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for the given hypothesis

\({H_0}:{\mu _1} - {\mu _2} = - 10 vs {H_a}:{\mu _1} - {\mu _2} < - 10\)

The random variable (standardized) is given two normal distributions.

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

has a \(t\) distribution with degrees of freedom \(\nu \), where \(\nu \) is the number of students.

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

The number must be rounded to the nearest integer.

To compute the test statistic, all values are provided.

\(\begin{array}{l}t = \frac{{\bar x - \bar y - \left( {{\mu _1} - {\mu _2}} \right)}}{{\sqrt {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} }} = \frac{{115.7 - 129.3 - ( - 10)}}{{\sqrt {\frac{{5.0{3^2}}}{6} + \frac{{5.3{8^2}}}{6}} }}\\ = \frac{{ - 3.6}}{{3.007}} = - 1.2\end{array}\)

The equivalent degrees of freedom can be calculated using the following formula.

\(\begin{array}{l}\nu = \frac{{{{\left( {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} \right)}^2}}}{{\frac{{{{\left( {s_1^2/m} \right)}^2}}}{{m - 1}} + \frac{{{{\left( {s_2^2/n} \right)}^2}}}{{n - 1}}}} = \frac{{{{\left( {\frac{{5.0{3^2}}}{6} + \frac{{5.3{8^2}}}{6}} \right)}^2}}}{{\frac{{{{\left( {5.0{3^2}/6} \right)}^2}}}{{6 - 1}} + \frac{{{{\left( {5.3{8^2}/6} \right)}^2}}}{{6 - 1}}}}\\ = \frac{{{{(4.2168 + 4.8241)}^2}}}{{\frac{{{{(4.2168)}^2}}}{5} + \frac{{4.824{1^2}}}{5}}} = 9.96\end{array}\)

The number must be rounded down to \(9\) in order to acquire the matching degrees of freedom, implying that.

\(\nu = 9\)

\({\rm{d f}}\). The \(P\) value for such test hypotheses is the area to the left of the \(t\) value under the students \(t\)distribution with degrees of freedom \(9\), and the \(P\) value is.

\(P(T \le - 1.2) = 0.13\)

If \(T\) is a random variable with the student distribution given, and the probability was calculated using the appendix table.

With significance level \(0.01\).

Because \(0.01 < 0.13\)

do not reject null hypothesis