Practice problems for these concepts can be found at:
Estimation and Sample Size Determination Solved Problems for Beginning Statistics
When a small sample (n < 30) is taken from a normally distributed population and the population standard deviation, σ, is unknown, a confidence interval for the population mean, μ, is given by formula (8.6):
In formula (8.6), is a point estimate of the population mean, is the estimated standard error, and t is determined by the confidence level and the degrees of freedom. The degrees of freedom is given by formula (8.7), where n is the sample size.
EXAMPLE 8.7 The distance traveled, in hundreds of miles by automobile, was determined for 20 individuals returning from vacation. The results are given in Table 8.4
For the data in Table 8.4, = 12.375, s = 3.741, and = 0.837. To find a 90% confidence interval for the mean vacation travel distance of all such individuals, we need to find the value of t in formula (8.6). The degrees of freedom is df = n – 1 = 20 – 1 = 19. Using df = 19 and the t distribution table, we must find the t value for which the area under the curve between –t and t is .90. This means the area to the left of –t is .05 and the area to the right of t is also .05. Table 8.5, which is selected from the t distribution table, indicates that the area to the right of 1.729 is .05. This is the proper value of t for the 90% confidence level when df = 19.
The following technique is also often recommended for finding the t value for a confidence interval: To find the t value for a given confidence interval, subtract the confidence level from 1 and divide the answer by 2 to find the correct area in the righthand tail.In this example, the confidence level is 90% or .90. Subtracting .90 from 1, we get .10. Now, dividing .10 by 2, we get .05 as the righthand tail area. From Table 8.5, we see that the proper t value is 1.729. Many students prefer this technique for finding the proper t value for confidence intervals.
Using formula (8.6), the lower limit for the 90% confidence interval is – = 12.375– 1.729 × .837 = 10.928 and the upper limit is + = 12.375 + 1.729 × .837 = 13.822. A 90% confidence interval for μ extends from 1093 to 1382 miles. The distribution of all vacation travel distances are assumed to be normally distributed.
To find a 95% confidence interval for μ using STATISTIX, enter the distance values into the worksheet and use the pull down Statistics ⇒One, Two, Multi Sample Tests ⇒One Sample T Testand click OK in the one sample t test dialog box. The 95% confidence interval for μ is given.
To find the 95% confidence interval using SPSS, enter the vacation distance values into SPSS, and give the pull down Analyze ⇒Compare Means ⇒One Sample T Test.The 95% confidence interval for μ is given.
The MINITAB analysis is as follows. The data is entered into column C1 and the pull down Stat ⇒Basic Statistics ⇒1 Sample t is executed. The following output is given.

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