A sample of 40 observations is selected from one population

A sample of 40 observations is selected from one population with a population standard deviation of 3.9. The sample mean is 102.0. A sample of 53 observations is selected from a second population with a population standard deviation of 3.2. The sample mean is 101.0. Conduct the following test of hypothesis using the 0.05 significance level.

H0 : 1 = 2 H1 : 1 2

a. This is a one-tail or two tailed test?

b. State the decision rule.

c. Compute the value of the test statistic.

d. What is your decision regarding H₀?

e. What is the p-value?

Solution

Set Up Hypothesis
Null Hypothesis ,there Is No-Significance between them Ho: u1 = u2
Alternate Hypothesis , there Is Significance between them H1: u1 != u2
Test Statistic
X(Mean)=102
Standard Deviation(s.d1)=3.9
Number(n1)=40
Y(Mean)=101
Standard Deviation(s.d2)=3.2
Number(n2)=53
we use Test Statistic (Z) = (X-Y)/Sqrt(s.d1^2/n1)+(s.d2^2/n2)
Zo=102-101/Sqrt((15.21/40)+(10.24/53))
Zo =1.32
| Zo | =1.32
Critical Value
The Value of |Z | at LOS 0.05% is 1.96
We got |Zo | =1.321 & | Z | =1.96
Make Decision
Hence Value of | Zo | < | Z | and Here we Do not Reject Ho
P-Value: Two Tailed ( double the one tail ) - Ha : ( P != 1.32 ) = 0.18666
Hence Value of P0.05 < 0.18666,Here We Do not Reject Ho