Question

In: Computer Science

A random variable X is said to follow the Weibull distribution with shape parameter

A random variable \(X\) is said to follow the Weibull distribution with shape parameter \(\alpha\) and scale parameter \(\beta\), written \(W(\alpha, \beta)\) if its p.d.f. is given by

$$ f(x)=\frac{\alpha}{\beta^{\alpha}} x^{\alpha-1} e^{-\left(\frac{g}{3}\right)^{\alpha}} $$

for \(x>0\). The Weibull distribution is used to model lifetime of item subject to failure. If \(\alpha \in(0,1),\) it is used to model decreasing failure rate overtime, whereas if \(\alpha>1,\) one models increasing failure rate over time. It is easy to show that the c.d.f. of \(X\) is given by

$$ F(x)=1-e^{-\left(\frac{x}{b}\right)^{\alpha}} $$

(a) Show that \(Y=X^{\alpha} \sim \operatorname{EXP}\left(\beta^{\alpha}\right),\) that is \(E(Y)=\beta^{\alpha}\).

(b) Let \(X_{1}, \ldots, X_{n}\) be a random sample of size \(n\) from \(X\). What is the distribution of

$$ Q=\frac{2 n \sum_{i=1}^{n} Y_{i}}{\beta^{\alpha}} $$

Conclude that \(Q\) is a pivot.

(c) It is of interest to compare the failure rate of two different types of electronics whose lifetimes \(X\) and \(Y\) are assumed to follow \(W\left(\alpha_{1}, 2\right)\) and \(W\left(\alpha_{2}, 2\right)\) respectively. To compare the two failure times, one can compare the failure rates \(2^{\alpha_{1}}\) and \(2^{\alpha_{2}}\) via their ratio or difference. To that end, consider a random sample of size \(n_{1}\left(X_{1}, \ldots X_{n_{2}}\right)\) and \(n_{2}\left(Y_{1}, \ldots, Y_{m_{2}}\right)\) from Population 1 and 2 of electronics respectively. Show that

$$ \frac{2^{\infty_{2}} 2 n_{1} \sum_{i=1}^{n_{1}} X_{i}^{\alpha_{1}}}{2^{\alpha_{1}} 2 n_{2} \sum_{i=1}^{m_{2}} Y_{i}^{\alpha_{2}}} $$

is a pivotal quantity and give its distribution.

(d) Based on your answer in (c), derive a \((1-\alpha) \%\) confidence interval for \(\frac{2^{\infty_{2}}}{2^{\prime \prime}}\).

(e) From your answer in (d), deduce a confidence interval for \(\left(\alpha_{2}-\alpha_{1}\right)\) by passing to logarithm and interpret it.

(f) Application: A random sample of size \(n_{1}=20\) and \(n_{2}=17\) electronics from Type 1 and Type 2 electronics are taken respectively. Derive and interpret the \(95 \%\) confidence interval for \(\left(\alpha_{2}-\alpha_{1}\right)\) and interpret it. What can you really say about that interval in terms of the rate of failure of one type versus the other?

Solutions

Expert Solution

(a) The cdf of \(Y\) is

$$ F_{Y}(y)=P(Y \leq y)=P\left(X^{\alpha} \leq y\right)=P\left(X \leq y^{\frac{1}{\alpha}}\right)=1-\exp \left(-\frac{y}{\beta^{\alpha}}\right) ; y>0 $$

The pdf of \(Y\) is \(f_{Y}(y)=\frac{\mathrm{d} F_{Y}(y)}{\mathrm{d} y}=\frac{1}{\beta^{\alpha}} \exp \left(-\frac{y}{\beta^{\alpha}}\right) ; y>0\)

\(=0\) otherwise

Hence \(Y \sim\) Exponential distribution with \(E(Y)=\beta^{a}\)

(b) Now transform \(Y \rightarrow Z\) where \(z=\frac{2 y}{\beta^{\alpha}} ; \frac{\mathrm{d} z}{\mathrm{~d} y}=\frac{2}{\beta^{\alpha}} ; \mid\) Jacobian \(\mid=\frac{\beta^{\alpha}}{2}\)

Then the pdf of \(Z\) is \(f_{Z}(z)=\frac{1}{2} e^{-\frac{z}{2}} ; z>0\)

\(=0\) otherwise Hence \(Z=\frac{2 Y}{\beta^{\alpha}} \sim \chi_{2}^{2}\)

Since \(Y_{i}^{\prime}\) s are independent so from additive property of \(\chi^{2}\), \(Q=\frac{2 \sum_{i=1}^{n} Y_{i}}{\beta^{\alpha}}=\frac{2 n \bar{Y}_{n}}{\beta^{\alpha}} \sim \chi_{2 n}^{2}\) which is independent of \(\alpha\) and \(\beta\)

Hence \(Q\) is pivot. (Note: I think the expression of \(\left.Q=\frac{2 \sum_{i=1}^{n} Y_{i}}{\beta^{\alpha}}=\frac{2 n \bar{Y}_{n}}{\beta^{\alpha}}\right)\)

