Comparing Two Means - Matched Pairs

1.

a)

MINITAB Analysis: 

	MATCHED PAIRS   -    MOTHER DAUGHTER HEIGHT EXAMPLE
 MTB > name c12 'dheight'  (dheight is the daughter's height in inches)
 MTB > name c13 'mheight'  (mheight is the mother's height in inches)
 MTB > let c20 = c12- c13
 MTB > name c20 'd-m' ( d-m is the column of differences- the daughter's
			 height minus her mother's height.)
 

 MTB > desc c12 c13
 
                 N     MEAN   MEDIAN   TRMEAN    STDEV   SEMEAN
 dheight        30   65.000   65.000   64.962    2.807    0.512
 mheight        30   64.050   63.500   63.827    2.621    0.478
 
               MIN      MAX       Q1       Q3
 dheight    60.000   71.000   63.000   67.000
 mheight    60.000   73.000   62.000   66.000
 
 
 Here are the data:
  ROW  dheight  mheight     d-m
 
    1     67.0     67.0     0.0
    2     67.0     63.0     4.0
    3     64.0     62.0     2.0
    4     60.0     63.0    -3.0
    5     71.0     68.0     3.0
    6     60.0     63.0    -3.0
    7     64.0     66.0    -2.0
    8     68.0     64.0     4.0
    9     65.0     66.0    -1.0
   10     62.0     65.0    -3.0
   11     68.0     65.0     3.0
   12     66.0     64.0     2.0
   13     70.0     73.0    -3.0
   14     63.0     61.0     2.0
   15     64.0     62.0     2.0
   16     63.0     62.0     1.0
   17     61.0     62.0    -1.0
   18     66.5     62.0     4.5
   19     67.0     66.0     1.0
   20     64.0     64.0     0.0
   21     61.0     63.0    -2.0
   22     62.0     61.0     1.0
   23     66.0     63.0     3.0
   24     68.0     66.0     2.0
   25     64.0     60.0     4.0
   26     66.0     62.0     4.0
   27     64.0     64.0     0.0
   28     65.5     62.5     3.0
   29     65.0     65.0     0.0
   30     68.0     67.0     1.0
 

 MTB > boxplot c20  (this is a boxplot of the differences d - m)
	If there were no difference between daughter's heights and mother's
 	heights  we would expect to see a boxplot centered on zero.  That is
 	not the case here.  The median is above zero.

	The boxplot is fairly symmetric and there are no outliers.
 
                            ----------------------------
               -------------I             +            I----------
                            ----------------------------
           ----+---------+---------+---------+---------+---------+--d-m  
            -3.0      -1.5       0.0       1.5       3.0       4.5
 
 MTB > hist c20
 
 Histogram of d-m   N = 30
 
 Midpoint   Count
       -3       4  ****
       -2       2  **
       -1       2  **
        0       4  ****
        1       4  ****
        2       5  *****
        3       4  ****
        4       4  ****
        5       1  *
 

 
 
 
 MTB > nscores c20 c30
 MTB > plot c20 c30
 
          -                                                      *
  d-m     -                                           4
          -
          -                                   4
       2.5+
          -                             5
          -
          -                        4
          -
       0.0+                   4
          -
          -               2
          -
          -           2
      -2.5+
          -    4
          -
            ----+---------+---------+---------+---------+---------+--C30     
            -1.40     -0.70      0.00      0.70      1.40      2.10
 
 
 MTB > desc c20 ( c20 is the column of differences d - m)
 
                 N     MEAN   MEDIAN   TRMEAN    STDEV   SEMEAN
 d-m            30    0.950    1.000    1.000    2.372    0.433
 
               MIN      MAX       Q1       Q3
 d-m        -3.000    4.500   -1.000    3.000
 
 MTB > ttest 0 c20;
 SUBC> alt = +1. (This specifies that we want to do a test with greater than
			in the alternate hypothesis.)
 
 TEST OF MU = 0.000 VS MU G.T. 0.000
 
              N      MEAN    STDEV   SE MEAN        T    P VALUE
 d-m         30     0.950    2.372     0.433     2.19      0.018

There is strong evidence against the null hypothesis that there is no
difference on average between the heights of the mothers and their daughters.

At alpha = .05 we would conclude that on average, the daughters are taller
than their mothers.


