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Calculators and computers can easily calculate any Student's-t probabilities. The TI-83,83+,84+ have a tcdf function to find the probability for given values of t. The grammar for the tcdf command is tcdf(lower bound, upper bound, degrees of freedom). However for confidence intervals, we need to use inverse probability to find the value of t when we know the probability.
A probability table for the Student's-t distribution can also be used. The table gives t-scores that correspond to the confidence level (column) and degrees of freedom (row). (The TI-86 does not have an invT program or command, so if you are using that calculator, you need to use a probability table for the Student's-t distribution.) When using t-table, note that some tables are formatted to show the confidence level in the column headings, while the column headings in some tables may show only corresponding area in one or both tails.
A Student's-t table (See the Table of Contents
Tables ) gives t-scores
given the degrees of freedom and the right-tailed probability. The table is very limited.
Calculators and computers can easily calculate any Student's-t probabilities.
If the population standard deviation is not known , the margin of error for a population mean is:
The format for the confidence interval is:
$(\overline{x}-\mathrm{ME},\overline{x}+\mathrm{ME})$ .
Suppose you do a study of acupuncture to determine how
effective it is in relieving pain. You measure sensory rates for 15 subjects withthe results given below. Use the sample data to construct a 95% confidence
interval for the mean sensory rate for the population (assumed normal) fromwhich you took the data.
The solution is shown step-by-step and by using the TI-83, 83+ and 84+ calculators.
To find the confidence interval, you need the sample mean, $\overline{x}$ , and the ME.
$\overline{x}=8.2267\phantom{\rule{20pt}{0ex}}s=1.6722\phantom{\rule{20pt}{0ex}}n=15$
$\text{df}=15-1=14$
$\text{CL}=0.95\phantom{\rule{5pt}{0ex}}$ so $\phantom{\rule{5pt}{0ex}}\alpha =1-\text{CL}=1-0.95=0.05$
$\frac{\alpha}{2}=0.025\phantom{\rule{20pt}{0ex}}{t}_{\frac{\alpha}{2}}={t}_{.025}$
The area to the right of ${t}_{.025}$ is 0.025 and the area to the left of ${t}_{.025}$ is 1−0.025=0.975
${t}_{\frac{\alpha}{2}}={t}_{.025}=2.14$ using invT(.975,14) on the TI-84+ calculator.
$\mathrm{ME}={t}_{\frac{\alpha}{2}}\cdot \left(\frac{s}{\sqrt{n}}\right)$
$\mathrm{ME}=2.14\cdot \left(\frac{\mathrm{1.6722}}{\sqrt{\mathrm{15}}}\right)=0.924$
$\overline{x}-\text{ME}=8.2267-0.9240=7.3$
$\overline{x}+\text{ME}=8.2267+0.9240=9.15$
The 95% confidence interval is (7.30, 9.15) .
We estimate with 95% confidence that the true population mean sensory rate is between 7.30 and 9.15.
**With contributions from Roberta Bloom
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