{Exercise 13.09 (Algorithmic)} To study the effect of temperature on yield in a
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Question
{Exercise 13.09 (Algorithmic)}
To study the effect of temperature on yield in a chemical process, five batches were produced at each of three temperature levels. The results follow.
Temperature
Use a .05 level of significance to test whether the temperature level has an effect on the mean yield of the process.
Calculate the value of the test statistic (to 2 decimals).
The p-value is Selectless than .01between .01 and .025between .025 and .05between .05 and .10greater than .10Item 11
What is your conclusion?
SelectConclude that the mean yields for the three temperatures are not all equalDo not reject the assumption that the mean yields for the three temperatures are equalItem 12
{Exercise 13.09 (Algorithmic)}
To study the effect of temperature on yield in a chemical process, five batches were produced at each of three temperature levels. The results follow.
Temperature
50°C 60°C 70°C 36 31 26 26 32 31 38 35 31 41 24 33 34 28 34 Construct an analysis of variance table (to 2 decimals, if necessary).Source of Variation Sum of Squares Degrees of Freedom Mean Square F Treatments Error Total
Use a .05 level of significance to test whether the temperature level has an effect on the mean yield of the process.
Calculate the value of the test statistic (to 2 decimals).
The p-value is Selectless than .01between .01 and .025between .025 and .05between .05 and .10greater than .10Item 11
What is your conclusion?
SelectConclude that the mean yields for the three temperatures are not all equalDo not reject the assumption that the mean yields for the three temperatures are equalItem 12
Explanation / Answer
Step 1
Null Hypothesis Ho : µ1 =µ2 =µ3
Alternative Hypothesis : µ1 µ2 µ3
Step 2
Degrees of freedom between = k - 1 = 3 - 1 = 2
Degrees of freedom Within = n - k = 15 - 3 = 12
Degrees of freedom Total F( k-1,n - k,) at 0.05 is = F Crit = 3.885
Step 3
Grand Mean = G / N = 35+30+31 / 3 = 32
SST = ( Xi - GrandMean)^2 = (36-32)^2 + (26-32)^2 + (38-32)^2 + ……..& so on = 306
SS Within = (Xi - Mean of Xi ) ^2 =,(36-35)^2 + (26-35)^2 + (38-35)^2 + ……..& so on = 236
SS Between = SST - SS Within = 306 - 236 = 70
Step 4
Mean Square Between = SS Between / df Between = 70/2 = 35
Mean Square Within = SS Within / df Within = 236/12 = 19.667
Step 5
F Cal = MS Between / Ms Within = 35/19.667 = 1.78
We got |F cal| = 1.78 & |F Crit| =3.885
MAKE DECISION
Hence Value of |F cal| < |F Crit|and Here We Accept Ho
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