A sample of 44 observations is selected from one population

A sample of 44 observations is selected from one population with a 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 0.04 significance level. Wo: p 1 = M2 Hi:Mi * M2 This is a two 0-tailed test. State the decision rule. (Negative amounts should be indicated by a minus sign. Round your answer to 2 decimal places.) The decision rule is to reject H0 if z is outside fl the interval ( , ). Compute the value of the test statistic. (Round your answer to 2 decimal places.)Value of the test statistic What is your decision regarding Hq? Do not reject Q Hq. What is the /3-value? (Round your answer to 4 decimal places.) 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)=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 0.04% is 2.054
We got |Zo | =1.611 & | Z | =2.054
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 P0.04 < 0.10723,Here We Do not Reject Ho

[ANSWERS]
a)two
b) -2.054, 2.054
c)Zo =1.61
d) Do not Reject Ho
e) ( P != 1.61 ) = 0.10723