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Chapter 8: Q12 E (page 473)

Consider again the situation described in Example 8.2.3. How small would蟽2 need to be in order for Pr(Y鈮�0.09)鈮�0.9?

Short Answer

Expert verified

\({\sigma ^2} \le 0.0563\)

Step by step solution

01

Given information

The concentration of lactic acid in several chunks of cheese is independent of normal random variables with mean 渭and variance\({\sigma ^2}\).

02

Calculate the probability 

Generally, for \({\sigma ^2}\),

\(P\left( {Y \le 0.09} \right) = P\left( {W \le \frac{{10 \times 0.09}}{{{\sigma ^2}}}} \right)\)

Where \(W = \frac{{10Y}}{{{\sigma ^2}}}\)has \({\chi ^2}\)distribution with d.f 10.

\(P\left( {Y \le 0.09} \right)\) is at least 0.9 if \(\frac{{0.9}}{{{\sigma ^2}}}\) is at least the 0.9 quantile of the \({\chi ^2}\)distribution with d.f 10.

This quantile is 15.99, so\(\frac{{0.9}}{{{\sigma ^2}}} \ge 15.99\)is equivalent to\({\sigma ^2} \le 0.0563\).

Hence, \({\sigma ^2} \le 0.0563\).

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Most popular questions from this chapter

Consider the analysis performed in Example 8.6.2. This time, use the usual improper before computing the parameters' posterior distribution.

Suppose that the random variables \({X_1},{X_2}\,\,\,\,{\rm{and}}\,\,\,{X_3}\) are i.i.d., and that each has the standard normal distribution. Also, suppose that

\(\begin{align}{Y_1} &= 0.8{X_1} + 0.6{X_2},\\{Y_2} &= \sqrt 2 \left( {0.3{X_1} - 0.4{X_2} - 0.5{X_3}} \right),\\{Y_3} &= \sqrt 2 \left( {0.3{X_1} - 0.4{X_2} + 0.5{X_3}} \right)\end{align}\)

Find the joint distribution of \({Y_1},{Y_2},{Y_3}\).

Suppose that\({{\bf{X}}_{\bf{1}}}{\bf{, \ldots ,}}{{\bf{X}}_{\bf{n}}}\)form a random sample from the normal distribution with mean渭and variance\({{\bf{\sigma }}^{\bf{2}}}\), and let\({{\bf{\hat \sigma }}^{\bf{2}}}\)denote the sample variance. Determine the smallest values ofnfor which the following relations are satisfied:

  1. \({\bf{Pr}}\left( {\frac{{{{{\bf{\hat \sigma }}}^{\bf{2}}}}}{{{{\bf{\sigma }}^{\bf{2}}}}} \le {\bf{1}}{\bf{.5}}} \right) \ge {\bf{0}}{\bf{.95}}\)
  2. \({\bf{Pr}}\left( {\left| {{{{\bf{\hat \sigma }}}^{\bf{2}}}{\bf{ - }}{{\bf{\sigma }}^{\bf{2}}}} \right| \le \frac{{\bf{1}}}{{\bf{2}}}{{\bf{\sigma }}^{\bf{2}}}} \right) \ge {\bf{0}}{\bf{.8}}\)

Question: Prove the limit formula Eq. (8.4.6).

Suppose that\({{\bf{X}}_{\bf{1}}}{\bf{,}}...{\bf{,}}{{\bf{X}}_{\bf{n}}}\)form a random sample from the normal distribution with unknown mean 渭 and unknown standard deviation 蟽, and let\({\bf{\hat \mu }}\,\,{\bf{and}}\,\,{\bf{\hat \sigma }}\)denote the M.L.E.鈥檚 of 渭 and 蟽. For the sample size n = 17, find a value of k such that

\({\bf{Pr}}\left( {{\bf{\hat \mu > \mu + k\hat \sigma }}} \right){\bf{ = 0}}{\bf{.95}}\)
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