Section 8.4 Confidence Intervals for a Standard Deviation 393 EXAMPLE 8.20 Constructing a confidence interval The compressive strengths of seven concrete blocks, in pounds per square inch, are measured, with the following results. 1989.9 1993.8 2074.5 2070.5 2070.9 2033.6 1939.6 Assume these values are a simple random sample from a normal population. Construct a 95% confidence interval for the population standard deviation ��. Solution Step 1: Find s2. s2 = Σ (x − ̄x)2 7 − 1 = 2699.8648 Step 2: Find the critical values. We have 7 − 1 = 6 degrees of freedom. Since the confidence level is 95%, the critical values are ��2 0.975 and ��2 0.025. From Table A.4, we find ��2 0.975 = 1.237 ��2 0.025 = 14.449 Step 3: Compute the lower and upper confidence bounds. Lower bound = (n − 1)s2 ��2 ��∕2 = (7 − 1)(2699.8648) 14.449 = 1121.129 Upper bound = (n − 1)s2 ��2 1−��∕2 = (7 − 1)(2699.8648) 1.237 = 13,095.545 Step 4: The 95% confidence interval for ��2 is 1121.129 < ��2 < 13,095.545 √ To find the confidence interval for ��, we take square roots. We find that 1121.129 = 33.48, and √ 13,095.545 = 114.44. The 95% confidence interval for �� is 33.48 < �� < 114.44 Step 5: Interpret the result. We are 95% confident that the population standard deviation of the strengths of the concrete blocks is between 33.48 and 114.44. Check Your Understanding 3. Construct a 95% confidence interval for the population standard deviation �� if a sample of size 10 has standard deviation s = 6. (4.13, 10.95) 4. Construct a 99% confidence interval for the population standard deviation �� if a sample of size 23 has standard deviation s = 12. (8.60, 19.15) Answers are on page 395. Justification for the method Confidence intervals for the variance of a normal distribution are based on the fact that when a sample of size n is drawn from a normal distribution, the quantity (n − 1)s2∕��2 follows a chi-square distribution with n−1 degrees of freedom. Therefore, for a proportion 1 − �� of all possible samples, ��2 ��∕2 < (n − 1)s2 ��2 < ��2 1−��∕2
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