Chapter 4.6, Problem 97E

The following failure time observations (1000s of hours) resulted from accelerated life testing of 16 integrated circuit chips of a certain type:

82.8	11.6	359.5	502.5	307.8	179.7
242.0	26.5	244.8	304.3	379.1	212.6
229.9	558.9	366.7	204.6

Use the corresponding percentiles of the exponential distribution with λ = 1 to construct a probability plot. Then explain why the plot assesses the plausibility of the sample having been generated from any exponential distribution.

To determine

Construct exponential probability plot for the data with $\lambda =1$.

Answer to Problem 97E

The exponential probability plot for the data with $\lambda =1$ is given below:

Explanation of Solution

Given info:

The observations are based on the failure time of the circuit chips.

Calculation:

The cumulative distribution function (cdf) for the exponential distribution with parameter $\lambda =1$ is given as:

$F\left(x\right)=\{\begin{array}{l}0,x\le 0\\ 1-e^{-x},x>0.\end{array}$

Denote the $\left(100p\right)\text{th}$ percentile of the exponential distribution as $\eta \left(p\right)$. It is known that if the $\left(100p\right)\text{th}$ percentile of the distribution of a random variable is $\eta \left(p\right)$, then, $P\left(X\le \eta \left(p\right)\right)=p$, that is, $F\left(\eta \left(p\right)\right)=p$.

For the exponential distribution with parameter $\lambda =1$, the $\left(100p\right)\text{th}$ percentile will be:

$\begin{array}{c}1-e^{-\eta \left(p\right)}=p\\ e^{-\eta \left(p\right)}=1-p\\ \ln \left(e^{-\eta \left(p\right)}\right)=\ln \left(1-p\right),\left(\text{taking logarithm on both sides}\right)\\ -\eta \left(p\right)=\ln \left(1-p\right)\end{array}$

$\begin{array}{c}\eta \left(p\right)=-\ln \left(1-p\right)\\ =\ln \left(\frac{1}{1-p}\right).\end{array}$

In order to construct a probability plot using the corresponding percentiles of the exponential distribution with $\lambda =1$, use the following steps:

• Arrange the sample observations in ascending order and assign serial numbers, i to the observations from 1 to $n=16$.
• Find the cumulative proportion up to each sample observation using the formula $p=\frac{\left(i-0.5\right)}{n}$, for each $i=1,2,\mathrm{...},n\left(=16\right)$.

[Multiplying the above quantity by 100 would yield the corresponding percentile. However, as the following calculations make use of the cumulative proportion p rather than the percentile $\left(100p\right)$, it is not required to multiply 100 in these calculations.]

• For each of the 16 values of p obtained, calculate $\eta \left(p\right)=\ln \left(\frac{1}{1-p}\right)$.

The observations in ascending order calculations are shown below:

Serial (i)	Failure Time	$p=\frac{\left(i-0.5\right)}{n}$	$1-p$	$\frac{1}{1-p}$	Percentile $\ln \left(\frac{1}{1-p}\right)$
1	11.6	0.03125	0.96875	1.0323	0.0317
2	26.5	0.09375	0.90625	1.1034	0.0984
3	82.8	0.15625	0.84375	1.1852	0.1699
4	179.7	0.21875	0.78125	1.2800	0.2469
5	204.6	0.28125	0.71875	1.3913	0.3302
6	212.6	0.34375	0.65625	1.5238	0.4212
7	229.9	0.40625	0.59375	1.6842	0.5213
8	242	0.46875	0.53125	1.8824	0.6325
9	244.8	0.53125	0.46875	2.1333	0.7577
10	304.3	0.59375	0.40625	2.4615	0.9008
11	307.8	0.65625	0.34375	2.9091	1.0678
12	359.5	0.71875	0.28125	3.5556	1.2685
13	366.7	0.78125	0.21875	4.5714	1.5198
14	379.1	0.84375	0.15625	6.4000	1.8563
15	502.5	0.90625	0.09375	10.6667	2.3671
16	558.9	0.96875	0.03125	32.0000	3.4657

The last column gives the values of $\eta \left(p\right)=\ln \left(\frac{1}{1-p}\right)$, when the underlying distribution is assumed to be exponential with parameter $\lambda =1$, for each value of p obtained from the corresponding data points.

Probability plot:

Software procedure:

Step by step procedure to obtain the probability plot using the MINITAB software is given below:

• Choose Graph > Scatterplot.
• Choose Single, and then click OK.
• In Y variables, enter the column of Failure Time.
• In X variables, enter the column of Percentile.
• Click OK.

Justification:

If the probability plot of the observed values and the corresponding percentiles of the assumed underlying probability distribution form a straight line or at least close to a straight line, then the assumed distribution can be considered as a good approximation of the observed data.

The probability plot for the exponential distribution with parameter $\lambda =1$ shows curvature. The plots are not on a straight line, nor are they very close to a straight line.

Hence, the exponential distribution with parameter $\lambda =1$ is not a very good approximation of the data.

The parameter $\lambda$ is the scale parameter of the exponential distribution. The general form of the exponential distribution, with both location parameter $\alpha$ and scale parameter $\lambda$ is given as:

$f\left(x\right)=\{\begin{array}{l}{e}^{-\frac{x-\alpha }{\lambda }},x>\alpha \\ \text{0, }x\le \alpha .\end{array}$

The scale parameter $\lambda$ is the one which determines the shape of the exponential curve. Thus, a change in the value of $\lambda$ would affect the decision regarding whether or not the data has an exponential distribution with the particular scale parameter $\lambda$.

However, for a particular value of $\lambda$, such as, $\lambda =1$ here, a change in the value of the location parameter $\alpha$ would result in the same shape of the exponential distribution. For a particular scale parameter ($\lambda$) value, there may be infinite possible values of the location parameter ($\alpha$).

Hence, the plot assesses the plausibility regarding whether the sample has been generated from any exponential distribution whatsoever, when $\lambda =1$.