Question

Determine the number of degrees of freedom for the two-sample t test or CI in each of the following.

\(\begin{array}{l}a. m = 10, n = 10, {s_1} = 5.0, {s_2} = 6.0\\b. m = 10, n = 15,{s_1} = 5.0, {s_2} = 6.0\\c. m = 10, n = 15, {s_1} = 2.0, {s_2} = 6.0\\d. m = 12, n = 24, {s_1} = 5.0, {s_2} = 6.0\\\\\end{array}\)

Short answer

the solution is

\(\begin{array}{l}a. \nu = 17; b. \nu = 21; c. \nu = 18; d. \nu = 26. \\\end{array}\)

Step-by-step solution

determine the number of degrees

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

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

has a t distribution for students with degrees of freedom v, where v 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.

a) \({\rm{m = 10, n = 10, }}{s_1}{\rm{ = 5, }}{s_2}{\rm{ = 6}}\)

The degrees of freedom v can be calculated using the formula above.

\(\begin{array}{l}\nu = \frac{{{{\left( {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} \right)}^2}}}{{\frac{{{{\left( {s_1^2/m} \right)}^2}}}{{m - 1}} + \frac{{{{\left( {s_2^2/n} \right)}^2}}}{{n - 1}}}} = \frac{{{{\left( {\frac{{{5^2}}}{{10}} + \frac{{{6^2}}}{{10}}} \right)}^2}}}{{\frac{{{{\left( {{5^2}/10} \right)}^2}}}{{10 - 1}} + \frac{{{{\left( {{6^2}/10} \right)}^2}}}{{10 - 1}}}}\\ = \frac{{37.21}}{{0.694 + 1.44}} = 17.43,\end{array}\)

The degrees of freedom for the two-sample t test, also known as Cl, are rounded down to the next integer.

\(\nu = 17.\)

step 2:

\(b. m = 10, n = 15,{s_1} = 5.0, {s_2} = 6.0\)

The degrees of freedom v can be calculated using the formula above.

\(\begin{array}{l}\nu = \frac{{{{\left( {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} \right)}^2}}}{{\frac{{{{\left( {s_1^2/m} \right)}^2}}}{{m - 1}} + \frac{{{{\left( {s_2^2/n} \right)}^2}}}{{n - 1}}}} = \frac{{{{\left( {\frac{{{5^2}}}{{10}} + \frac{{{6^2}}}{{15}}} \right)}^2}}}{{\frac{{{{\left( {{5^2}/10} \right)}^2}}}{{10 - 1}} + \frac{{{{\left( {{6^2}/15} \right)}^2}}}{{15 - 1}}}}\\ = \frac{{24.01}}{{0.694 + 0.411}} = 21.7,\end{array}\)

The degrees of freedom for the two-sample t test, also known as Cl, are rounded down to the next integer.

\(\nu = 21\)

step 3:

\(c. m = 10, n = 15, {s_1} = 2.0, {s_2} = 6.0\)

The degrees of freedom v can be calculated using the formula above.

\(\begin{array}{l}\nu = \frac{{{{\left( {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} \right)}^2}}}{{\frac{{{{\left( {s_1^2/m} \right)}^2}}}{{m - 1}} + \frac{{{{\left( {s_2^2/n} \right)}^2}}}{{n - 1}}}} = \frac{{{{\left( {\frac{{{2^2}}}{{10}} + \frac{{{6^2}}}{{15}}} \right)}^2}}}{{\frac{{{{\left( {{2^2}/10} \right)}^2}}}{{10 - 1}} + \frac{{{{\left( {{6^2}/15} \right)}^2}}}{{15 - 1}}}}\\ = \frac{{7.84}}{{0.018 + 0.411}} = 18.27,\end{array}\)

The degrees of freedom for the two-sample t test, also known as Cl, are rounded down to the next integer.

\(\nu = 18\)

step 4:

\(d. m = 12, n = 24, {s_1} = 5.0, {s_2} = 6.0\)

The degrees of freedom v can be calculated using the formula above.

\(\begin{array}{l}\nu = \frac{{{{\left( {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} \right)}^2}}}{{\frac{{{{\left( {s_1^2/m} \right)}^2}}}{{m - 1}} + \frac{{{{\left( {s_2^2/n} \right)}^2}}}{{n - 1}}}} = \frac{{{{\left( {\frac{{{5^2}}}{{12}} + \frac{{{6^2}}}{{24}}} \right)}^2}}}{{\frac{{{{\left( {{5^2}/12} \right)}^2}}}{{12 - 1}} + \frac{{{{\left( {{6^2}/24} \right)}^2}}}{{24 - 1}}}}\\ = \frac{{12.84}}{{0.395 + 0.098}} = 26.05\\\end{array}\)

The degrees of freedom for the two-sample t test, also known as Cl, are rounded down to the next integer.

\(\nu = 26\)

conclusion

\(\begin{array}{l}a. \nu = 17; b. \nu = 21; c. \nu = 18; d. \nu = 26. \\\end{array}\)