(c) \(\frac{2 \sum_{i=1}^{n_{1}} X_{i}}{2^{\alpha_{1}}}=\frac{2 n_{1} \bar{X}_{n_{1}}}{2^{\alpha_{1}}} \sim \chi_{2 n_{1}}^{2} ; \frac{2 \sum_{i=1}^{n_{2}} Y_{i}}{2^{\alpha_{2}}}=\frac{2 n_{2} \bar{Y}_{n_{2}}}{2^{\alpha_{2}}} \sim \chi_{2 n_{2}}^{2}\)

and \(X\) and \(Y^{\prime}\) s are independent \(\mathrm{so}\) \(\frac{\frac{2 n_{1} \bar{X}_{n_{1}}}{2^{\alpha_{1}}} /\left(2 n_{1}\right)}{\frac{2 n_{1} \bar{Y}_{n_{2}}}{2^{\alpha_{2}}} /\left(2 n_{2}\right)}=\frac{2^{\alpha_{2}} 2 n_{1} \sum_{i=1}^{n_{1}} X_{i}^{\alpha_{1}}}{2^{\alpha_{2}} 2 n_{2} \sum_{i=1}^{n_{2}} Y_{i}^{\alpha_{2}}} \sim F_{2 n_{1}, 2 n_{2}}\) and it is al so free from \(\left(\alpha_{1}, \alpha_{2}\right),\) so it is pivot.

$$ \begin{aligned} &\text { (d) } P\left(F_{1-\alpha / 2,2 n_{1}, 2 n_{2}} \leq \frac{2^{\alpha_{2}} 2 n_{1} \sum_{i=1}^{n_{1}} X_{i}^{\alpha_{1}}}{2^{\alpha_{2}} 2 n_{2} \sum_{i=1}^{n_{2}} Y_{i}^{\alpha_{2}}} \leq F_{\alpha / 2,2 n_{1}, 2 n_{2}}\right)=1-\alpha\\ &\text { So } 100(1-\alpha) \% \text { C.I. for } \frac{2^{\alpha_{2}}}{2 \alpha_{1}} \text { is }\\ &\left(F_{1-\alpha / 2,2 n_{1}, 2 n_{2}} \times \frac{\bar{Y}_{n}}{\bar{X}_{n}}, F_{\alpha / 2,2 n_{1}, 2 n_{2}} \times \frac{\bar{Y}_{n}}{\bar{X}_{n}}\right) \end{aligned} $$

(e) \(100(1-\alpha) \%\) C.I. for \(\alpha_{2}-\alpha\) is \(\left(\frac{1}{\log 2} \times \log \left(F_{1-\alpha / 2,2 n_{1}, 2 n_{2}} \times \frac{\bar{Y}_{n}}{\bar{X}_{n}}\right), \frac{1}{\log 2} \times \log \left(F_{\alpha / 2,2 n_{1}, 2 n_{2}} \times \frac{\bar{Y}_{n}}{\bar{X}_{n}}\right)\right)\)


     

Related Solutions

A sampling distribution refers to the distribution of:
A sampling distribution refers to the distribution of: repeated populations repeated samples a sample statistic a population parameter
Absorption and variable costing:
The Dorset Corporation produces and sells a single product. The following data refer to the year just completed:Beginning inventory 0Units produced 9,000Units sold 7,000Selling price per unit $ 47Selling and administrative expenses:Variable per unit $ 4Fixed per year $ 58,000Manufacturing costs:Direct materials cost per unit $ 10Direct labor cost per unit $ 6Variable manufacturing overhead cost per unit $ 5Fixed manufacturing overhead per year $ 90,000Assume that direct labor is a variable cost.Required:a. Prepare an income statement for the year...
Determine whether the distribution is a discrete probability distribution. If not, state why.
Determine whether the distribution is a discrete probability distribution. If not, state why. x P(x) 0 0.1 1 0.5 2 0.05 3 0.25 4 0.1
if f and g are the functions whose graphs are shown, let u(x) = f(x)g(x) and v(x) = f(x)/g(x).
if f and g are the functions whose graphs are shown, let u(x) = f(x)g(x) and v(x) = f(x)/g(x) (a) Find u'(1) (b) Find v'(5).
Let P(x) = F(x)G(x) and Q(x) = F(x)/G(x), where F and G are the functions whose graphs are shown.
Let P(x) = F(x)G(x) and Q(x) = F(x)/G(x), where F and G are the functions whose graphs are shown.(a) Find P ' (2)(b) Find Q ' (7)
Suitable Example of Law of Variable Proportions in Economics.
What do you understand by Law of Variable Proportions in Economics? Explain with suitable example.
Find equations of the tangent lines to the curve y = (x-1)/(x+1) that are parallel to the line x − 2y = 5.
Find equations of the tangent lines to the curve y = (x-1)/(x+1) that are parallel to the line x − 2y = 5.
Y varies directly as the square of X. If Y = 144 when X = 6, what is the value of y when x = 3
Y varies directly as the square of x. If y = 144 when x = 6, what is the value of ywhen x = 3
What is the total period cost for the month under variable costing?
The following information for the next 2 questions Davison Corporation, which has only one product , has provided the following data concerning its most recent month of operations: Selling price 95 ,Units in beginning inventory 0 ,Units produced 5000, Units sold 4,900, Units in ending inventory 100. Variable costs per unit Direct materials. 26 ,Direct labor 40 , Variable manufacturing overhead 1, Variable selling and administrative expense 4, Fixed costs Fixed manufacturing overhead $40,000, Fixed selling and administrative expense $73,500 ...
What can be said about an Endothermic reaction with a negative entropy change?
What can be said about an Endothermic reaction with a negative entropy change? The reaction is a. spontaneous at all temperatures. b. spontaneous at high temperatures. c. spontaneous at low temperatures. d. spontaneous in the reverse direction at all temperatures. e. nonspontaneous in either direction at all temperatures.   What can be said about an Exothermic reaction with a negative entropy change? The reaction is a. spontaneous at all temperatures. b. spontaneous at high temperatures. c. spontaneous at low temperatures....