Section 6-4 – Confidence Intervals for the Population Variance and Standard Deviation Estimating Population Parameters

Section 6-4 – Confidence Intervals for Variance and Standard Deviation In manufacturing, it is necessary to control the amount that a process varies. For instance, an automobile part manufacturer must produce thousands of parts to be used in the manufacturing process. It is important that the parts vary little or not at all. How can you measure, and thus control, the amount of variation in the parts? You can start with a point estimate. The point estimate for 𝜎 2 is 𝑠 2 and the point estimate for 𝜎 is s. 𝑠 2 is the most unbiased estimate for 𝜎 2 . You can use a chi-square distribution to construct confidence intervals for the variance and standard deviation. If the random variable x has a normal distribution, then the distribution of 𝑋 2 = (𝑛−1) 𝑠 2 𝜎 2 as long as n > 1.

Section 6-4 – Confidence Intervals for Variance and Standard Deviation   There are 4 properties of the chi-square distribution: 1) All chi-square values 𝑋 2 are greater than or equal to zero. 2) The chi-square distribution is a family of curves, each determined by the degrees of freedom. To form a confidence interval for 𝜎 2 , use the 𝑋 2 -distribution with degrees of freedom equal to one less than the sample size. d.f. = n – 1. 3) The area under each curve of the chi-square distribution equals one. 4) Chi-square distributions are positively skewed.

Section 6-4 – Confidence Intervals for Variance and Standard Deviation   There are two critical values for each level of confidence. The value 𝑋 𝑅 2 represents the right-tail critical value and 𝑋 𝐿 2 represents the left-tail critical value. Area to the right of 𝑋 𝑅 2 = 1−𝑐 2 and the area to the right of 𝑋 𝐿 2 = 1+𝑐 2 Table 6 in Appendix B lists critical values of 𝑋 2 for various degrees of freedom and areas. (I gave you a copy of this). Each area in the table represents the region under the chi-square curve to the right of the critical value.

Section 6-4 – Confidence Intervals for Variance and Standard Deviation   Confidence Intervals for 𝜎 2 and 𝜎. You can use the critical values 𝑋 𝑅 2 and 𝑋 𝐿 2 to construct confidence intervals for a population variance and standard deviation. The formula for the confidence interval for a population variance 𝜎 2 is: (𝑛−1) 𝑠 2 𝑋 𝑅 2 < 𝜎 2 < (𝑛−1) 𝑠 2 𝑋 𝐿 2 . Remember that the population standard deviation 𝜎 is simply the square root of the variance. (𝑛−1) 𝑠 2 𝑋 𝑅 2 <𝜎< (𝑛−1) 𝑠 2 𝑋 𝐿 2

Constructing a Confidence Interval for Variance and Standard Deviation. There is no magic calculator button to do this for you. 1) Verify that the population has a normal distribution. 2) Identify the sample statistic n and the degrees of freedom. 3) Find the point estimate 𝑠 2 . 𝑠 2 = (𝑥− 𝑥 ) 2 𝑛−1 . (s is usually given to you) 4) Find the critical values 𝑋 𝑅 2 and 𝑋 𝐿 2 that correspond to the given level of confidence c. Use Table 6 in Appendix B or the chart I gave you. 5) Find the left and right endpoints and form the confidence interval for the population variance. 6) Find the confidence interval for the population standard deviation by taking the square root of each endpoint.

Section 6-4 – Confidence Intervals for Variance and Standard Deviation  EXAMPLE 1 (Page 345) Find the critical values 𝑋 𝑅 2 and 𝑋 𝐿 2 for a 90% confidence interval when the sample size is 20. Because n = 20, the degrees of freedom = 19. 𝑋 𝑅 2 = 1−𝑐 2 = 1−.9 2 =.05 𝑋 𝐿 2 = 1+𝑐 2 = 1+.9 2 =.95 Look in Table 6 in Appendix B (or on the hand-out) Find the row for the d.f. = 19. Find the columns for .05 and .95.

