T-Statistic

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PAS

Cegep/2

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251 words

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2 minutes

Distance of a random variable from its mean in standard errors
Similar to z-score, but used when ==the sample size is small== or ==the population standard deviation is unknown==

$$\begin{array}{rl}t& =\frac{x-\mu }{s}\\ df& =n-1\end{array}$$

The higher the degrees of freedom, the closer the t-distribution to the z-distribution.

Examples

Using t-statistic, derive a $1-a$ confidence interval for $\mu$ by taking $n$ random samples.

$$\begin{array}{rl}P(-t_{\frac{\alpha }{2}}<T_{n-1}<t_{\frac{\alpha }{2}})& =1-\alpha \\ ⟺P(-t_{\frac{\alpha }{2}}<\frac{\bar{X}-\mu }{\frac{S}{\sqrt{n}}}<t_{\frac{a}{2}})& =1-\alpha \\ ⟺P(-\bar{X}-t_{\frac{\alpha }{2}}\frac{S}{\sqrt{n}}<-\mu <-\bar{X}+t_{\frac{\alpha }{2}}\frac{S}{\sqrt{n}})& =1-\alpha \\ ⟺P(\bar{X}-t_{\frac{\alpha }{2}}\frac{S}{\sqrt{n}}<\mu <\bar{X}+t_{\frac{\alpha }{2}}\frac{S}{\sqrt{n}})& =1-\alpha \end{array}$$

The confidence interval is $(\bar{X}-t_{\frac{\alpha }{2}}\frac{S}{\sqrt{n}},\bar{X}+t_{\frac{\alpha }{2}}\frac{S}{\sqrt{n}})$.

A manufacturer of printers wishes to estimate the mean number of characters printed before the printhead fails. The printer manufacturer tests n = 15 printheads and finds $\bar{x}$ = 1.24 million characters printed until failure and s = 0.19 million characters printed until failure. Form a 99% confidence interval for the mean number of characters printed before the printhead fails. Interpret the result.

$$\begin{array}{rl}n=15⟹df& =15-1=14\\ t_{\frac{\alpha }{2}}& =t_{0.005}=2.977\\ (\bar{x}-t_{\frac{\alpha }{2}}\frac{s}{\sqrt{n}},\bar{x}+t_{\frac{\alpha }{2}}\frac{s}{\sqrt{n}})& =(1.09,1.39)\end{array}$$

Therefore, $(1.09,1.39)$ is a 99% C.I. for $\mu$.

Contributors

Ajitani Hifumi

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