Question

Suppose you needed to test the claim that the two samples described below come from populations with the same mean. Assume that the samples are independent simple random samples.

Sample 1: n1=17, x¯¯¯1=24.7, s1=3.05n1=17, x¯1=24.7, s1=3.05

Sample 2: n2=7, x¯¯¯2=22.5, s2=4.62n2=7, x¯2=22.5, s2=4.62

Compute:

(a) the degrees of freedom:

(b) the test statistic (use Sample 1 −− Sample 2):

(c) he P-value:

Answer 1

Given

Sample 1: n1=17, x1=24.7, s1=3.05

Sample 2: n2=7, x2=22.5, s2=4.62

To test

H0 : 1 = 2      ( two populations have same mean )

H1 : 1 2      ( two populations mean differ significantly )

we have n1 = 17   ,  1 =24.7      , s1 = 3.05

            n2=7       , 2=22.5       , s2=4.62

Test statistics T.S :

T.S =

where

S.E =

To calculate sd² we have

sd²   =

Using above samples information

sd²   =

after calculation

sd²   =

Hence

S.E = =    = ( 2.734156 ) ^1/2= 1.653528

Hence S.E = 1.653528

Thus

T.S =

       = = 1.186761

Hence calculated test statistics is T.S = 1.330488

a) Degree of freedom df = n1+n2-2 = 17+7-2 = 22

b) the test statistic is T.S = 1.186761

c) To calculate P-value we have

P-value = P ( t < -T.S ) + P (t > T.S )

               = 2 * P ( t < -T.S )

P-value =    2 * P ( t < -1.330488)           

where t is t-distributed with n1-n2-2 = 22 degree of freedom

Required probability can be computed from statistical book or more accurately from any software like R,Excel

From R       

> 2*pt( -1.330488,22)             # 2 * P ( t < -1.186761 )   
[1] 0.1969851

So 2 * P ( t < -1.330488 )     = 0.1969851

Hence

P-value =    2 * P ( t < -1.330488 ) =0.1969851

i.e P-value = 0.1969851

Conclusion

Now P-value = 0.1969851 > 0.05 then we can not reject null hypothesis at 5% of level of significance ,

hence we conclude that two samples described below may come from populations with the same mean.