Consider the following competing hypotheses and accompanying

Consider the following competing hypotheses and accompanying sample data drawn independently from normally distributed populations. Use Table 2.

Calculate the value of the test statistic under the assumption that the population variances are unknown but equal. (Negative values should be indicated by a minus sign. Round intermediate calculations to 4 decimal places and final answer to 2 decimal places.)

Calculate the critical value at the 5% level of significance. (Negative value should be indicated by a minus sign. Round your answer to 3 decimal places.)

Calculate the value of the test statistic under the assumption that the population variances are unknown and are not equal. (Negative values should be indicated by a minus sign. Round intermediate calculations to 4 decimal places and final answer to 2 decimal places.)

Calculate the critical value at the 5% level of significance. (Negative value should be indicated by a minus sign. Round your answer to 3 decimal places.)

Solution

I) WHEN UNKNOWN & NOT EQUAL
Set Up Hypothesis
Null Hypothesis , There Is No-Significance between them Ho: u1 - u2 >=0
Alternate Hypothesis, There Is Significance between them - H1: u1 - u2<0
Test Statistic
X(Mean)=222
Standard Deviation(s.d1)=32 ; Number(n1)=12
Y(Mean)=253
Standard Deviation(s.d2)=26; Number(n2)=12
we use Test Statistic (t) = (X-Y)/Sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =222-253/Sqrt((1024/12)+(676/12))
to =-2.6
| to | =2.6
Critical Value
The Value of |t | with Min (n1-1, n2-1) i.e 11 d.f is 1.796
We got |to| = 2.60452 & | t | = 1.796
Make Decision
Hence Value of | to | > | t | and Here we Reject Ho
P-Value:Left Tail - Ha : ( P < -2.6045 ) = 0.01225
Hence Value of P0.05 > 0.01225,Here we Reject Ho

a.1 -2.6
a.2 1.796
a-3. Yes, since the value of the test statistic is not less than the critical value.

Set Up Hypothesis
Null Hypothesis, There Is NoSignificance between them Ho: u1 - u2 >=0
Alternative Hypothesis, There Is Significance between themH1: u1 - u2 < 0
Test Statistic
X (Mean)=222; Standard Deviation (s.d1)=32
Number(n1)=12
Y(Mean)=253; Standard Deviation(s.d2)=26
Number(n2)=12
Value Pooled variance S^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
S^2 = (11*1024 + 11*676) / (24- 2 )
S^2 = 850
we use Test Statistic (t) = (X-Y)/Sqrt(S^2(1/n1+1/n2))
to=222-253/Sqrt((850( 1 /12+ 1/12 ))
to=-31/11.9024
to=-2.6045
| to | =2.6045
Critical Value
The Value of |t | with (n1+n2-2) i.e 22 d.f is 1.717
We got |to| = 2.6045 & | t | = 1.717
Make Decision
Hence Value of | to | > | t | and Here we Reject Ho
P-Value: Left Tail - Ha : ( P < -2.6045 ) = 0.00809
Hence Value of P0.05 > 0.00809,Here we Reject Ho

b.1 -2.6045
b.2 1.717
b-3. Yes, since the value of the test statistic is not less than the critical value.