91Ó°ÊÓ

(a):

The hypotheses of interest are \({H_0}:{\mu _1} - {\mu _2} = 0\)versus\({H_a}:{\mu _1} - {\mu _2} \ne 0\), where\({\mu _1}\), and \({\mu _2}\)are the true average weights for operations 1 and 2, respectively.

Given two normal distributions, the random variable (standardized)

\(T = \frac{{\bar X - \bar Y - \left( {{\mu _1} - {\mu _2}} \right)}}{{\sqrt {\frac{{S_1^2}}{m} + \frac{{S_2^2}}{n}} }}\)

has approximately students t distribution with degrees of freedom\(\nu \), where \(\nu \) is

\(\nu = \frac{{\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}}}{{\frac{{{{\left( {s_1^2/m} \right)}^2}}}{{m - 1}} + \frac{{{{\left( {s_2^2/n} \right)}^2}}}{{n - 1}}}}\)

has to be rounded down to the nearest integer.

The two-sample t test for testing \({H_0}:{\mu _1} - {\mu _2} = {\Delta _0}\)uses the following value of test statistic

\(t = \frac{{\bar x - \bar y - {\Delta _0}}}{{\sqrt {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} }}\)

For the adequate alternative hypothesis the adequate area under the \({t_\nu }\)curve is the P value.

The value of the test statistic is

\(t = \frac{{1402.24 - 1419.63 - 0}}{{\sqrt {\frac{{{{10.97}^2}}}{{30}} + \frac{{{{9.96}^2}}}{{30}}} }} = \frac{{ - 17.39}}{{\sqrt {4.0114 + 3.307} }} = - 6.43.\)

The degrees of freedom are

\(\nu = \frac{{\frac{{s_1^2}}{m} + \frac{s}{2}{{_2^2}^2}}}{{\frac{{{{\left( {s_1^2/m} \right)}^2}}}{{m - 1}} + \frac{{{{\left( {s_2^2/n} \right)}^2}}}{{n - 1}}}} = \frac{{\frac{{{{10.97}^2}}}{{30}} + {{\frac{{{{9.96}^2}}}{{30}}}^2}}}{{\frac{{{{\left( {{{10.97}^2}/30} \right)}^2}}}{{30 - 1}} + \frac{{{{\left( {{{9.96}^2}/30} \right)}^2}}}{{30 - 1}}}} = 57.5,\)

and the round down to the nearest integer, which actually represent the degrees of freedom is

\(\nu = 57.\)

The P value is two times the area under the \({t_{58}}\)curve to the right of |t|

\(P = 2 \cdot P(T > 6.43) \approx 2 \cdot 0 = 0\)

which indicates to

Reject null hypothesis

\(\begin{array}{l}{\rm{at any reasonable significance level, as well as }}\alpha = 0.05.{\rm{ There is a statistically significant }}\\{\rm{difference in the means}}{\rm{. }}\end{array}\)

(b):

The one-sample t test should be used to the hypotheses \({H_0}:{\mu _1} = 1400\)versus\({H_a}:{\mu _1} > 1400\). The test statistic value is

\(\begin{array}{l}t = \frac{{\bar x - 1400}}{{{s_1}/\sqrt m }} = \frac{{1402.24 - 1400}}{{10.97/\sqrt {30} }}\\ = \frac{{2.24}}{2} = 1.1.\end{array}\)

The degrees of freedom are\(m - 1 = 30 - 1 = 29\). The P value is the area under \({t_{29}}\)curve to the right of t value

\(P = P(T > 1.1) = 0.14\)

where the value was computed using software (you can estimate it using table in the appendix). Since

\(P = 0.14 > 0.05\)

Do not reject null hypothesis

The true average does not exceed 1400.

\(\begin{array}{l}{\rm{a}}{\rm{. Reject null hypothesis; }}\\{\rm{b}}{\rm{. Do not reject null hypothesis}}{\rm{. }}\end{array}\)