Testing a Claim about a Standard Deviation or

Assumption • In testing hypothesis made about a population standard deviation s or variance s 2, we assume that the population has values that are normally distributed. Copyright © 1998, Triola, Elementary Statistics Addison Wesley Longman 2

Chi-Square Distribution Copyright © 1998, Triola, Elementary Statistics Addison Wesley Longman 3

Chi-Square Distribution Test Statistic X 2 = (n – 1) s 2 Copyright ©

Chi-Square Distribution Test Statistic X 2 = n (n – 1) s 2 = sample size s 2 = sample variance s 2 = population variance (given in null hypothesis) Copyright © 1998, Triola, Elementary Statistics Addison Wesley Longman 5

Chi-Square Distribution for 10 and 20 Degrees of Freedom Properties of the Chi-Square Distribution Not symmetric 0 c 2 All values are nonnegative Figure 7 -14 c 2 All values are nonnegative 0 Figure 7 -15 There is a different distribution for each number of degrees of freedom. Copyright © 1998, Triola, Elementary Statistics Addison Wesley Longman 6

Critical Value for Chi-Square Distribution Table A-4 v Formula card v Appendix v Degrees

When using Table A-4, it is essential to note that each critical value separates an area to the right that responds to the value given in the top row. Copyright © 1998, Triola, Elementary Statistics Addison Wesley Longman 8

When using Table A-4, it is essential to note that each critical value separates an area to the right that responds to the value given in the top row. a Right-tailed test a Copyright © 1998, Triola, Elementary Statistics Addison Wesley Longman 9

When using Table A-4, it is essential to note that each critical value separates an area to the right that responds to the value given in the top row. a Right-tailed test a a 1– a Left-tailed test Copyright © 1998, Triola, Elementary Statistics Addison Wesley Longman 10

When using Table A-4, it is essential to note that each critical value separates an area to the right that responds to the value given in the top row. a Right-tailed test a a 1– a Left-tailed test a/2 Two-tailed test 1 – a/2 Copyright © 1998, Triola, Elementary Statistics Addison Wesley Longman a/2 11

All three methods 1) Traditional method 2) P-value method 3) Confidence intervals and the testing procedure Step 1 to Step 8 in Section 7 -3 are still valid, except that the test statistic is a chisquare test statistic Copyright © 1998, Triola, Elementary Statistics Addison Wesley Longman 12

P-Value Method Use Table A-4 to identify limits that contain the P-value, similar to

Figure 717 Testing a Claim about a Mean, Proportion, Standard Deviation, or Variance Copyright

Example Shown below are birth weights (in kilograms) of male babies born to mothers on a special vitamin supplement. Test the claim that this sample comes from a population with a standard deviation equal to 0. 470 kg (which is the standard deviation for male birth weights in general). 3. 73 4. 37 3. 73 4. 33 3. 39 3. 68 4. 68 3. 52 3. 02 4. 09 2. 47 4. 13 4. 47 3. 22 3. 42 2. 54 Copyright © 1998, Triola, Elementary Statistics Addison Wesley Longman 15

Example Solution Step 1: s = 0. 470 Step 2: s = 0. 470 Step 3: H 0: s = 0. 470 versus H 1: s = 0. 470 Step 4: Select a = 0. 05 (significance level) Step 5: The sample standard deviation s is relevant to this test ---- Use chi-square test statistic Copyright © 1998, Triola, Elementary Statistics Addison Wesley Longman 16

X 2 (Step 6) Assume the conjecture is true! n = 16, Test Statistic X 2 = Test Statistic: x = 3. 675 (n – 1) s 2 s 2 =. 432 Critical Region: X 2 < 6. 262 or: X 2 > 27. 488 Critical values: 6. 262 & 24. 788 a/2 = 0. 025 a/2 Two-tailed test 2 a/2 1 – a/2 X =6. 262 X 2 =27. 488 Copyright © 1998, Triola, Elementary Statistics Addison Wesley Longman 17

(Step 6) Assume the conjecture is true! X 2 Test Statistic X 2 = Test Statistic: n = 16 x = 3. 675 (n – 1) s 2 s 2 =. 432 X 2 = 15 *. 432 /. 4702 = 29. 339 Critical values: 6. 262 & 24. 788 Critical Region: X 2 < 6. 262 or: X 2 > 27. 488 a/2 = 0. 025 Sample data: X 2= 29. 339 Two-tailed test 2 a/2 1 – a/2 X =6. 262 X 2 =27. 488 Copyright © 1998, Triola, Elementary Statistics Addison Wesley Longman 18

Example Step 8: Conclusion: Based on the reported sample measurements, there is enough evidence to rejection H 0: s = 0. 470. Therefore, the vitamin supplement does appear to affect the variation among birth weights. (The effect on babies are not the same!) Copyright © 1998, Triola, Elementary Statistics Addison Wesley Longman 19

P-Value Method When n is large, X 2 center (50% quantile) is close to df = (n -1). So, in a two side test if a p-value method is used, then (i) if sample X 2 > n - 1, p-value = 2 * (right tail area) (ii) if sample X 2 < n - 1, p-value = 2 * (left tail area) Copyright © 1998, Triola, Elementary Statistics Addison Wesley Longman 20

Example (p-value method) Solution Step 1: s = 0. 470 Step 2: s = 0. 470 Step 3: H 0: s = 0. 470 versus H 1: s = 0. 470 Step 4: Select a = 0. 05 (significance level) Step 5: The sample standard deviation s is relevant to this test ---- Use chi-square test statistic Copyright © 1998, Triola, Elementary Statistics Addison Wesley Longman 21

(Step 6) Assume the conjecture is true! X 2 Test Statistic: n = 16 x = 3. 675 s 2 =. 432 X 2 = (n – 1) s 2 X 2 = 15 *. 432 /. 4702 = 29. 339 Two-tailed test !! X 2= 29. 339 > df =15 p-value = 2 *( right tail area) 15 Sample data: X 2= 29. 339 Copyright © 1998, Triola, Elementary Statistics Addison Wesley Longman 22

(Step 6) Assume the conjecture is true! X 2 Test Statistic: X 2 = (n – 1) s 2 Two-tailed test !! p-value = 2*( right tail area) 15 p-value limits: 2 * 0. 01 = 0. 02 (>) 2 * 0. 025 = 0. 05 (<) P-value < 0. 05 Sample data: X 2= 29. 339 Reject Null Hypothesis Copyright © 1998, Triola, Elementary Statistics Addison Wesley Longman 23

Example Step 8: Conclusion: Based on the reported sample measurements, there is enough evidence to rejection H 0: s = 0. 470. Therefore, the vitamin supplement does appear to affect the variation among birth weights. (The effect on babies are not the same!) Copyright © 1998, Triola, Elementary Statistics Addison Wesley Longman 24