Surfstat.australia: an online text in introductory Statistics

STATISTICAL INFERENCE

COMPARING MEANS OF TWO CONTINUOUS VARIABLES

Paired Samples

Sometimes there is a natural pairing between observations in one sample and the other, e.g. shoes worn by the same boy. In this case the assumption that the 2 samples are independent is likely to be wrong so the method described above is not correct. Instead make use of the pairing and use the differences between the paired values

Di = Xi - Yi for i=1, 2, ... , n

If E(Xi) = µX and E(Yi) = µ Y then

E(Di) = E(Xi) - E(Yi) = µX - µY

Also var(Di) = var(Xi) + var(Yi) - 2cov(Xi,Yi)

Although we could try to estimate each of these terms, this is not necessary and instead we let var (Di) = s2D

Inferences about (µX - µY) are based on the single sample of Di values with sD estimated by sD (as usually sD is unknown).

To test H0 : µX = µY versus H1 : µX µY
use where

~ N (µx - µy, s2D / n)

hence

For the boys shoes, using the differences d = A - B

= - 0.410, s = 0.387, n = 10.

If H0 is true then (µX - µY) = 0 

so the observed value of the t-statistic
t9  is








p-value = P(t9< - 3.35 or t> 3.35)
      < 2 × P(t9> 3.250)
      < 2 × 0.005
      < 0.01
As this p value is "small", reject H0 and conclude the result is statistically significant (because the p-value is much less than 0.05).

The data do not support the assumption that µX = µ Y.

Conclusion - There is strong evidence that the materials wear differently.

The unpaired analysis did not detect this difference because it assumed that Xi and Yi were independent (it used an inappropriate model).

An analysis, using MINITAB, is shown over the page. The graph shows that for most boys the wear for material B was greater than for material A. There was considerable variation between boys and this was greater than the variation between materials for each boy individually.

To calculate the 95% CI for µX - µY using these data, we need to find c such that

Pr (-c < t9< c) = 0.95.

From Table 2, c = 2.262.

Hence 95% CI = ( - ) ± 2.262

= (-0.410) ± 2.262

= (-0.687, -0.33).

As the value 0 is not contained in the interval, the data do not support µX - µY = 0. There is a difference in mean wear between material A and material B.

The paired samples analysis is shown below the graph. 

MTB> let c4=c1-c2


MTB> MPlot 'materA' 'student' 'materB' 'student'.


     15.0+
         -                     2
         -      B                                        B
         -      A                                        A
         -
     12.0+                          B
         -                B                        B
         -                A         A              A
         -
         -                                    2
      9.0+           B                                  2
         -           A
         -
         -
         -                               2
      6.0+
         -
           +---------+---------+---------+---------+---------+------
         0.0       2.0       4.0       6.0       8.0      10.0


        A = materA vs. student    B = materB vs. student


MTB> desc c4


                N     MEAN   MEDIAN   TRMEAN    STDEV   SEMEAN


C4             10   -0.410   -0.400   -0.400    0.387    0.122


              MIN      MAX       Q1       Q3
C4         -1.100    0.200   -0.650   -0.200


MTB> ttest c4


TEST OF MU = 0.000 VS MU N.E. 0.000


             N      MEAN    STDEV   SE MEAN        T    P VALUE
C4          10    -0.410    0.387     0.122    -3.35     0.0086

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