4 The following data recorded in days represent the length o
Solution
Set Up Hypothesis
Null Hypothesis, There Is NoSignificance between them Ho: u1 > u2
Alternative Hypothesis, There Is Significance between themH1: u1 < u2
Test Statistic
X (Mean)=17; Standard Deviation (s.d1)=1.5
Number(n1)=14
Y(Mean)=19; Standard Deviation(s.d2)=1.8
Number(n2)=16
Value Pooled variance S^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
S^2 = (13*2.25 + 15*3.24) / (30- 2 )
S^2 = 2.7804
we use Test Statistic (t) = (X-Y)/Sqrt(S^2(1/n1+1/n2))
to=17-19/Sqrt((2.7804( 1 /14+ 1/16 ))
to=-2/0.6102
to=-3.2775
| to | =3.2775
Critical Value ( at 0.05)
The Value of |t ?| with (n1+n2-2) i.e 28 d.f is 1.7
We got |to| = 3.2775 & | t ? | = 1.7
Make Decision
Hence Value of | to | > | t ?| and Here we Reject Ho
P-Value: Left Tail - Ha : ( P < -3.2775 ) = 0.0014
Hence Value of P0.05 > 0.0014,Here we Reject Ho
We claim that First is smaller
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at 0.01 LOS
Critical Value
The Value of |t ?| with (n1+n2-2) i.e 28 d.f is 2.47
We got |to| = 3.2775 & | t ? | = 2.47
Make Decision
Hence Value of | to | > | t ?| and Here we Reject Ho
P-Value: Left Tail - Ha : ( P < -3.2775 ) = 0.0014
Hence Value of P0.01 > 0.0014,Here we Reject Ho
We claim that First is smaller
