Consider the following sample data drawn independently from

Consider the following sample data drawn independently from normally distributed populations with equal population variances. Use Table 2.

Construct the relevant hypotheses to test if the mean of the second population is greater than the mean of the first population.

Calculate the value of the test statistic. (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 1% level of significance. (Negative value should be indicated by a minus sign. Round your answer to 3 decimal places.)

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)=10.975; Standard Deviation (s.d1)=2.408
Number(n1)=8
Y(Mean)=11.8; Standard Deviation(s.d2)=2.2633
Number(n2)=8
Value Pooled variance S^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
S^2 = (7*5.7985 + 7*5.1225) / (16- 2 )
S^2 = 5.4605
we use Test Statistic (t) = (X-Y)/Sqrt(S^2(1/n1+1/n2))
to=10.975-11.8/Sqrt((5.4605( 1 /8+ 1/8 ))
to=-0.825/1.1684
to=-0.7061
| to | =0.7061
Critical Value
The Value of |t | with (n1+n2-2) i.e 14 d.f is 1.761
We got |to| = 0.7061 & | t | = 1.761
Make Decision
Hence Value of |to | < | t | and Here we Do not Reject Ho
P-Value: Left Tail - Ha : ( P < -0.7061 ) = 0.24585
Hence Value of P0.05 < 0.24585,Here We Do not Reject Ho

a) H0: 1 2 0; HA: 1 2 < 0
b) to =-0.71
c) 1.761
d) No, since the p-value is more than .
e) No, since the value of the test statistic is not less than the critical value of -1.761.