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Copyright © 1998, Triola, Elementary Statistics Addison Wesley Longman 1 Testing a Claim about a Standard Deviation or Variance Section 7-6 M A R I O F. T R I O L A Copyright © 1998, Triola, Elementary Statistics Addison Wesley Longman
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Copyright © 1998, Triola, Elementary Statistics Addison Wesley Longman 2 Assumption In testing hypothesis made about a population standard deviation or variance 2, we assume that the population has values that are normally distributed.
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Copyright © 1998, Triola, Elementary Statistics Addison Wesley Longman 3 Chi-Square Distribution
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Copyright © 1998, Triola, Elementary Statistics Addison Wesley Longman 4 Chi-Square Distribution X 2 = ( n – 1) s 2 2 Test Statistic
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Copyright © 1998, Triola, Elementary Statistics Addison Wesley Longman 5 Chi-Square Distribution n = sample size s 2 = sample variance 2 = population variance (given in null hypothesis) X 2 = ( n – 1) s 2 2 Test Statistic
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Copyright © 1998, Triola, Elementary Statistics Addison Wesley Longman 6 Figure 7-14Figure 7-15 There is a different distribution for each number of degrees of freedom. Properties of the Chi-Square Distribution Not symmetric All values are nonnegative 0 Chi-Square Distribution for 10 and 20 Degrees of Freedom All values are nonnegative 0
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Copyright © 1998, Triola, Elementary Statistics Addison Wesley Longman 7 Critical Value for Chi-Square Distribution Table A-4 Formula card Appendix Degrees of freedom ( df ) = n – 1
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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.
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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. Right-tailed test
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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. Right-tailed test Left-tailed test 1 –
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Copyright © 1998, Triola, Elementary Statistics Addison Wesley Longman 11 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. Right-tailed test Left-tailed test Two-tailed test 1 – 2 1 – /2 /2/2
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Copyright © 1998, Triola, Elementary Statistics Addison Wesley Longman 12 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 chi- square test statistic
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Copyright © 1998, Triola, Elementary Statistics Addison Wesley Longman 13 P -Value Method Use Table A-4 to identify limits that contain the P -value, similar to method in section 7-4.
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Copyright © 1998, Triola, Elementary Statistics Addison Wesley Longman 14 Figure 7-17 Testing a Claim about a Mean, Proportion, Standard Deviation, or Variance
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Copyright © 1998, Triola, Elementary Statistics Addison Wesley Longman 15 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 0.470kg (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
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Copyright © 1998, Triola, Elementary Statistics Addison Wesley Longman 16 Example Solution Step 1: = 0.470 Step 2: = 0.470 Step 4: Select = 0.05 (significance level) Step 5: The sample standard deviation s is relevant to this test ---- Use chi-square test statistic Step 3: H 0: = 0.470 versus H 1: = 0.470
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Copyright © 1998, Triola, Elementary Statistics Addison Wesley Longman 17 X 2 Test Statistic Assume the conjecture is true! Test Statistic: (Step 6) X 2 = ( n – 1) s 2 2 Two-tailed test 2 2 = 0.025 1 – /2 /2/2 n = 16, x = 3.675 s 2 =.432 Critical values: 6.262 & 24.788 Critical Region: X 2 < 6.262 or: X 2 > 27.488 X 2 =27.488 X 2 =6.262
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Copyright © 1998, Triola, Elementary Statistics Addison Wesley Longman 18 X 2 Test Statistic Assume the conjecture is true! Test Statistic: (Step 6) X 2 = ( n – 1) s 2 2 Two-tailed test 2 2 = 0.025 1 – /2 /2/2 n = 16 x = 3.675 s 2 =.432 X 2 = 15 *.432 /.470 2 = 29.339 Critical values: 6.262 & 24.788 Critical Region: X 2 < 6.262 or: X 2 > 27.488 X 2 =27.488 X 2 =6.262 Sample data: X 2 =29.339
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Copyright © 1998, Triola, Elementary Statistics Addison Wesley Longman 19 Example Conclusion: Based on the reported sample measurements, there is enough evidence to rejection H 0: = 0.470. Therefore, the vitamin supplement does appear to affect the variation among birth weights. (The effect on babies are not the same!) Step 8:
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Copyright © 1998, Triola, Elementary Statistics Addison Wesley Longman 20 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)
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Copyright © 1998, Triola, Elementary Statistics Addison Wesley Longman 21 Example (p-value method) Solution Step 1: = 0.470 Step 2: = 0.470 Step 4: Select = 0.05 (significance level) Step 5: The sample standard deviation s is relevant to this test ---- Use chi-square test statistic Step 3: H 0: = 0.470 versus H 1: = 0.470
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Copyright © 1998, Triola, Elementary Statistics Addison Wesley Longman 22 X 2 Test Statistic Assume the conjecture is true! Test Statistic: (Step 6) X 2 = ( n – 1) s 2 2 Two-tailed test !! n = 16 x = 3.675 s 2 =.432 X 2 = 15 *.432 /.470 2 = 29.339 Sample data: X 2 =29.339 15 X 2 =29.339 > df =15 p-value = 2 *( right tail area)
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Copyright © 1998, Triola, Elementary Statistics Addison Wesley Longman 23 X 2 Test Statistic Assume the conjecture is true! Test Statistic: (Step 6) X 2 = ( n – 1) s 2 2 Two-tailed test !! Sample data: X 2 =29.339 15 p-value = 2*( right tail area) p-value limits: 2 * 0.01 = 0.02 (>) 2 * 0.025 = 0.05 (<) P-value < 0.05 Reject Null Hypothesis
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Copyright © 1998, Triola, Elementary Statistics Addison Wesley Longman 24 Example Conclusion: Based on the reported sample measurements, there is enough evidence to rejection H 0: = 0.470. Therefore, the vitamin supplement does appear to affect the variation among birth weights. (The effect on babies are not the same!) Step 8:
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