that Cox's rule in this case leads to a power somewhat above
0.8. This could however easily change if we modified some of the
settings for the test (i.e., the significance level and the two-sided
alternative).

The sample size in a) is based on a confidence interval with a margin of
error of 1cm. This means that if the observed mean differs from the
species average by 1cm, then one limit of the confidence interval will
equal the species average. We can test the null hypothesis against a two-sided
alternative at a significance level of 0.05 by checking whether the
hypothesized value is inside the confidence interval. As in this case it
is on its boundary, P=0.05. The assumption for this is that a two-sided
alternative hypothesis is used.

The confidence interval gives a statement about an observed mean, but
power calculations give statements about the datasets that might result
under the alternative hypothesis. Even if the true difference was 1cm,
the observed difference would not (most unlikely) be equal to 1cm. As the
observed difference can be both less and more extreme, and we need a 
probability (power) of getting significance across all scenarios, it
requires a larger sample size to ensure that the means we could get in 
the study are extreme enough.

---

Appendix: output from power calculation using the Stata menu:

. power onemean 0 1, power(0.7 0.8 0.9) sd(2.1)

Performing iteration ...

Estimated sample size for a one-sample mean test
t test
Ho: m = m0  versus  Ha: m != m0

  +---------------------------------------------------------+
  |   alpha   power       N   delta      m0      ma      sd |
  |---------------------------------------------------------|
  |     .05      .7      30   .4762       0       1     2.1 |
  |     .05      .8      37   .4762       0       1     2.1 |
  |     .05      .9      49   .4762       0       1     2.1 |
  +---------------------------------------------------------+


---

Appendix: output from power calculation using R function:

> power.t.test(delta=1, sd=2.1, sig.level=0.05, power=0.7, type="one.sample", alternative="two.sided")

     One-sample t test power calculation 

              n = 29.19071
          delta = 1
             sd = 2.1
      sig.level = 0.05
          power = 0.7
    alternative = two.sided

> power.t.test(delta=1, sd=2.1, power=0.8, type="one.sample")

     One-sample t test power calculation 

              n = 36.58076
          delta = 1
             sd = 2.1
      sig.level = 0.05
          power = 0.8