1. 1 Section 8.5 Testing a claim about a mean (σ unknown) Objective For a population with mean µ (with σ unknown), use a sample to test a claim about the mean. Testing a mean (when σ known) uses the t-distribution

2. 2 Notation

3. 3 (1) The population standard deviation σ is unknown (2) One or both of the following: Requirements The population is normally distributed or The sample size n > 30

4. 4 Test Statistic Denoted t (as in t-score) since the test uses the t-distribution.

5. 5 People have died in boat accidents because an obsolete estimate of the mean weight (of 166.3 lb.) was used. A random sample of n = 40 men yielded the mean x = 172.55 lb. and standard deviation s = 26.33 lb. Do not assume the population standard deviation  is known. Test the claim that men have a mean weight greater than 166.3 lb. using 90% confidence. What we know: µ 0 = 166.3 n = 40 x = 172.55 s = 26.33 Claim: µ > 166.3 using α = 0.1 Note: Conditions for performing test are satisfied since n >30 Example 1

6. 6 What we know: µ 0 = 166.3 n = 40 x = 172.55 s = 26.33 Claim: µ > 166.3 using α = 0.1 H 0 : µ = 166.3 H 1 : µ > 166.3 right-tailed test Initial Conclusion: Since t in critical region, Reject H 0 Final Conclusion: Accept the claim that the mean weight is greater than 166.3 lb. t in critical region (df = 39) Using Critical Regions Example 1 t α = 1.304 t = 1.501 Test statistic: Critical value:

7. 7 Stat → T statistics → One sample → with summary Calculating P-value for a Mean (σ unknown)

8. 8 Then hit Next Enter the Sample mean (x) Sample std. dev. (s) Sample size (n) Calculating P-value for a Mean (σ unknown)

9. 9 Then hit Calculate Select Hypothesis Test Enter the Null:mean (µ 0 ) Select Alternative (“ ”, or “≠”) Calculating P-value for a Mean (σ unknown)

10. 10 Test statistic (t) P-value Calculating P-value for a Mean (σ unknown) The resulting table shows both the test statistic (t) and the P-value Initial Conclusion Since P-value < α (α = 0.1), reject H 0 Final Conclusion Accept the claim the mean weight greater than 166.3 Ib

11. 11 Using StatCrunch Using the P-value Example 1 Stat → T statistics→ One sample → With summary Null: proportion= Alternative Sample mean: Sample std. dev.: Sample size: ● Hypothesis Test 172.55 37.8 40 166.3 > P-value = 0.0707 What we know: µ 0 = 166.3 n = 40 x = 172.55 s = 26.33 Claim: µ > 166.3 using α = 0.1 Initial Conclusion: Since P-value < α, Reject H 0 Final Conclusion: Accept the claim that the mean weight is greater than 166.3 lb. H 0 : µ = 166.3 H 1 : µ > 166.3

12. 12 P-Values A useful interpretation of the P-value: it is observed level of significance Thus, the value 1 – P-value is interpreted as observed level of confidence Recall: “Confidence Level” = 1 – “Significance Level” Note: Only useful if we reject H 0 If H 0 accepted, the observed significance and confidence are not useful.

13. 13 P-Values From Example 1: P-value = 0.0707 1 – P-value = 0.9293 Thus, we can say conclude the following: The claim holds under 0.0707 significance. or equivalently… We are 92.93% confident the claim holds

14. 14 Loaded Die When a fair die (with equally likely outcomes 1-6) is rolled many times, the mean valued rolled should be 3.5 Your suspicious a die being used at a casino is loaded (that is, it’s mean is a value other than 3.5) You record the values for 100 rolls and end up with a mean of 3.87 and standard deviation 1.31 Using a confidence level of 99%, does the claim that the dice are loaded? What we know: µ 0 = 3.5 n = 100 x = 3.87 s = 1.31 Claim: µ ≠ 3.5 using α = 0.01 Note: Conditions for performing test are satisfied since n >30 Example 2

15. 15 H 0 : µ = 3.5 H 1 : µ ≠ 3.5 What we know: µ 0 = 3.5 n = 100 x = 3.87 s = 1.31 Claim: µ ≠ 3.5 using α = 0.01 two-tailed test Example 2 t in critical region (df = 99) Test statistic: Critical value: z = 3.058 z α = -2.626 z α = 2.626 Using Critical Regions Initial Conclusion: Since P-value < α, Reject H 0 Final Conclusion: Accept the claim the die is loaded.

