Supplementary Exercises 7.73 and 7.74 of IPS7e
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Phosphorus levels (mg/dl) in the blood of one patient, 
measured at n=6 occasions.

Data: X1,...,X6.

Model: X1,...,X6 are i.i.d. (a SRS) and normally distributed
N(mu,sigma), where mu and sigma are unknown parameters, corresponding
to the values of this patient.


7.73:
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(a) Estimation of mu: Xmean=5.367, and of sigma: s=0.6653. Therefore,
      SE=sd(Xmean)=s/sqrt(6)=0.2716.

(b) The tstar value for a 90% CI is taken from t(5): tstar=2.015. Then,
      90% CI for mu: Xmean +- tstar*s/sqrt(6) = Xmean +- tstar*SE
                     = 5.367 +- 2.015*0.2716 = 5.367 +- 0.547 = (4.820,5.914).


7.74:
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Normal range for phosphorus levels is 2.6-4.8. In order to assess
whether the data indicate an elevated phosphorus level, we should test
  null hypothesis H0: mu=4.8
  against one-sided alternative Ha: mu>4.8

Note that the question asked refers to a phosphorus level exceeding 4.8,
however with the potentially invalid justification that the patient's
mean level is high. It is NOT valid to look at the data and decide the 
alternative based on the observed values! There may however be valid
reasons to be particularly concerned about elevated phophorus levels.

Test statistic: t=(Xmean-4.8)/sd(Xmean) = (5.367-4.8)/0.2716 = 2.088

A t-distribution table (Table C of PSLS, Table 3 of S, Table D of IPS) 
gives the percentiles of t(5):
  95% percentile = 2.015, 97.5% percentile = 2.571
  (note that the percentiles are found in the column corresponding to
   1 minus the one-tailed alpha, or P)

The observed t-value falls between these limits, therefore the P-value
is between 0.025 and 0.05 (and presumably closest to 0.05). Statistical
software gives P=0.046. Formally this constitutes evidence (at the 5%
level) that the patient's phosphorus is higher than normal, however
because the P-value is so close to 0.05 we should consider this as weak
evidence only. In practice the strength of evidence required will often
depend on the consequences. For example, if the follow-up procedure is
a more detailed examination of the patient, no further evidence may be
required. 

The assumptions of the analysis are:
  - independent observations,
  - normal distribution of the phosphorus measurements (and for such a 
small sample the robustness of the t-procedure against violations is not
great, so some justification of the normality assumption is needed)
  - same mean and standard deviation of all observations.

The small sample size means that it is almost pointless to try assess
the normal distribution from the data. The justification of assuming a
normal distribution should come from knowledge about the distribution of
phosphorus measurements in a larger sample.


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Analysis by Minitab:

MTB > WOpen "H:\VHM\VHM801\Datasets\Minitab\Chapter 7\ex07_073.mtw".
Retrieving worksheet from file: 'H:\VHM\VHM801\Datasets\Minitab\Chapter
7\ex07_073.mtw'
Worksheet was saved on 02/10/2014

MTB > Describe 'phos'.
Descriptive Statistics: phos 

Variable  N  N*   Mean  SE Mean  StDev  Minimum     Q1  Median     Q3  Maximum
phos      6   0  5.367    0.272  0.665    4.600  4.750   5.350  5.875    6.400

MTB > Dotplot 'phos'.
Dotplot of phos 

MTB > PPlot 'phos';
SUBC>   Normal;
SUBC>   Symbol;
SUBC>   FitD;
SUBC>   Grid 2;
SUBC>   Grid 1;
SUBC>   MGrid 1.
Probability Plot of phos 
The P-value of the A-D normality test is 0.76.

MTB > OneT 'phos';
SUBC>   Confidence 90;
SUBC>   Alternative 0.
One-Sample T: phos 

Variable  N   Mean  StDev  SE Mean      90% CI
phos      6  5.367  0.665    0.272  (4.819, 5.914)

MTB > OneT 'phos';
SUBC>   Test 4.8;
SUBC>   Confidence 95;
SUBC>   Alternative 1.
One-Sample T: phos 

Test of mu = 4.8 vs > 4.8

Variable  N   Mean  StDev  SE Mean  95% Lower Bound     T      P
phos      6  5.367  0.665    0.272            4.819  2.09  0.046

Comment:
--------
The last Minitab listing with the t-test against a one-sided alternative
includes a one-sided CI instead of the usual two-sided CI. In VHM 801
and also in general statistical practice, all confidence intervals are 
two-sided. One practical implication of this Minitab setting is if one
wants both a confidence interval and a test against a one-sided
alternative, the t-test menu needs to be run twice.