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MartinY

Community Trekker

Joined:

Feb 5, 2017

question about Linear model

Hi, Martin here,

I am using linear model to evaluate two types of concrete substrate impact on the adhesion of adhesives.

 

so basically we have " old concrete substrate" and " new concrete substrate", upon which adhesives is applied.

we have three curing condtions: STD (standard); water immersion and heat aging.

Under each curing condition, two types of concrete substrate is evaluated by the value of adhesion.

 

so far so good, basically the attached linear model could tell whether the two types of concrete substrate are significantly different in terms of the

average adhesion value.

 

But I am also interested in the standard deviation of old concrete slabs, and the standard deviation of new concrete slabs, and be able to see which one gives smaller standard deviation.

I think the linear model I attached is deficient in terms of this concern. what is the best way to incorporate standard deviation comparison ( old concrete substrate; new concrete substrate; and comparison of the two) in my linear model?

Thanks!

 

Martin

 

 

 

 

 

7 REPLIES 7
MartinY

Community Trekker

Joined:

Feb 5, 2017

Re: question about Linear model

I should mention how we do adhesion test. for one concrete slab, we apply two adhesives, each with 10 testing point.
see the picture. 

when I say I want to get information about standard deviation, it means the variation between two piceses of concrete slabs. 

 

markbailey

Staff

Joined:

Jun 23, 2011

Re: question about Linear model

I assume that you will use Fit Least Squares to perform this two-way ANOVA. Instead, select Analyze > Fit Model and then Personality > Loglinear Variance. Now define the hypothesized effects for both the mean and the variance of the response.

Learn it once, use it forever!
MartinY

Community Trekker

Joined:

Feb 5, 2017

Re: question about Linear model

@markbailey 

thank you very much for introducing another JMP tool to me.

 

Just to confirm: if my goal is to measure 

A: adhesion variation within one concrete slab ( 10 data points)

B: adhesioin variation between two concrete slabs (same type)  ( 10 data points on each slab)

C:  compare the two types of variations for old and new concrete slabs, show wether old and new concrete differ significantly in terms of the two types of variation

 

Can Loglinear Variance method calculate A and B, C? If it can do all these, it will be the ideal tool for this experiment!

 

Thanks!

Martin

 

 

 

markbailey

Staff

Joined:

Jun 23, 2011

Re: question about Linear model

You apparently have 10 replicates for each treatment. Is a replicate a slab or measurements within the same slab? If it is the former case, then you have replication. If it is the latter, then you have something else.

 

Requirement A leads me to believe that you are measuring the same slab 10 times. That is not replication but homogeneity or uniformity. These 10 measurements as a group will represent the lowest level of variation. The loglinear model should determine if it depends on the Curing Condition and Concrete Substrate levels.

 

Requirement B will require replication. That is, more than 1 slab at each treatment defined by Curing Condition and Concrete Substrate levels.

 

Requirement C will be tested by the Concrete Substrate effect in the mean and variance models.

 

So the loglinear variance model should be able to deliver all the required information with a replicated design.

Learn it once, use it forever!
MartinY

Community Trekker

Joined:

Feb 5, 2017

Re: question about Linear model

@markbailey 

 

So I need 2 concrete slabs for each curing conditions:

 

28d SC:  Old concrete #1 (10 data points), Old concrete #2 (10 data points); new concrete #1 ( 10 data points), new concrete #2 ( 10 data points)

 

for next curing conditon, say water immersion, do we still go with 

water immersion:   Old concrete #1 (10 data points), Old concrete #2 (10 data points); new concrete #1 ( 10 data points), new concrete #2 ( 10 data points)

or 

water immersion:   Old concrete #3 (10 data points), Old concrete #4 (10 data points); new concrete #3 ( 10 data points), new concrete #4 ( 10 data points)

?

 

The thing is: can I artifically divide old concrete to two groups: old concrete #1 VS old concrete #2; or do I need to name them as Old concrete #3 ,Old concrete #4? same for the new concrete

Thanks for taking time answer my questions!

 

Martin

 

markbailey

Staff

Joined:

Jun 23, 2011

Re: question about Linear model

I am not sure what you intend by "can I artificially divide old concrete into two groups." If you mean to re-use a slab or use two halves of one slab, that is not replication. You must prepare at least two slabs to be able to estimate the within and the between slab variance. And 2 slabs are not going to give you much information about variance. The confidence interval will be very long.

 

Your design will be 2 levels for concrete type x 3 levels for adhesion method x N replicate slabs for a full factorial. The 10 measurements are not replication but provide a single response. I would make two responses, one for the mean and one for the standard deviation. I would then add 10 new continuous numeric columns to record the measures for each run. I would add a formula to the two response columns to compute the mean and standard deviation, respectively. (Note: you might want to apply the log transform to the standard deviation formula as they are not normally distributed.)

 

I would skip the loglinear variance platform and use standard least squares platform after all.

Learn it once, use it forever!
MartinY

Community Trekker

Joined:

Feb 5, 2017

Re: question about Linear model

@markbailey 

I wonder how do you incorporate N replicate in full factorical design.

2 level for concrete and 3 level for adhesion method are easy to understand, for the N replicate, let's say 3 replicates, do you simply do full factorial design ( 2 level concrete and 3 level of adhesion method) three times on different slabs?

 

I do not get the concept of N replicate in factorical design. 

 

Thanks

 

Martin