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MANE 3332.04

Lecture 22, April 16

  • Topics: Chapter 8, Case 3

  • Assignments:

    • New: Chapter 8, Case 2 Quiz (assigned 4/16/2025, due 4/21/2025)
    • New: Chapter 8, Case 3 Practice Problems (assigned 4/16/2025, due 4/21/2025)
    • New: Tech Report Assignment 2 (assigned 4/16/2025, due 4/21/2025)
  • Attendance
  • Questions?

Handouts

Class Schedule

Monday Lecture Wednesday Lecture
4/14: Chapter 8: Case 2 4/16: Chapter 8: Case 3
4/21: Chapter 9, case 1 4/23: Chapter 9, Case 2
4/28: Chpater 9, Case 3 4/30: Chapter 11
5/5: Chapter 11 5/7: Review

7 Sessions plus final exam

Final Exam: Tuesday May 12, 2025 8:00 - 9:45 am

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Confidence Intervals for \(\sigma^2\) and \(\sigma\)

If \(s^2\) is the sample variance from a random sample of \(n\) observations from a normal distribution with unknown variance \(\sigma^2\), then a \(100(1-\alpha)\%\) confidence interval on \(\sigma^2\) is

\[ \frac{(n-1)s^2}{\chi_{\alpha/2,n-1}^2}\leq\sigma^2\leq\frac{(n-1)s^2}{\chi_{1-\alpha/2,n-1}^2} \]

where \(\chi^2_{\alpha/2,n-1}\) and \(\chi^2_{1-\alpha/2,n-1}\) are the upper and lower \(100\alpha/2\) percentage points of the \(\chi^2\)-distribution with \(n-1\) degrees of freedom

Problem 8-36 (6th edition)

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One-sided confidence bounds

  • Are easy to construct

  • Use only the appropriate upper or lower bound

  • Change \(\chi^2_{\alpha/2,n-1}\) to \(\chi^2_{\alpha,n-1}\) or \(\chi^2_{1-\alpha/2,n-1}\) to \(\chi^2_{1-\alpha,n-1}\)

  • See eqn (8-20) on page 184

Chapter 8, Case 3 Practice Problems

Large-Sample CI for a Population Proportion (Case 4)

  • Recall from chapter 4, that the sampling distribution of \(\widehat{P}\) is approximately normal with mean \(p\) and variance \(p(1-p)/p\), if \(n\) is not too close to either 0 or 1 and if \(n\) is relatively large.

  • Typically, we require both \(np\geq 5\) and \(n(1-p)\geq 5\)

f \(n\) is large, the distribution of

\[ Z=\frac{X-np}{\sqrt{np(1-p)}}=\frac{\widehat{P}-p}{\sqrt{\frac{p(1-p)}{n}}} \]

is approximately standard normal.

If \(\hat{p}\) is the proportion of observations in a random sample of size \(n\) that belongs to a class of interest, an approximate \(100(1-\alpha)\%\) confidence interval on the proportion \(p\) of the population that belongs to this class is

\[ \hat{p}-z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\leq p\leq \hat{p}+z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \]

where \(z_{\alpha/2}\) is the upper \(\alpha/2\) percentage point of the standard normal distribution

Other Considerations

  • We can select a sample so that we are \(100(1-\alpha)\%\) confident that error \(E=|p-\widehat{P}|\) using
\[ n=\left(\frac{z_{\alpha/2}}{E}\right)^2p(1-p) \]
  • An upper bound on is given by
\[ n=\left(\frac{z_{\alpha/2}}{E}\right)^2(0.25) \]
  • One-sided confidence bounds are given in eqn (8-26) on page 187

Guidelines for Constructing Confidence Intervals

  • Review excellent guide given in Table 8-1

Other Interval Estimates

  • When we want to predict the value of a single value in the future, a prediction interval is used

  • A tolerance interval captures \(100(1-\alpha)\%\) of observations from a distribution

Prediction Interval for a Normal Distribution

  • Excellent discussion on pages 189 - 190

  • A \(100(1-\alpha)\%\) PI on a single future observation from a normal distribution is given by

\[ \bar{x}-t_{\alpha/2,n-1}s\sqrt{1+\frac{1}{n}}\leq X_{n+1}\leq \bar{x}+t_{\alpha/2,n-1}s\sqrt{1+\frac{1}{n}} \]

Tolerance Intervals for a Normal Distribution

  • A tolerance interval to contain at least \(\gamma\%\) of the values in a normal population with confidence level \(100(1-\alpha)\%\) is
\[ \bar{x}-ks,\bar{x}+ks \]

where \(k\) is a tolerance interval factor for the normal distribution found in appendix A Table XII. Values are given for \(1-\alpha\)=0.9, 0.95 and 0.99 confidence levels and for \(\gamma=.90,\,.95,\,\mbox{and }.99\%\) probability of coverage

  • One-sided tolerance bounds can also be computed. The factors are also in Table XII