 MTB > tint c20

	     N      MEAN     STDEV   SE MEAN   95.0 PERCENT C.I.
d-m	    30     0.950     2.372    0.433    ( 0.064,  1.836)

The 95% confidence interval tells us that the mean difference in height
 between daughters and their mothers is between 0.064 inches and 1.836 inches.


What if we took the difference  as mother's height minus daughter's height?
The results are shown below.
 MTB > let c40 = c13-c121
 MTB > name c40 'm-d'
 
 MTB > ttest 0 c40;
 SUBC> alt = -1.  (This specifies that we want to do a test with less than in
			the alternate hypothesis.
 
 TEST OF MU = 0.000 VS MU L.T. 0.000
 
              N      MEAN    STDEV   SE MEAN        T    P VALUE
 m-d         30    -0.950    2.372     0.433    -2.19      0.018

The p-value is the same, but we had to remember to write the null hypothesis
as the mean of the mother's heights is less than the mean of the daughter's
heights.

 MTB > tint c40
 
              N      MEAN    STDEV  SE MEAN   95.0 PERCENT C.I.
 m-d         30    -0.950    2.372    0.433  (  -1.836,  -0.064)
 
The confidence interval has the same values but they are minus values
because the differences are mothers height minus daughter's height, and
the mothers are on average shorter.
 
 The 95% confidence interval tells us that the mean difference in height
 between daughters and their mothers is between 0.064 inches and 1.836 inches.

di = daughter's height - mother's height

$H_{0}:\ \mu_{D} = \mu_{M}$ or $\mu_{D} - \mu_{M} = 0$ or $\mu_{d} = 0$

$H_{A}:\ \mu_{D} > \mu_{M}$ or $\mu_{D} - \mu_{M} > 0$ or $\mu_{d} > 0$

Where

$\mu_{D}$ = true mean height of daughters

$\mu_{M}$ = true mean height of mothers

$\mu_{d}$ = true mean difference in height between daughters and mothers

$\overline{x}_{d} = 0.950$, sd = 2.372, nd = 30

$t = \frac{0.950 - 0}{\frac{2.372}{\sqrt{30}}} = 2.19$

t has 29 df, 0.01 < p < 0.02

There is very strong evidence against the null hypothesis of equal height. The null hypothesis will be rejected for any $\alpha \ge 0.02$.

b)

$90\%$ confidence interval for $\mu_{D} - \mu_{M}$.

$0.950 \pm 1.699\left(\frac{2.372}{\sqrt{30}}\right)$

t has 29 df, t* = 1.699

$0.950 \pm 0.736$

(0.21 inches, 1.69 inches)


2.

$\overline{d} = \overline{x}_{d} = 349.7$, sd = 264.65, nd = 6

a)

$H_{0}: \ \mu{E} = \mu_{S}$ or $\mu_{E} - \mu{S} = 0$

$H_{A}: \ \mu{E} > \mu_{S}$ or $\mu_{E} - \mu{S} > 0$

Where

$\mu_{E}$ = mean weight gain on enriched formula

$\mu_{S}$ = mean weight gain on standard formula

$t = \frac{349.7 - 0}{\frac{264.65}{\sqrt{6}}} = 3.232$, t has 5 df

0.01 < p < 0.02

At $\alpha$ = 0.05 reject H0. At $\alpha$ = 0.01 do not reject H0.

b)

$H_{0}: \ \mu{E} = \mu_{S}$ or $\mu_{E} - \mu{S} =300$

$H_{A}: \ \mu{E} > \mu_{S}$ or $\mu_{E} - \mu{S} > 300$

$t = \frac{349.7 - 300}{\frac{264.65}{\sqrt{6}}} = 0.46$

t has 5 df, p > 0.25.

There is little or no evidence against the null hypothesis.

c)

$95\%$ C.I. for $\mu_{E} - \mu_{S}$

t has 5 df, t* = 2.571

$349.7 \pm 2.571\left(\frac{264.65}{\sqrt{6}}\right)$

$349.7 \pm 277.78$

(71.92 g, 627.48 g)

The babies on enriched formula gain, on average, between 71.92 g and 627.48 g more than those on standard formula.

d)

m = 100 g, $95\%$ C.I., z* = 1.96, s = 264.65 (from pilot study of six pairs)

$n = \left(\frac{(1.96)(264.65)}{100}\right)^{2} = 26.9$, need at least 27 pairs.


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Jonathan Payne
1999-04-06