Calculating Delta Method Confidence Intervals for Quantiles at Ambient Stress

The following demonstration includes:

• Description of Delta Method Confidence Intervals
• Calculation of Delta Method Confidence Intervals for quantiles at ambient stress.

Calculation of Delta Method Confidence Intervals for quantiles at ambient stress.

The formula for q-th quantile estimation of a weibull or lognormal distribution is: 
where if the distribution is weibull,
and where if the distribution is lognormal.
is the inverse function of the standard normal cdf.

Examples of values for a_q:
 

q	weibull	lognormal
0.5	-0.367	0
0.1	-2.250	-1.282
0.01	-4.600	-2.362

In this tutorial the 1-st quantile of a weibull distribution at voltage = 35, and temperature = 85 will be calculated, so -4.600 will be used.

Remember our regression model is a no-interaction model:


For a description of the no-interaction model, see tutorial "Weibull Regression with voltage and temperature stresses."

Given this model, the formula for the 1st quantile, or b1,  is:

Notice that mu was replaced with its' model formula.
Note:  the term b1, is being used for both the estimate of the model parameter beta1, and the 1st quantile.  The context will clarify which meaning is to be used.

In tutorial "Using JMP's nonlinear fit platform to do weibull regression with voltage and temperature stresses" the following model parameter estimates were found:

The estimate of b1 (1st quantile) at voltage = 35, and temperature = 85 is: exp(12.84445762) = 378684.10 hours.

To calculate a confidence interval for b1 (1st quantile) one needs to calculate a delta method 95% confidence interval for ln(b1), (log of the 1st quantile), then exponentiate.  A 95% confidence interval for b1 could be calculated directly, but such a direct calculation can produce a negative value for the lower limit.  Calculating a confidence interval for ln(b1), then exponentiating will always produce a  confidence interval for b1 that will have a positive lower limit.

The formula for ln(b1) is: 
A formula for the standard error of ln(b1) is needed:
Below is the formula for the variance of ln(b1):

To calculate the standard error, the variance formula the variance is calculated using the above formula, then the standard error is calculated by taking the square root.

To calculate the variance of ln(b1), the variance-covariance matrix of the estimates is needed.  The JMP output produces the correlation matrix, and the standard errors.  Using this information, the variance-covariance matrix can be calculated.

We need the following formulas to convert the correlation matrix of the estimates to a variance-covariance matrix:
var(bi) = standard error of bi squared.
cov(bi,bj) = (stderror of bi)*(stderror of bj)*( correlation between bi and bj).

so the Variance-Covariance matrix is:
 

	b0	b1	b2	sigma
b0	var(b0) = 184.756	cov(b0,b1) = -56.398	cov(b0,b2) = 1.322	cov(b0,sigma) = 3.568
b1	cov(b0,b1) = -56.398	var(b1) = 19.774	cov(b1,b2) = -0.657	cov(b1,sigma) = -0.929
b2	cov(b0,b2) = 1.322	cov(b1,b2) = -0.657	var(b2) = 0.035	cov(b2,sigma) = 0.014
sigma	cov(b0,sigma) = 3.568	cov(b1,sigma) = -0.929	cov(b2,sigma) = 0.014	var(sigma) = 0.129

Using this Variance-Covariance matrix, and the stress levels of 35 volts, and 85 degrees C:

 var(ln(b1)) = 0.843, and the standard error of ln(b1) = 0.711 hours.
The estimate of ln(b1)  is 12.84445762, so the estimate of b1 is exp(12.84445762) = 378684.10 hours ( 43.2 yrs).
 

Now a delta method 95% CI for ln(b1) at 35 volts and 85 degrees C can be calculated:
Approx. 95% CI for ln(b1)=(12.84445762 +/- 1.96*0.711) = (11.451,14.238)

By exponentiating the CI for ln(b1), an approx. 95% CI for b1 at 35 volts and 85 degrees C is calculated:
Approx. 95% CI for b1 = [exp(11.451),exp(14.238)] = (93995, 1525755) hrs or (10.7 yrs, 174.2 yrs).
The estimate of b1 is exp(12.84445762) = 378684.10 hours ( 43.2 yrs).

With about 95% confidence, we can say b1 at 35 volts and 85 degrees C is between 93995 hrs (10.7 yrs) and 1525755 hrs (174.2 yrs).

Still assuming weibull errors, what if a 95% confidence interval for b10 at 35 volts and 85 degrees C was needed?
If one wanted to calculate such a delta method confidence interval for b10 replace -4.600 with -2.25 (see table of a_q values above) in the variance formula for ln(b10) and recalculate the estimated variance: var(ln(b10)) = 2.145
Calculate the standard of ln(b10): stderror(ln(b10)) = 1.465
Calculate the estimate of ln(b10): ln(b10) = 18.32615555
One would also want an estimate of b10 even if it is not used to produce the confidence interval:
b10 = 90980481.28 hrs.  (10385.9 yrs)

Therefore, an approx. 95% CI for ln(b10) is (18.326 +/- 1.96*1.465) = (15.455, 21.197), so an approx. 95% CI for b10 is (exp(15.455),exp(21.197)) = (5152538 hrs, 1606622522 hrs.) or (58.8 yrs, 183404 yrs).

The next tutorial changes focus, and addresses how one can use JMP to perform Reliability analysis using multiple censoring types.  For this tutorial, go to "Analysis of failure data with multiple types of censoring" tutorial.