Suppose we are studying a normally distributed population with unknown variance s^2. We can estimate s^2 with the sample variance S^2 from a random sample of size n. Then we can construct a confidence interval for s^2. However in this case, the interval will not be of the form S^2 +/- e.
The data can be used to form a Chi Square Distribution with n - 1 degrees of freedom. Unlike the Normal and t-Distributions, the Chi Square Distributions are not symmetric. Therefore if we desire a level of confidence r, then we must find separate endpoints c and d such that P(c <= s^2 <= d) = r, with P(s^2 < c) = (1 - r) / 2 = P(d < s^2). In other words, the probability that the variance s^2 lies outside of a confidence interval [c, d] will be 1 - r, with s^2 being equally likely to fall below c or above d.
To find these endpoints, we first find two chi-square scores Q and T as follows. If X ~ X2(n - 1), then Q is the value such that P(X <= Q) = (1 - r) / 2 and T is the value such that P(X <= T) = (1 + r) / 2. The endpoints of the confidence interval are then given by c = (n - 1)(S^2) / T and d = (n - 1)(S^2) / Q. If we desire a confidence interval for the standard deviation s, then we use [Sqrt(c), Sqrt(d)].
To execute the program, we need only enter the sample size, the sample variance, and the level of confidence. The program displays confidence intervals for both the variance and the standard deviation. Be patient (especially on the TI-89). The program may take a minute or so to find the chi-square scores Q and T.
Using a Data Set: If we have a collection of data points, we can enter the data into the STAT Edit screen to calculate the sample deviation. This value, along with the sample size, then can be entered into the program from the VARS screen.
Example. In the article "The malleability of IQ as judged from adoption studies " in the journal Intelligence, 14, (1990), the IQ's of a group of 29 adopted children yielded an average IQ of 97 with a sample deviation of S =13. Find 95% confidence intervals for true variance and standard deviation of the IQ scores of all such children.
Solution. After calling up the VARCONFI program, enter 29 for SAMPLE SIZE, enter 13^2 for SAMPLE VAR. (square it since 13 was the deviation), and enter .95 for CONF. LEVEL. We find a 95% confidence interval for s^2 to be [106.4309, 309.1222], and the interval for the true standard deviation to be and [10.3165, 17.5819].
1. Below is a random sample of combined SAT scores from a group of second year college students. Assuming that SAT scores are normally distributed, find a 98% confidence interval for the standard deviation of SAT scores among second year students.
2. Independent random samples of Verbal SAT scores from 21 girls and 21 boys yielded a sample deviation for the 21 girls' scores of Sx ~ 110.2789 and a sample deviation for the boys' scores of Sy ~ 65.6868.
Find 90% confidence intervals for the variances of the scores. Is it at all possible that girls' Verbal SAT scores and boys' Verbal SAT scores have the same variance?
1. Enter the data into the STAT Edit screen (LIST EDIT on the TI-86, APPS 6 on the TI-89), and use the 1-Var Stats command (OneVar on the TI-86 and 89) to compute the statistics. We see that n = 32 and S = 177.9531925.
Call up the VARCONFI program. After entering the necessary values of n and S^2, we find that a 98% interval for the standard deviation is [137.1473, 250.4112].
2. If we first execute the VARCONFI program for the 21 girls' scores with S^2 = 110.2789^2, we obtain a 90% confidence interval of [7743.5646, 22415.7168]. If we then execute the program for the 21 boys' scores with S^2 = 65.6868^2, we obtain a 90% confidence interval of [2747.3392, 7952.872].
Because there is some small overlap in these intervals, it is barely possible for the true variances to be equal. (For instance, they could both equal 7800.)
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