Section 6-4 – Confidence Intervals for Variance and Standard Deviation

Section 6-4 – Confidence Intervals for Variance and Standard Deviation  EXAMPLE 1 (Page 345) Now find the critical values 𝑋 𝑅 2 and 𝑋 𝐿 2 for a 95% confidence interval when the sample size is 25. d.f. = 24. 𝑋 𝑅 2 = 1−.𝑐 2 = 1−.95 2 =.025 𝑋 𝐿 2 = 1+𝑐 2 = 1+.95 2 =.975 Look in Table 6 in Appendix B (or on the hand-out) Find the row for the d.f. = 24 Find the columns for .025 and .975. 𝑋 𝐿 2 =12.401 and 𝑋 𝑅 2 =39.364

EXAMPLE 2 (Page 347) You randomly select and weigh 30 samples of an allergy medicine. The sample standard deviation is 1.2 milligrams. Assuming the weights are normally distributed, construct 99% confidence intervals for the population variance and standard deviation. d.f. = 29. Look in Table 6 in Appendix B (or on the hand-out) Find the row for the d.f. = 29 Find the columns for .005 and .995. 𝑋 𝑅 2 = 1−𝑐 2 = 1−.99 2 =.005 𝑋 𝐿 2 = 1+𝑐 2 = 1+.99 2 =.995 𝑋 𝐿 2 =13.121; 𝑋 𝑅 2 =52.336

EXAMPLE 2 (Page 347) You randomly select and weigh 30 samples of an allergy medicine. The sample standard deviation is 1.2 milligrams. Assuming the weights are normally distributed, construct 99% confidence intervals for the population variance and standard deviation. (𝑛−1) 𝑠 2 𝑋 𝑅 2 < 𝜎 2 < (𝑛−1) 𝑠 2 𝑋 𝐿 2 (29)( 1.2 2 ) 52.336 < 𝜎 2 < (29)( 1.2 2 ) 13.121 0.8< 𝜎 2 <3.2

EXAMPLE 2 (Page 347) Now that you have the interval for the variance, take the square roots of the endpoints to find the interval for the standard deviation. Remember, round the interval to the same number of decimal points as your data has. You were given the standard deviation (1.2) to one decimal, so state your interval to one decimal. We are 99% confident that the actual population variance of weights of allergy medicines is between 0.8 and 3.2, and that the actual population standard deviation of weights of allergy medicines is between 0.9 and 1.8. 0.80 <𝜎< 3.18 .89<𝜎<1.78

EXAMPLE 2 (Page 347) Find the 90% and 95% confidence intervals for the population variance and standard deviation of the medicine weights. Rounded to one decimal, your 90% intervals should be: 1.0< 𝜎 2 <2.4 ; 1.0<𝜎<1.54 𝑋 𝑅 2 = 1−.90 2 =.05 𝑋 𝐿 2 = 1+.90 2 =.95 𝑋 𝑅 2 =42.557 𝑋 𝐿 2 =17.708 (𝑛−1) 𝑠 2 𝑋 𝑅 2 < 𝜎 2 < (𝑛−1) 𝑠 2 𝑋 𝐿 2 (29)( 1.2 2 ) 42.557 < 𝜎 2 < (29)( 1.2 2 ) 17.708 0.98< 𝜎 2 <2.36 0.99<𝜎<1.54

EXAMPLE 2 (Page 347) Find the 90% and 95% confidence intervals for the population variance and standard deviation of the medicine weights. Rounded to one decimal, your 95% intervals should be: 0.9< 𝜎 2 <2.6 ; 1.0<𝜎<1.6 𝑋 𝑅 2 = 1−.95 2 =.025 𝑋 𝐿 2 = 1+.95 2 =.975 𝑋 𝐿 2 =16.047 𝑋 𝑅 2 =45.722 (29) (1.2 2 ) 45.722 < 𝜎 2 < (29)( 1.2 2 ) 16.047 0.91< 𝜎 2 <2.60 0.95<𝜎<1.61

Helpful Hints for Calculator: 1) Use “Alpha y=“ to access fraction function. 2) Put (n–1)(s2) in the numerator, and your 𝑋 𝑅 2 value in the denominator. This gives you the left end of the interval for the variance. 3) Press 2nd x2 and then 2nd and the negative sign to take the square root of the answer. This gives you the left end of the interval for the standard deviation. 4) Press 2nd Enter twice to get back to the fraction. Use the arrow key to put your cursor in the denominator, press Clear and enter the 𝑋 𝐿 2 value. This gives you the right end of the interval for the variance. 5) Repeat step 3 to find the right end of the interval for the standard deviation. 6) Now that you have the exact answers, round to the appropriate number of decimals.

ASSIGNMENTS Classwork: Page 348; #1-6 All Homework: Pages 348-350; #7-20 All