1925 Equip male and female students with a small device that

(19.25) Equip male and female students with a small device that secretly records sound for a random 30 seconds during each 12.5-minute period over two days. Count the words each subject speaks during each recording period, and from this, estimate how many words per day each subject speaks. The published report includes a table summarizing six such studies. Here are two of the six: Readers are supposed to understand that, for example, the 56 women in the first study had x = 16,177 and s = 7520.

Solution

STUDY 1
Set Up Hypothesis
Null, Ho: uF = uM
Alternate , - H1: uF > uM
Test Statistic
X(Mean)=16177
Standard Deviation(s.d1)=7520 ; Number(n1)=56
Y(Mean)=16569
Standard Deviation(s.d2)=9108; Number(n2)=56
we use Test Statistic (t) = (X-Y)/Sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =16177-16569/Sqrt((56550400/56)+(82955664/56))
to =-0.248
| to | =0.248
Critical Value
The Value of |t | with Min (n1-1, n2-1) i.e 55 d.f is 1.673
We got |to| = 0.24836 & | t | = 1.673
Make Decision
Hence Value of |to | < | t | and Here we Do not Reject Ho
P-Value:Right Tail -Ha : ( P > -0.2484 ) = 0.59761
Hence Value of P0.05 < 0.59761,Here We Do not Reject Ho

STUDY 2
t Test of Difference Means for Unequal Variance
Set Up Hypothesis
Null, Ho: uF = uM
Alternate , - H1: uF > uM
Test Statistic
X(Mean)=16496
Standard Deviation(s.d1)=7914 ; Number(n1)=27
Y(Mean)=12867
Standard Deviation(s.d2)=8343; Number(n2)=20
we use Test Statistic (t) = (X-Y)/Sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =16496-12867/Sqrt((62631396/27)+(69605649/20))
to =1.507
| to | =1.507
Critical Value
The Value of |t | with Min (n1-1, n2-1) i.e 19 d.f is 1.729
We got |to| = 1.50686 & | t | = 1.729
Make Decision
Hence Value of |to | < | t | and Here we Do not Reject Ho
P-Value:Right Tail -Ha : ( P > 1.5069 ) = 0.07415
Hence Value of P0.05 < 0.07415,Here We Do not Reject Ho

ANS:
B.Ho: uF = uM vs H1: uF > uM
to =-0.248, to =1.507
Right Tail -Ha : ( P > -0.2484 ) = 0.59761, Right Tail -Ha : ( P > 1.5069 ) = 0.07415
Study 1: P > 0.25, study 2: 0.05 < P < 0.10.