MANE 3332.05

Lecture 21

Agenda

Have worked through partial credit requests
Continue Chapter 8 lecture
Chapter 8, Case 1 Practice Problems (assigned 11/6/2025, due 11/11/2025)
New: Chapter 8, Case 1 Quiz (assigned 11/11/2025, due 11/13/2025)
New: Chapter 8, Case 2 Practice Problems (assigned 11/11/2025, due 11/13/2025)
Attendance
Questions?

Handouts

Week	Tuesday Lecture	Thursday Lecture
11	11/11 - Chapter 8 (part 2)	11/13 - Chapter 8 (part 3)
12	11/18 - Chapter 8 (part 4)	11/25 - Chapter 9 (part 1)
13	11/25 - Chapter 9 (part 2)	11/27 - Thanksgiving Holiday (no class)
14	12/2 - Chapter 9 (part 3)	12/4 - Linear Regression
15	12/9 - Review Session	12/11 - Study Day (no class)

The final exam for MANE 3332.01 is Thursday December 18, 2025 at 10:15 AM - 12:00 PM.

Confidence Interval for the mean of Normal distribution with variance unknown (Case 2)

The \(t\) distribution

Definition.

Let \(X_1, X_2,\ldots,X_n\) be a random sample from a normal distribution with unknown mean \(\mu\) and unknown variance \(\sigma^2\). The random variable

\[ T=\frac{\overline{X}-\mu}{S/\sqrt{n}} \]

has a \(t\) distribution with \(n-1\) degrees of freedom.

A table of percentage points (quantiles) is given in Appendix A Table 5
Figure 8-4 on page 180 shows the relationship between the \(t\) and normal distributions.
Figure 8-5 explains the percentage points of the \(t\) distribution

Confidence interval definition

Using the \(t\) distribution it is possible to construct CIs

If \(\bar{x}\) and \(s\) are the mean and standard deviation of a random sample from a normal distribution with unknown variance \(\sigma^2\), a \(100(1-\alpha)\%\) confidence interval on \(\mu\) is given by

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

where \(t_{\alpha/2,n-1}\) is the upper \(100(\alpha/2)\) percentage point of the \(t\) distribution with \(n-1\) degrees of freedom.

Problem 8-30 (6th edition)

One-sided confidence bounds

Are easy to construct
Use only the appropriate upper or lower bound
Change \(t_{\alpha/2,n-1}\) to \(t_{\alpha,n-1}\)

Chapter 8, Case 2 Practice Problems

Confidence Interval for \(\sigma^2\) and \(\sigma\) (Case 3)

Section 8-3 presents a CI for \(\sigma^2\) or \(\sigma\)
Requires the \(\chi^2\) (chi-squared) distribution

Let \(X_1,X_2,\ldots,X_n\) be a random sample from a normal distribution with mean \(\mu\) and variance \(\sigma^2\) and let \(S^2\) be the sample variance. Then the random variable

\[ \chi^2=\frac{(n-1)S^2}{\sigma^2} \]

has a chi-square (\(\chi^2\)) distribution with \(n-1\) degrees of freedom

A table of the upper percentage points of the \(\chi^2\) distribution are given in Table 4 in the appendix
Figure 8-9 on page 183 explains the percentage points of the \(\chi^2\) distribution

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)

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