$t$-distribution

• Central Limit Theorem (Theorem 8.2) assumes $\sigma$ is known in
  $$Z=\frac{\bar{X}-\mu}{\sigma/\sqrt{n}}$$

• However, $\sigma$ might not be known. Then, consider the random variable
  $$T=\frac{\bar{X}-\mu}{S/\sqrt{n}}$$

• The value of sample variance $S^2$ fluctuate considerably from sample to sample, $T$ does not follow the standard normal distribution but follows $t$-distribution with the degrees of freedom $n-1$.
• In developing the sampling distribution of $T$, we shall assume that our random sample was selected from a normal population.
  $$T=\frac{(\bar{X}-\mu)/(\sigma/\sqrt{n})}{\sqrt{S^2/\sigma^2}}=\frac{Z}{\sqrt{V/(n-1)}}$$

• where $Z=\frac{\bar{X}-\mu}{\sigma/\sqrt{n}}$ and $V=\frac{(n-1)S^2}{\sigma^2}$

• Theorem 8.5:
  

• Corollary 8.1:
  
• Student $t$-distribution
  • The probability distribution of $T$ was first published in 1908 in a paper by W. S. Gosset.
  • Employed by an Irish brewery, but disallowed publication.
  • Published his work secretly under the name ``Student''.

• The shape of $T$ looks like the standard normal ($Z$) (depending on the degrees of freedom, $n-1$). Symmetric about $\mu=0$, bell-shaped.
• Difference between $T$ and $Z$: variance of $T \geq 1$ and depends on $n$
• $T$ and $Z$ are the same as $n \rightarrow \infty$
• $t_{\alpha}$ represents the $t$-value above which we find an area of $\alpha$ to the right.
• $t$-distribution is symmetric about 0: $t_{1-\alpha}=-t_{\alpha}$

  Figure 7.16: The $t$-distribution curves for $\nu = 2,5$, and $\infty$

  

  Figure 7.17: Symmetry property of the $t$-distribution.

  

• Example 8.11: The $t$-value with $\nu = 14$ degrees of freedom that leaves an area of 0.025 to the left, and therefore an area of 0.975 to the right, is
  $$t_{0.975}=-t_{0.025}=-2.145$$
  Look up $t_{0.025}$, and then place a negative sign.
• Example 8.12: Find $P(-t_{0.025}<T < t_{0.05})=?$
  $$P(-t_{0.025}<T < t_{0.05})=1-0.05-0.025=0.925$$
  Since $t_{0.05}$ leaves an area of 0.05 to the right, and $-t_{0.025}$ leaves an area of 0.025 to the left, we find a total area of 0.925.
• Find $k$ such that $P(k < T < -1.761) = 0.045$, for a random sample of size 15 selected from a normal distribution and
  $$T=\frac{\bar{X}-\mu}{S/\sqrt{n}}$$

• Solution:
  $$\nu=15-1=14$$
  From Table A.4, $-t_{0.05}=-1.761$ Let $k=-t_{\alpha}$, $0.045=0.05-\alpha \Rightarrow \alpha=0.005$ see Fig. 7.18
  $$k=-t_{0.05}=-2.977 ~(Table~ A.4)$$

  Figure 7.18: The $t$-values for Example 8.13.

  

• Example 8.14: A engineer claims that the population mean of a process is 500 grams. To check this claim he samples 25 batches each month.
• If the computed $t$-value falls between $-t_{0.05}$ and $t_{0.05}$, he is satisfied with his claim.
• What conclusion should he draw from a sample that has a mean $\bar{x}=518$ grams and a sample standard deviation $s = 40$ grams? Assume the distribution of yields to be approximately normal.

• Solution: From Table A.4,
  $$t_{0.05}=1.711 (\nu=24)$$
  Assumption $\mu=500 \Rightarrow$
  $$t=\frac{518-500}{40/\sqrt{25}}=2.25$$
  $$2.25>1.711 \rightarrow error$$
  if $\mu>500$, $t$-value would be more reasonable. The process produces a better product than he thought.

• Exactly 95% of the values of a $t$-distribution with $\nu = n -1$ degrees of freedom lie between $-t_{0.025}$ and $t_{0.025}$.
• A $t$-value that falls below $-t_{0.025}$ or above $t_{0.025}$ would tend to make us believe that either a very rare event has taken place or perhaps our assumption about $\mu$ is in error.
• What is the $t$-distribution used for?. The $t$-distribution is used extensively in problems that deal with
  • Inference about the population mean.
  • Comparative samples (two sample means).
• Use of the $t$-distribution for the statistic
  $$T=\frac{\bar{X}-\mu}{S/\sqrt{n}}$$
  requires that $X_1, X_2,\ldots ,X_n$ be normal.