1. 10.5 Testing Claims about the Population Standard Deviation

2. for testing claims about a population standard deviation or variance 1) The sample is a simple random sample. 2) The population has values that are normally distributed (a strict requirement).

3. Test Statistic X 2 = ( n - 1) s 2  2

4. n = sample size s 2 = sample variance  2 = population variance (given in null hypothesis) Test Statistic X 2 = ( n - 1) s 2  2

5.  Found in Table VI  Degrees of freedom = n -1

6.  All values of X 2 are nonnegative, and the distribution is not symmetric.  There is a different distribution for each number of degrees of freedom.  The critical values are found in Table VI using n-1 degrees of freedom.

7. Figure 7-12 All values are nonnegative Not symmetric x2x2 Properties of the Chi-Square Distribution

8. 0 510 15 202530354045 Figure 7-13 df = 10 df = 20 Figure 7-12 Not symmetric x2x2 There is a different distribution for each number of degrees of freedom. Properties of the Chi-Square Distribution Chi-Square Distribution for 10 and 20 Degrees of Freedom All values are nonnegative

9. Claim:   43.7 H 0 :  = 43.7 H 1 :   43.7 Example: Aircraft altimeters have measuring errors with a standard deviation of 43.7 ft. With new production equipment, 81 altimeters measure errors with a standard deviation of 52.3 ft. Use the 0.05 significance level to test the claim that the new altimeters have a standard deviation different from the old value of 43.7 ft.

10. Claim:   43.7 H 0 :  = 43.7 H 1 :   43.7   2  = 0.025 0.025   = 0.05 Example: Aircraft altimeters have measuring errors with a standard deviation of 43.7 ft. With new production equipment, 81 altimeters measure errors with a standard deviation of 52.3 ft. Use the 0.05 significance level to test the claim that the new altimeters have a standard deviation different from the old value of 43.7 ft.

11. Claim:   43.7 H 0 :  = 43.7 H 1 :   43.7 106.629 0.025 n = 81 df = 80 Table A-4 0.025   = 0.05 Example: Aircraft altimeters have measuring errors with a standard deviation of 43.7 ft. With new production equipment, 81 altimeters measure errors with a standard deviation of 52.3 ft. Use the 0.05 significance level to test the claim that the new altimeters have a standard deviation different from the old value of 43.7 ft.   2  = 0.025

12. Claim:   43.7 H 0 :  = 43.7 H 1 :   43.7 57.153 0.025 n = 81 df = 80 Table A-4 0.975 0.025   = 0.05 Example: Aircraft altimeters have measuring errors with a standard deviation of 43.7 ft. With new production equipment, 81 altimeters measure errors with a standard deviation of 52.3 ft. Use the 0.05 significance level to test the claim that the new altimeters have a standard deviation different from the old value of 43.7 ft. 106.629   2  = 0.025

13. x 2 = =  114.586 (81 -1) (52.3) 2 ( n - 1)s 2  2 2 43.7 2

14. x 2 = 114.586 x 2 = =  114.586 (81 -1) (52.3) 2 ( n - 1)s 2  2 43.7 2 Reject H 0 57.153106.629

15. x 2 = 114.586 x 2 = =  114.586 (81 -1) (52.3) 2 ( n - 1)s 2  2 43.7 2 Reject H 0 57.153 106.629 The sample evidence supports the claim that the standard deviation is different from 43.7 ft.

16. x 2 = 114.586 x 2 = =  114.586 (81 -1) (52.3) 2 ( n - 1)s 2  2 43.7 2 Reject H 0 57.153 106.629 The new production method appears to be worse than the old method. The data supports that there is more variation in the error readings than before.

17.  Table VI includes only selected values of   Specific P -values usually cannot be found  Use Table to identify limits that contain the P -value  Some calculators and computer programs will find exact P -values

18. Use the Chi-square distribution with (n -1) s 2 22 x 2 = St. Dev  or Variance  2 Which parameter does the claim address ? Proportion P Use the normal distribution where P = x/n z = P - P ˆ pq n ˆ Yes Mean (µ) Is n > 30 ? Use the normal distribution with (If  Is unknown use s instead.) z = x - µ x  n Start

19. Yes Is n > 30 ? Use the normal distribution with (If  is unknown use s instead.) z = x - µ x  n No Yes No Yes No Is the distribution of the population essentially normal ? (Use a histogram.) Use nonparametric methods which don’t require a normal distribution. See Chapter 13. Use the normal distribution with z = x - µ x  n (This case is rare.) Is  known ? Use the Student t distribution with t = x - µ x s  n

20. Example: Tests in the author’s past statistics classes have scores with a standard deviation equal to 14.1. One of his current classes now has 27 test scores with a standard deviation of 9.3. Use a 0.01 significance level to test the claim that this current class has less variation that past classes. Does a lower standard deviation suggest that the current class is doing better?

21. Claim:  <14.1 H 0 :  > 14.1 H 1 :   14.1 Example: Tests in the author’s past statistics classes have scores with a standard deviation equal to 14.1. One of his current classes now has 27 test scores with a standard deviation of 9.3. Use a 0.01 significance level to test the claim that this current class has less variation that past classes. Does a lower standard deviation suggest that the current class is doing better?

22. 0.01   = 0.01 Example: Tests in the author’s past statistics classes have scores with a standard deviation equal to 14.1. One of his current classes now has 27 test scores with a standard deviation of 9.3. Use a 0.01 significance level to test the claim that this current class has less variation than past classes. Does a lower standard deviation suggest that the current class is doing better? Claim:  <14.1 H 0 :  > 14.1 H 1 :   14.1

23.   = 0.01 Example: Tests in the author’s past statistics classes have scores with a standard deviation equal to 14.1. One of his current classes now has 27 test scores with a standard deviation of 9.3. Use a 0.01 significance level to test the claim that this current class has less variation than past classes. Does a lower standard deviation suggest that the current class is doing better? Claim:  <14.1 H 0 :  > 14.1 H 1 :   14.1 n = 27 df = 26 Table A-4 12.198 0.01 0.99

24. x 2 = =  11.311 (27 -1) (9.3) 2 ( n - 1)s 2  2 2 14.1 2

25. x 2 = 11.311 x 2 = =  11.311 (27 -1) (9.3) 2 ( n - 1)s 2  2 14.1 2 Reject H 0 12.198

26. x 2 = 11.311 x 2 = =  11.311 (27 -1) (9.3) 2 ( n - 1)s 2  2 14.1 Reject H 0 12.198 The sample evidence supports the claim that the standard deviation is less than previous classes. A lower standard deviation means there is less variance in their scores.

27.  P 5563 – 5, 9, 10, 17, 18