91Ó°ÊÓ

The Infant data is given in the following table:

\(\begin{array}{*{20}{c}}i&{{\rm{ DD method, }}{x_i}}&{{\rm{ TW method, }}{y_i}}&{{\rm{ Difference, }}{d_i} = {x_i} - {y_i}}\\1&{1509}&{1498}&{11}\\2&{1418}&{1254}&{164}\\3&{1561}&{1336}&{225}\\4&{1556}&{1565}&{ - 9}\\5&{2169}&{2000}&{169}\\6&{1760}&{1318}&{442}\\7&{1098}&{1410}&{ - 312}\\8&{1198}&{1129}&{69}\\9&{1479}&{1342}&{137}\\{10}&{1281}&{1124}&{157}\\{11}&{1414}&{1468}&{ - 54}\\{12}&{1954}&{1604}&{350}\\{13}&{2174}&{1722}&{452}\\{14}&{2058}&{1518}&{540}\\{}&{}&{}&{}\end{array}\)

The normal probability plot indicated that the differences are normally distributed. It is plausible that the population distribution of differences is normal.

Because the experiment performed on the same\(14\)randomly selected infants, the pair\(t\)test should be used.

The Paired\(t\)Test:

When\(D = X - Y\)(difference between observations within a pair),\({\mu _D} = {\mu _1} - {\mu _2}\)

and null hypothesis

\({H_0}:{\mu _D} = {\Delta _0},\)

the test statistic value for testing the hypotheses is

\(t = \frac{{\bar d - {\Delta _0}}}{{{s_D}/\sqrt n }}\)

where \(\bar d\) and\({s_D}\) are the sample mean and the sample standard deviation of differences\({d_i}\), respectively. In order to use this test, assume that the differences \({D_i}\) are from a normal population. Depending on alternative hypothesis, the \(P\) value can be determined as the corresponding area under the \({t_{n - 1}}\)curve.

The hypotheses of interest are\({H_o}:{\mu _D} = 0\)versus\({H_a}:{\mu _D} \ne 0\). The value of test statistic can be obtained when the sample mean and the sample standard deviation of the differences is found.

The Sample Mean\(\bar x\)of observations\({x_1},{x_2}, \ldots ,{x_n}\)is given by

\(\bar x = \frac{{{x_1} + {x_2} + \ldots + {x_n}}}{n} = \frac{1}{n}\sum\limits_{i = 1}^n {{x_i}} \)

The sample mean\(\bar d\)for the differences is

\(\bar d = \frac{1}{{14}} \cdot (11 + 164 + \ldots + 540) = 167.2\)

The delta value is

\({\Delta _0} = 0\)

The only missing value to compute the test statistic value is the sample standard deviation of differences\({d_i}\).

The Sample Variance\({s^2}\)is

\({s^2} = \frac{1}{{n - 1}} \cdot {S_{xx}}\)

\({S_{xx}} = \sum {{{\left( {{x_i} - \bar x} \right)}^2}} = \sum {x_i^2} - \frac{1}{n} \cdot {\left( {\sum {{x_i}} } \right)^2}\)

The Sample Standard Deviation\(s\)is

\(s = \sqrt {{s^2}} = \sqrt {\frac{1}{{n - 1}} \cdot {S_{xx}}} \)

The sample variance is

\(\begin{array}{l}s_D^2 = \frac{1}{{14 - 1}} \cdot \left( {{{(11 - 167.2)}^2} + {{(164 - 167.2)}^2} + \ldots + {{(540 - 167.2)}^2}} \right)\\ = 52,080.18\end{array}\)

and the sample standard deviation\({s_D} = \sqrt {52,080.18} = 228\)

The test statistic value is

\(t = \frac{{\bar d - {\Delta _0}}}{{{s_D}/\sqrt n }} = \frac{{167.2 - 0}}{{228/\sqrt {14} }} = 2.74\)

The test is two-sided where the alternative hypothesis is\({H_a}:{\mu _D} \ne 0\), therefore the\(P\)value is two times the area under the the\({t_{n - 1}} = {t_{14 - 1}} = \$ \$ {t_1}3\)curve to the right of\(|t|\)value

\(P = 2 \cdot P(T > 2.74) = 2 \cdot 0.009 = 0.018\)

where\(T\)has student distribution with\(13\)degrees of freedom, and the probability as computed using software (you can use table in the appendix of the book). At significance level\(0.05\), because

\(P = 0.018 < 0.05\)

reject null hypothesis

at given significance level. There is enough evidence to conclude that the true average difference between the values of the two methods is different than \(0.\)