Question

Determine the number of degrees of freedom for the two-sample t test or CI in each of the following situations. (Round your answers down to the nearest whole number.)
(a) m = 12, n = 15, s1 = 4.0, s2 = 6.0
(b) m = 12, n = 21, s1 = 4.0, s2 = 6.0
(c) m = 12, n = 21, s1 = 3.0, s2 = 6.0
(d) m = 10, n = 24, s1 = 4.0, s2 = 6.0

Answer 1

Answer:

a

  $df = 24.32$

b

 $df = 30.10$

c

 $df = 30.7$

d

$df = 25.5$

Step-by-step explanation:

Generally degree of freedom is mathematically represented as

          $df = \frac{ [\frac{ s^2_i }{m} + \frac{ s^2_j }{n} ]^2 }{ \frac{ [ \frac{s^2_i}{m} ]^2 }{m-1 } +\frac{ [ \frac{s^2_j}{n} ]^2 }{n-1 } }$

Considering a

         a) m = 12, n = 15, s1 = 4.0, s2 = 6.0

           $df = \frac{ [\frac{ 4^2 }{12} + \frac{ 6^2 }{15} ]^2 }{ \frac{ [ \frac{4^2}{12} ]^2 }{12-1 } +\frac{ [ \frac{6^2}{15} ]^2 }{15-1 } }$

         $df = 24.32$

Considering b

       (b) m = 12, n = 21, s1 = 4.0, s2 = 6.0

          $df = \frac{ [\frac{ 4^2 }{12} + \frac{ 6^2 }{21} ]^2 }{ \frac{ [ \frac{4^4}{12} ]^2 }{12-1 } +\frac{ [ \frac{6^2}{21} ]^2 }{21-1 } }$

        $df = 30.10$

Considering c

      (c) m = 12, n = 21, s1 = 3.0, s2 = 6.0

           $df = \frac{ [\frac{ 3^2 }{12} + \frac{ 6^2 }{21} ]^2 }{ \frac{ [ \frac{3^4}{12} ]^2 }{12-1 } +\frac{ [ \frac{6^2}{21} ]^2 }{21-1 } }$

           $df = 30.7$

Considering c

        (d) m = 10, n = 24, s1 = 4.0, s2 = 6.0

                $df = \frac{ [\frac{ 4^2 }{10} + \frac{ 6^2 }{24} ]^2 }{ \frac{ [ \frac{4^2}{10} ]^2 }{10-1 } +\frac{ [ \frac{6^2}{24} ]^2 }{24-1 } }$

               $df = 25.5$