Question

Mean 55.533796

Std Dev 4.5528105

Std Err Mean 1.0180394

Upper 95% Mean 57.664577

Lower 95% Mean 53.403015

N 20

t Test Test Statistic Prob > t 0.9170

Answer the following questions by filling in the blanks for each situation.

Report each answer as a number and round to two (2) decimal places. Please note that answers that do not follow this exact specification will not receive any credit.

1. The value of the population mean under the null hypothesis equals

2. The number of degrees of freedom for the sampling distribution of the t-ratio in this problem equals

3. The p-value associated with this test equals

Answer 1

Given

Mean ( )- 55.533796

Std Dev ( Std) - 4.5528105

Std Err Mean ( S.E )- 1.0180394

Upper 95% Mean - 57.664577

Lower 95% Mean - 53.403015

N = 20

t Test Test Statistic Prob > t 0.9170

1. To find the value of the population mean under the null hypothesis equals

We need to find

Here test statistics is given by

T.S =

we have      = 55.533796 and S.E = 1.0180394 , also t-value is 0.9170

Thus , T.S = 0.9170

   and T.S =

Thus     = 0.9170

                         = - S.E * 0.9170

                         = 55.533796 - 1.0180394 *0.9170

                 = 54.60025

Hence , population mean under the null hypothesis equals 54.60025

2. to find the number of degrees of freedom for the sampling distribution of the t-ratio in this problem equals

Degree of freedom t-ratio has is n-1 degree of freedom.

here   n = 20  

thus    n -1 = 19    degree of freeom

Hence the number of degrees of freedom for the sampling distribution of the t-ratio in this problem equals 19

3. To find the p-value associated with this test equals

P - Value =   P ( - T.S < t-table ) + P ( T.s > t-table )

                  = P ( - 0.9170 < ) + P ( 0.9170 > )

                  = P ( - 0.9170 < )   + { 1 - P ( 0.9170 < )    }

                  = 0.18532 + { 1 - 0.81468 }

                  = 0.37064

{

calculating required probabilities from R -

> pt( - 0.9170 , 19 )                # P ( - 0.9170 < )
[1] 0.1853182

> 1-pt( 0.9170 , 19 )           # 1 - P ( 0.9170 < )
[1] 0.1853182

}

Hence

P ( - 0.9170 < )   = 0.18532

P ( 0.9170 > ) = { 1 - P ( 0.9170 < )    }

                    =   { 1 - 0.81468 } = 1 - 0.81468

Hence

P - Value = P ( - 0.9170 < ) + P ( 0.9170 > )

               = 0.18532 + 0.18532

P - Value = 0.37064