Two factor ANOVA example

TWO FACTOR ANOVA

USING BOWLING PIN FACTORY(FACTISM)

IME 435/L DESIGN OF EXPERIMENT

YOUNG KIM

2/19/2014

IME DEPARTMENT

CALIFORNIA STATE POLYTECHNIC UNIVERSITY

Title:
Two Factors ANOVA using bowling pin factory (FACTISM)

Statement of the problem:
Using statistical analysis of variance, ANOVA, using two factors out of 5, temperature and alloy, experiment will be conducted to see if any of those factor contributes to change in response.

Materials or Tools:
FACTISM
Microsoft Excel
Microsoft Words

Assumptions:
1. Other factors other than temperature and alloy will not have effect on response.
2. H_0 =  u1 =  u2  =  u3  =  u4  =  u5, temperature has no effect on weight
    H_a , temperature does have effect on weight

    H_0 =  u1 =  u2  =  u3  =  u4  =  u5, alloy has no effect on weight
    H_a , alloy does have effect on weight

    H_0 =  u1 =  u2  =  u3  =  u4  =  u5, interaction has no effect on weight
    H_a , interaction does have effect on weight
3. Alpha = 0.95

Procedure:
1. Collect responses from Factism by changing temperature and alloy value to 0, 25, 50, 75, 100, and 30, 35, 40, 45, 50 respectively.  Repeat each trial 8 times.
2. Perform two factor ANOVA by both manual and using Megastats.

Results:
Using manual method:

Performing two way ANOVA manually :

Using Megastats:

 

Findings & conclusion:

Since our calculated F value for temperature is less than our critical F value, we fail to reject our null hypothesis. Our p-value from Megastat also indicates that we fail to reject our null hypothesis since 0.2743 is greater than 0.05 which is our alpha value. However, we reject null hypothesis for alloy since its calculated F value is greater than critical F value, yet, without confident since both p values are so close to each other. Finally we can confidently say that we fail to reject our null hypothesis for interaction of temperature and alloy due to greater magnitude in calculated p value from critical p value.