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