Stat 407 Lab 12 Inference for the Mean Fall 2001 SOLUTION
In this lab we use SPlus to compute Hotellings T 2 , and simultaneous confidence regions, and Bonferroni
multiple comparisons.
Download the salsolinol data from class web site. Use Dr Cook’s version that is structured to be entered into
S-Plus. The data comes from Table 10.1 in the text. It comes from a study on the role the alkaloid salsolinol
plays on bodily dependence on alcohol. The patients were studied for 4 days following admission to an alcohol
treatment unit. Urine samples were taken daily. The patients were previously categorized into two groups,
moderate dependence (6) and severe dependence (8). The response variable is the concentration of salsolinol in
the urine samples.
1. Plot the data using histograms for each day, without conditioning by group. Comment on the normality
or lack of it based on these plots. If you detect lack of normality, suggest a remedy.
All the variables are highly skewed. We examine using a log transformation, which seems to work reasonably
well. All variables should be transformed in a similar manner since they are all measuring the same
quantities, salsolinol concentration.
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2. Calculate the overall mean, and overall variance of the data (or transformed data).
X̄ = (0.019 0.184 0.565

1.107 −0.038
 −0.038 0.682

S = 
 0.146
0.012
−0.559 0.187




S−1 = 
0.781)0
0.146 −0.560
0.012 0.187
0.594 0.059
0.059 0.720






1.69 −0.29 −0.55 1.44
−0.29 1.63
0.10 −0.66 


−0.55 0.10
1.88 −0.61 
1.44 −0.66 −0.61 2.73
3. Compute Hotellings T 2 to test the hypothesis:
Ho : µ = (2.0 2.0 2.5 2.5) vs HA : µ 6= (2.0 2.0 2.5 2.5)
at the α = 0.10 level of significance. Be sure to adapt the hypothesis if you have transformed your data.
Note also that F4,10 = 2.61.
Hotellings T 2 = 16.2 for logged data and the critical value is 13.5. Hence we would reject Ho .
4. Calculate the 90% simultaneous confidence intervals for the individual variables of population mean.
0.019 ±
0.184 ±
0.565 ±
0.781 ±
√
√
√
√
q
13.5 1.107/14 = (−1.01, 1.05)
q
13.5 0.682/14 = (−0.63, 0.99)
q
13.5 0.594/14 = (−0.19, 1.32)
q
13.5 0.719/14 = (−0.05, 1.61)
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5. Calculate 90% Bonferroni adjusted confidence intervals for each individual variables of population mean.
Note that t13 (0.1/(2 × 4)) = 2.53.
q
0.019 ± 2.53 1.107/14 = (−0.69, 0.73)
q
0.184 ± 2.53 0.682/14 = (−0.37, 0.74)
q
0.565 ± 2.53 0.594/14 = (−0.04, 1.09)
q
0.781 ± 2.53 0.719/14 = (0.21, 1.72)
6. Plot the simultaneous and Bonferroni confidence intervals for each variable, along with the hypothesized
mean on a histogram or scatterplot matrix view of the transformed data. Discuss the results.
The Bonferroni intervals are shorter than the simultaneous confidence intervals. This reflects the conservative contraction of the intervals to compensate for the lack of use of the variance-covariance structure.
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