Answer:
a. The 95% C.I. is approximately -5.83 < μ₁ - μ₂ < -0.57
b. Yes
c. The power of the test is 0.03836
d. The sample size can be considered adequate
Step-by-step explanation:
a. The given data in the study
Catalyst 1: 57.9, 66.2, 65.4, 65.4, 65.2, 62.6, 67.6, 63.7, 67.2, 71.0
Catalyst 2: 66.4 , 71.7, 70.3, 69.3, 64.8, 69.6, 68.6, 69.4, 65.3, 68.8
Using the Average, Standard Deviation function from Microsoft Excel, we have;
The mean for Catalyst 1, = 65.22
The standard deviation for catalyst 1, σ₁ = 3.444416
The mean for Catalyst 2, = 68.42
The standard deviation for catalyst 2, σ₂ = 2.22401
The 95% confidence interval on the difference in mean is given as follows;
The critical-z for a 95% confidence interval = 1.96
Therefore, we have;
Therefore, we have;
The 95% C.I. is approximately -5.83 < μ₁ - μ₂ < -0.57
The test statistics is given as follows;
Therefore we have;
The p-value = P(Z<-2.39 or Z > 2.39) = 2P(Z<-2.39) = 2 × 0.00842 = 0.01684
b. From the confidence interval which range from approximately -5.83 to -0.57 and does not include 0, therefore, there is a difference in mean active concentration which depends on the choice of catalyst
c. The power of the test
The sample mean difference is given as follows;
Therefore, we have;
The z-value is given as follows;
μ₁ - μ₂ = 5
When
When
The power of the test is given by P = P(Z<-5.69) + P(Z>-1.77) = 0 + 0.03836 = 0.03836
The power of the test = 0.03836
d. The sample size is statistically adequate because the confidence interval of -5.83 < μ₁ - μ₂ < -0.57 has a value of -5 as a possible population difference in mean
The assumption of normality seems adequate because the confidence interval obtained by using the sample standard deviation is given as follows;
(-5.74, -0.66) which also contains -5 which is the difference in the population mean