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Section 8.6 Testing a claim about a standard deviation. Objective For a population with standard deviation σ , use a sample too test a claim about the standard deviation. Tests of a standard deviation use the c 2 -distribution. Notation. Notation. (1) The sample is a simple random sample
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Section 8.6Testing a claim about a standard deviation Objective For a population with standard deviation σ, use a sample too test a claim about the standard deviation. Tests of a standard deviation use the c2-distribution
(1) The sample is a simple random sample (2) The population is normally distributed Very strict condition!!! Requirements
Test Statistic Denoted c2 (as in c2-score) since the test uses the c2 -distribution. n Sample size s Sample standard deviation σ0Claimed standard deviation
Critical Values Right-tailed test “>“Needs one critical value (right tail) Use StatCrunch: Chi-Squared Calculator
Critical Values Left-tailed test “<”Needs one critical value (left tail) Use StatCrunch: Chi-Squared Calculator
Critical Values Two-tailed test “≠“Needs two critical values (right and left tail) Use StatCrunch: Chi-Squared Calculator
Example 1 Statisics Test Scores Tests scores in the author’s previous statistic classes have followed a normal distribution with a standard deviation equal to 14.1. His current class has 27 tests scores with a standard deviation of 9.3. Use a 0.01 significance level to test the claim that this class has less variation than the past classes. Problem 14, pg 449 What we know: σ0= 14.1 n= 27 s= 9.3 Claim:σ < 14.1 using α= 0.01 Note: Test conditions are satisfied since population is normally distributed
Using Critical Regions Example 1 What we know: σ0= 14.1 n= 27 s= 9.3 Claim:σ < 14.1 using α= 0.01 H0:σ = 14.1 H1:σ < 14.1 Left-tailed Test statistic: c2L= 12.20 c2 = 11.31 Critical value: (df = 26) c2in critical region Initial Conclusion:Sincec2in critical region, Reject H0 Final Conclusion: Accept the claimthat the new class has less variance than the past classes
Calculating P-value for a Variance Stat → Variance → One sample → with summary
Calculating P-value for a Variance • Enter the Sample variance (s2) • Sample size (n) • NOTE: Must use Variance s2 = 9.32 = 86.49 Then hit Next
Calculating P-value for a Variance Select Hypothesis Test Enter the Null:variance(σ02) Select Alternative(“<“, “>”, or “≠”) σ02= 14.12 = 198.81 Then hit Calculate
Calculating P-value for a Variance The resulting table shows both the test statistic (c2) and the P-value Test statistic (c2) P-value
Using Critical Regions Example 1 What we know: σ0= 14.1 n= 27 s= 9.3 Claim:σ < 14.1 using α= 0.01 H0:σ = 14.1 H1:σ < 14.1 Stat → Variance→ One sample → With summary ● Hypothesis Test Sample variance: Sample size: 86.49 27 198.81 < Null: proportion= Alternative Using StatCrunch P-value = 0.0056 s2 = 86.49 σ02= 198.81 Initial Conclusion:Since P-value < α(α = 0.01), Reject H0 Final Conclusion: Accept the claimthat the new class has less variance than the past classes We are 99.44% confident the claim holds
Example 2 BMI for Miss America Listed below are body mass indexes (BMI) for recent Miss America winners. In the 1920s and 1930s, distribution of the BMIs formed a normal distribution with a standard deviation of 1.34. Use a 0.01 significance level to test the claim that recent Miss America winners appear to have variation that is different from that of the 1920s and 1930s. Problem 17, pg 449 19.5 20.3 19.6 20.2 17.8 17.9 19.1 18.8 17.6 16.8 Using StatCrunch: s= 1.1862172 What we know: σ0= 1.34 n= 10 s= 1.186 Claim:σ ≠ 1.34 using α= 0.01 Note: Test conditions are satisfied since population is normally distributed
Using Critical Regions Example 2 What we know: σ0= 1.34 n= 10 s= 1.186 Claim:σ ≠ 1.34 using α= 0.01 H0:σ = 1.34 H1:σ ≠ 1.34 two-tailed 0.005 Test statistic: c2L= 2.088 c2R= 26.67 c2 = 7.053 Critical values: (df = 26) c2notin critical region Initial Conclusion:Since c2not in critical region, Accept H0 Final Conclusion: Reject the claimrecent winners have a different variations than in the 20s and 30s Since H0 accepted, the observed significance isn’t useful.
Using P-value Example 2 What we know: σ0= 1.34 n= 10 s= 1.186 Claim:σ ≠ 1.34 using α= 0.01 H0:σ2 = 1.796 H1:σ2<1.796 Stat → Variation → One sample → With summary ● Hypothesis Test Sample variance: Sample size: 1.407 10 1.796 < Null: proportion= Alternative Using StatCrunch P-value = 0.509 s2 = 1.407 σ02= 1.796 Initial Conclusion:Since P-value ≥ α(α = 0.01), Accept H0 Final Conclusion: Reject the claimrecent winners have a different variations than in the 20s and 30s Since H0 accepted, the observed significance isn’t useful.