91影视

The hypotheses of interest are \({H_0}:{\mu _1} - {\mu _2} = 0\) versus \({H_a}:{\mu _1} - {\mu _2} \ne 0.\)

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

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

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

\(T = \frac{{\overline X - \overline Y - {\Delta _0}}}{{\sqrt {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} }}\)

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

Everything is given in the exercise; thus, the \(t\) value is

\(\begin{array}{c}T = \frac{{\overline X - \overline Y - {\Delta _0}}}{{\sqrt {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} }}\\ = \frac{{807 - 757}}{{\sqrt {\frac{{{{27}^2}}}{{10}} + \frac{{{{41}^2}}}{{10}}} }}\\ = \frac{{50}}{{\sqrt {241} }}\\ = 3.22.\end{array}\)

The degrees of freedom can be computed from

\(\begin{array}{c}\nu = \frac{{{{\left( {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} \right)}^2}}}{{\frac{{{{\left( {s_1^2/m} \right)}^2}}}{{m - 1}} + \frac{{{{\left( {s_2^2/n} \right)}^2}}}{{n - 1}}}}\\ = \frac{{{{241}^2}}}{{\frac{{{{72.9}^2}}}{9} + \frac{{{{168.1}^2}}}{9}}}\\ = 15.6\end{array}\)

which needs to be rounded down to the nearest integer; thus

\(\nu = 15.\)

The alternative hypothesis is two-sided, which indicates that the \(P\) value is two times the area under \({t_{15}}\)curve to the right of \(|t|\)

\(\begin{array}{c}P = 2P(T > 3.22)\\ = 2 \times 0.003\\ = 0.006\end{array}\)

Because the \(P\) value is

\(P < 0.01 < 0.05 < 0.1\)

Reject null hypothesis at given significance levels.

The difference between the compression strength using fixed and floating platen method is big (there is difference) compared to normal variation in compression strength between identical boxes. Which means that the done analysis does not agree with the authors. The prediction of normality was assumed.