Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Suppose that we will sample 20 chunks of cheese in Example 8.2.3. Let\({\bf{T = }}\sum\limits_{{\bf{i = 1}}}^{{\bf{20}}} {{{\left( {{{\bf{X}}_{\bf{i}}}{\bf{ - \mu }}} \right)}^{\bf{2}}}{\bf{/20}}} \)wherexiis the concentration of lactic acid in theith chunk. Assume thatσ2=0.09. What numbercsatisfies Pr(Tc)=0.9?

Short Answer

Expert verified

The value of c is 0.127854

Step by step solution

01

Given information

Xiis the concentration of lactic acid in theith chunk.

02

Calculating the value of c

The distribution of 20T/0.09 is \({\chi ^2}\) the distribution with 20 degrees of freedom

\(\begin{align}P\left( {T \le c} \right) &= 0.9\\P\left( {20T/0.09 \le 20c/0.09} \right) &= 0.9\\P\left( {{\chi ^2}\left( {20} \right) \le 20c/0.09} \right) &= 0.9\\20c/0.09 &= 28.41198\\c &= 0.127854\end{align}\)

So, the value of c is 0.127854

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Prove that the distribution of\({\hat \sigma _0}^2\)in Examples 8.2.1and 8.2.2 is the gamma distribution with parameters\(\frac{n}{2}\)and\(\frac{n}{{2{\sigma ^2}}}\).

Suppose that\({{\bf{X}}_{\bf{1}}}{\bf{,}}...{\bf{,}}{{\bf{X}}_{\bf{n}}}\)form a random sample from the Poisson distribution with unknown mean θ, and let

\({\bf{Y = }}\sum\nolimits_{{\bf{i = 1}}}^{\bf{n}} {{{\bf{X}}_{\bf{i}}}} \).

a. Determine the value of a constant c such that the estimator\({{\bf{e}}^{{\bf{ - cY}}}}\)is an unbiased estimator of\({{\bf{e}}^{{\bf{ - \theta }}}}\).

b. Use the information inequality to obtain a lower bound for the variance of the unbiased estimator found in part (a).

Suppose that \({X_1},...,{X_n}\) form a random sample from the normal distribution with mean μ and variance \({\sigma ^2}\) . Assuming that the sample size n is 16, determine the values of the following probabilities:

\(\begin{align}a.\,\,\,\,P\left( {\frac{1}{2}{\sigma ^2} \le \frac{1}{n}\sum\limits_{i = 1}^n {{{\left( {{X_i} - \mu } \right)}^2} \le 2{\sigma ^2}} } \right)\\b.\,\,\,\,P\left( {\frac{1}{2}{\sigma ^2} \le \frac{1}{n}\sum\limits_{i = 1}^n {{{\left( {{X_i} - {{\bar X}_n}} \right)}^2} \le 2{\sigma ^2}} } \right)\end{align}\)

Suppose that X1,…….,Xn form a random sample from the Bernoulli distribution with unknown parameter p. Show that the variance of every unbiased estimator of (1-p)2 must be at least 4p(1-p)3/n.

Question:Suppose that a random variable X has the geometric distribution with an unknown parameter p (0<p<1). Show that the only unbiased estimator of p is the estimator \({\bf{\delta }}\left( {\bf{X}} \right)\) such that \({\bf{\delta }}\left( {\bf{0}} \right){\bf{ = 1}}\) and \({\bf{\delta }}\left( {\bf{X}} \right){\bf{ = 0}}\) forX>0.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free