Probability of Difference in Means for Two Independent Paint Experiments

[tzx9633]

Homework Statement

Two independent experiments are being run in which two different types of paints are compared. Eighteen specimens are painted using type A and the drying time in hours is recorded on each. The same is done with type B. The population standard deviations are both known to be 1.0.

Assuming that the mean drying time is equal for the two types of paint, find P(XA–XB>0.3), where XA and XB are the average drying times for samples of size nA = nB = 20. XA and XB has the same mean.

Homework Equations

The Attempt at a Solution

σ between 2 means = sqrt ( 2x(1/sqrt(20))^2 ) = 0.316 P(XA–XB>0.3) = P ( z > ( 0.3 - 0) / 0.316) = 0.171 , but the an sprovided is 0.00135 , is my ans wrong ? [/B]

 

[Orodruin]

XA-XB is a sum of 36 stochastic variables. What is the distribution of XA-XB and how do you reach that conclusion?

Edit: Actually your question is unclear. It states eighteen samples for each paint but then jumps to nA = nB = 20. Which is it?

 

[tzx9633]

so

XA-XB is a sum of 36 stochastic variables. What is the distribution of XA-XB and how do you reach that conclusion?

Edit: Actually your question is unclear. It states eighteen samples for each paint but then jumps to nA = nB = 20. Which is it?

sorry , typo there . n1 = n2 = 20 , so XA-XB = (0, 0.316)

 

[Orodruin]

It is not so important what you call them as much as what they are. What I am asking about is the apparent discrepancy between

Eighteen specimens are painted

and

samples of size nA = nB = 20

 

[tzx9633]

It is not so important what you call them as much as what they are. What I am asking about is the apparent discrepancy between

and

ignore the 18 , take n1 = n2 = 20

 

[Ray Vickson]

Homework Statement

Two independent experiments are being run in which two different types of paints are compared. Eighteen specimens are painted using type A and the drying time in hours is recorded on each. The same is done with type B. The population standard deviations are both known to be 1.0.

Assuming that the mean drying time is equal for the two types of paint, find P(XA–XB>0.3), where XA and XB are the average drying times for samples of size nA = nB = 20. XA and XB has the same mean.

Homework Equations

The Attempt at a Solution

σ between 2 means = sqrt ( 2x(1/sqrt(20))^2 ) = 0.316 P(XA–XB>0.3) = P ( z > ( 0.3 - 0) / 0.316) = 0.171 , but the an sprovided is 0.00135 , is my ans wrong ? [/B]

Your answer is correct.

 

[WWGD]

I would suggest, tzx, more a matter of taste, I guess, you mention at least slightly your use of the CLT here.

 

[Orodruin]

I would suggest, tzx, more a matter of taste, I guess, you mention at least slightly your use of the CLT here.

I second this. There is no mention of the drying times having a normal distribution so it is an important point.

Edit: Along the same lines, giving the answer with three significant digits is probably quoting a bit more precision than available.

 

[WWGD]

I second this. There is no mention of the drying times having a normal distribution so it is an important point.

.

Together with the assumed independence, to avoid covariance issues, and maybe some reference to adding i.i.d RVs..