Answer: 0.31207
Step-by-step explanation: from the question, the distribution follows a normal distribution and population mean of both data set are equal.
I will also assume a 5% level of significance since any level of significance was not given in the question.
Population standard deviation of the first population (σ1) = 6.40
Population variance of the first population (σ1)² = 40.96
Population standard deviation of the second population (σ2) = 7.20
Population variance of the second population (σ2)² = 51.84
Sample size of first population (n1) = 64
Sample size of second population (n2) = 64
Difference between sample mean (x1 - x2) = 0.6.
Since population standard deviation is known, z test is used in getting the z score and a normal distribution table is used in getting the probability attached to the z score.
Z = (x1 - x2) /√{(σ1)²/n1 + (σ2)²/n2}
By substituting parameters, we have
Z = 0.6/√(40.96/64) + (51.84/64)
Z = 0.6/ √0.64 + 0.81
Z = 0.6/√1.45
Z = 0.6/1.2041
Z = 0.49
Since our difference in sample mean is greater than zero, it means that the probability of the z score (z=0.49) must be that of the one greater than 0.49 which is area towards the right of the distribution.
Do note that the probability value under z score to the left + probability value under z score to the right = 1
The table I'm using here is only giving probability of z scores toward the left (z<0.49)
Hence z>0.49 = 1 - z<0.49
From the table, z<0.49 = 0.68793
z>0.49 = 1 - 0.68793
z>0.49 = 0.31207