16. 16 Using StatCrunch Using the P-value Example 2 Null: proportion= Alternative Sample mean: Sample std. dev.: Sample size: ● Hypothesis Test 3.87 1.31 100 3.5 ≠ P-value = 0.0057 Initial Conclusion: Since P-value < α, Reject H 0 Final Conclusion: Accept the claim the die is loaded. H 0 : µ = 3.5 H 1 : µ ≠ 3.5 What we know: µ 0 = 3.5 n = 100 x = 3.87 s = 1.31 Claim: µ ≠ 3.5 using α = 0.01 We are 99.43% confidence the die are loaded Stat → T statistics→ One sample → With summary

17. 17

18. 18 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  2 -distribution

19. 19 Notation

20. 20 Notation

21. 21 (1) The sample is a simple random sample (2) The population is normally distributed Very strict condition!!! Requirements

22. 22 Test Statistic Denoted  2 (as in  2 -score) since the test uses the  2 -distribution. nSample size sSample standard deviation σ 0 Claimed standard deviation

23. 23 Critical Values Right-tailed test“>“ Needs one critical value (right tail) Use StatCrunch: Chi-Squared Calculator

24. 24 Critical Values Left-tailed test“<” Needs one critical value (left tail) Use StatCrunch: Chi-Squared Calculator

25. 25 Critical Values Two-tailed test“≠“ Needs two critical values (right and left tail) Use StatCrunch: Chi-Squared Calculator

26. 26 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. Example 1 What we know: σ 0 = 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 Problem 14, pg 449

27. 27 What we know: σ 0 = 14.1 n = 27 s = 9.3 Claim: σ < 14.1 using α = 0.01 H 0 : σ = 14.1 H 1 : σ < 14.1 Left-tailed Using Critical Regions Example 1  2 in critical region (df = 26) Initial Conclusion: Since  2 in critical region, Reject H 0 Final Conclusion: Accept the claim that the new class has less variance than the past classes  2   2 L  Test statistic: Critical value:

28. 28 Calculating P-value for a Variance Stat → Variance → One sample → with summary

29. 29 Then hit Next Enter the Sample variance (s 2 ) Sample size (n) Calculating P-value for a Variance s 2 = 9.3 2 = 86.49 NOTE: Must use Variance

30. 30 Then hit Calculate Select Hypothesis Test Enter the Null:variance (σ 0 2 ) Select Alternative (“ ”, or “≠”) Calculating P-value for a Variance σ 0 2 = 14.1 2 = 198.81

31. 31 Test statistic (  2 ) P-value The resulting table shows both the test statistic (  2 ) and the P-value Calculating P-value for a Variance

32. 32 What we know: σ 0 = 14.1 n = 27 s = 9.3 Claim: σ < 14.1 using α = 0.01 Using Critical Regions Example 1 Using StatCrunch Initial Conclusion: Since P-value < α (α = 0.01), Reject H 0 Final Conclusion: Accept the claim that the new class has less variance than the past classes We are 99.44% confident the claim holds Stat → Variance → One sample → With summary Null: proportion= Alternative Sample variance: Sample size: 86.49 27 198.81 < ● Hypothesis Test P-value = 0.0056 s 2 = 86.49 σ 0 2 = 198.81 H 0 : σ 2 = 198.81 H 1 : σ 2 < 198.81

33. 33 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. Example 2 What we know: σ 0 = 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 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

34. 34 Using Critical Regions Example 2  2 not in critical region (df = 26) Test statistic: Critical values:  2  0.005  2 R  2 L  Initial Conclusion: Since  2 not in critical region, Accept H 0 Final Conclusion: Reject the claim recent winners have a different variations than in the 20s and 30s Since H 0 accepted, the observed significance isn’t useful. What we know: σ 0 = 1.34 n = 10 s = 1.186 Claim: σ ≠ 1.34 using α = 0.01 H 0 : σ = 1.34 H 1 : σ ≠ 1.34 two-tailed

35. 35 Using P-value Example 2 Using StatCrunch Null: proportion= Alternative Sample variance: Sample size: 1.407 10 1.796 < ● Hypothesis Test P-value = 0.509 Initial Conclusion: Since P-value ≥ α (α = 0.01), Accept H 0 Final Conclusion: Reject the claim recent winners have a different variations than in the 20s and 30s Since H 0 accepted, the observed significance isn’t useful. s 2 = 1.407 σ 0 2 = 1.796 What we know: σ 0 = 1.34 n = 10 s = 1.186 Claim: σ ≠ 1.34 using α = 0.01 H 0 : σ 2 = 1.796 H 1 : σ 2 < 1.796 Stat → Variance → One sample → With summary