In: Statistics and Probability
PLEASE USE SPSS!!!
A metallurgy company wanted to investigate the effect of the percentage of ammonium and the stir rate on the density of the power produced. The results are as follows:
STIR RATE
AMMONIUM(%) 100 150
2 10.95 7.54
2 14.68 6.66
2 17.68 8.03
2 15.18 8.84
30 12.65 12.46
30 15.12 14.96
30 17.48 14.96
30 15.96 12.62
At the 0.05 level of significance
A, is there an interaction between the percentage of ammonium and the stir rate?
B, is there an effect due to the percentage of ammonium?
C, is there an effect due to the stir rate?
D, Plot the mean density for each percentage of ammonium for each stir rate
SOLUTION A: Null and Alternative Hypotheses
The following null and alternative hypotheses need to be tested:
Ho: There is no interaction between the percentage of ammonium and the stir rate.
Ha: There is an interaction between the percentage of ammonium and the stir rate.
F statistic= 7.710
P value=0.017
Using the P-value approach: The p-value is p = 0.017, and since p=0.017<0.05, it is concluded that the null hypothesis is rejected.
Conclusion: It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that ,there is an interaction between the percentage of ammonium and the stir rate.at the 0.05 significance level.
SOLUTION B: Null and Alternative Hypotheses
The following null and alternative hypotheses need to be tested:
Ho: μ1 = μ2
Ha: μ1 μ2
Where
μ1 = Population mean for 2% AMMONIUM
μ2= Population mean for 30% AMMONIUM
F statistic= 12.173
P value=0.004
Using the P-value approach: The p-value is p = 0.004, and since p=0.004<0.05, it is concluded that the null hypothesis is rejected.
Conclusion: It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that ,there is an effect due to the percentage of ammonium on density .at the 0.05 significance level.
SOLUTION C: Null and Alternative Hypotheses
The following null and alternative hypotheses need to be tested:
Ho: μ1 = μ2
Ha: μ1 μ2
Where
μ1 = Population mean for STIR RATE 100
μ2= Population mean for STIR RATE 150
F statistic= 19.384
P value=0.001
Using the P-value approach: The p-value is p = 0.001, and since p=0.004<0.05, it is concluded that the null hypothesis is rejected.
Conclusion: It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that ,there is an effect due to the STIR RATE on density .at the 0.05 significance level.
SOLUTION D: