# 8.4 Confidence intervals: confidence interval, single population  (Page 2/9)

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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.

## The notation for the student's-t distribution is (using t as the random variable) is

• $T$ ~ ${t}_{\text{df}}$ where $\text{df}=n-1$ .
• For example, if we have a sample of size n=20 items, then we calculate the degrees of freedom as df=n−1=20−1=19 and we write the distribution as $T$ ~ ${t}_{\text{19}}$

If the population standard deviation is not known , the margin of error for a population mean is:

• $\mathrm{ME}={t}_{\frac{\alpha }{2}}\cdot \left(\frac{s}{\sqrt{n}}\right)\phantom{\rule{20pt}{0ex}}$
• ${t}_{\frac{\alpha }{2}}$ is the t-score with area to the right equal to $\frac{\alpha }{2}$
• use $\text{df}=n-1$ degrees of freedom
• $s$ = sample standard deviation

The format for the confidence interval is:

$\left(\overline{x}-\mathrm{ME},\overline{x}+\mathrm{ME}\right)$ .

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.

• 8.6
• 9.4
• 7.9
• 6.8
• 8.3
• 7.3
• 9.2
• 9.6
• 8.7
• 11.4
• 10.3
• 5.4
• 8.1
• 5.5
• 6.9
• Check the assumptions and conditions.
• The first solution is step-by-step (Solution A).
• The second solution uses an two column model(Solution B).

## Solution a

Start by discussing the assumptions and conditions that support you model.
• Randomization-The problem does not state that the sample was randomly selected. I will assume that it is in order to meet the conditions necessary to calculate a confidence interval.
• Independence- The problem does not state that the 15 subjects are independent. I will assume that they are in order to meet the conditions necessary to calculate a confidence interval.
• 10% condition-A sample size of 15 must be less than 10% of the total population receiving acupuncture to relieving pain. (A reasonable assumption).
• Nearly Normal-The problem states that the population should be assumed to be nearly normal (skewed to the left in histogram but somewhat evenly distributed about the median). The histogram and boxplot below support that statement.  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.

When calculating the margin of error, a probability table for the Student's-t distribution can be used to find the value of t. The table gives t-scores that correspond to the confidence level (column) and degrees of freedom (row); the t-score is found where the row and column intersect in the table.

**With contributions from Roberta Bloom

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