A sample of 44 observations is selected from one population

A sample of 44 observations is selected from one population standard deviation of 4.9. The sample mean is 100.0. A sample of 44 observations is selected from a second population with a population standard deviation of 5.0. The sample mean is 98.3. conduct the following test of hypothesis using the o.04 significance level. H_o:mu = mu_2 H_1 : mu is not equal to mu_2 this is a two tailed test atate the decision rule compute the value of the test statistic. what is your decision regarding H_0? what is the p-value?

Solution

Set Up Hypothesis
Null, Ho: u1 = u2
Alternate, H1: u1 != u2
Test Statistic
X(Mean)=100
Standard Deviation(s.d1)=4.9
Number(n1)=44
Y(Mean)=98.3
Standard Deviation(s.d2)=5
Number(n2)=44
we use Test Statistic (Z) = (X-Y)/Sqrt(s.d1^2/n1)+(s.d2^2/n2)
Zo=100-98.3/Sqrt((24.01/44)+(25/44))
Zo =1.61
| Zo | =1.61
Critical Value
The Value of |Z | at LOS 0.02% is 2.326
We got |Zo | =1.611 & | Z | =2.326
Make Decision
Hence Value of | Zo | < | Z | and Here we Do not Reject Ho
P-Value: Two Tailed ( double the one tail ) - Ha : ( P != 1.61 ) = 0.10723
Hence Value of P0.02 < 0.10723,Here We Do not Reject Ho

[ANSWERS]
1. Two Tailed
2. Reject Ho if Z<-2.33, OR Z>2.33
3. Zo =1.611
4. Do not Reject Ho
5. P-Value: Two Tailed ( double the one tail ) - Ha : ( P != 1.61 ) = 0.10723