Lecture 04 Inferential Statisitcs 2
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The document covers inferential statistics, focusing on hypothesis testing, including single and two sample t-tests and the chi-squared (c2) test. It provides detailed examples related to car fuel efficiency and machine part sizes, illustrating how to set hypotheses, calculate statistics, and interpret significance levels. By the end, students will be able to conduct these tests and evaluate their results appropriately.
Lecture 04 Inferential Statisitcs 2
- 2. This topic will cover: ◦ Hypothesis testing with a sample CI, fixed level, significance testing ◦ Two sample t-test ◦ Significance, errors and power ◦ Frequency data and the c2 test
- 3. By the end of this topic students will be able to: ◦ Perform a single sample t-test of the mean ◦ Perform a two sample t-test ◦ Interpret significance probabilities ◦ Perform a c2 goodness of fit test
- 4. ◦ A car manufacturer releases a new car and claims that its urban cycle fuel efficiency is 18.5 km per litre. A car enthusiast magazine decides to test this claim. Null hypothesis H0: m = 18.5 Alternative hypothesis is H1: m ≠ 18.5
- 6. 𝑥 = 17.79 𝑠 = 0.4782 𝜇−, 𝜇+ = 𝑥 − 𝑡 𝛾 𝑠 𝑛 , 𝑥 + 𝑡 𝛾 𝑠 𝑛 = (17.20, 18.38) a1 5.00% 2.50% 1.00% 0.50% a2 10.00% 5.00% 2.00% 1.00% g 90.00% 95.00% 98.00% 99.00% n = n - 1 1 6.3138 12.7062 31.8205 63.6567 2 2.9200 4.3027 6.9646 9.9248 3 2.3534 3.1824 4.5407 5.8409 4 2.1318 2.7764 3.7469 4.6041
- 8. ◦ Process: State null and alternative hypotheses Decide test statistic and its distribution given H0 is true Set rejection region (the significance level for the test) Calculate test statistic from sample Does test statistic fall in rejection region? ◦ Under H0 the statistic T 𝑇 = 𝑥 − 𝜇 𝑠 𝑛 ◦ is distributed as T ~ t(n- 1)
- 10. H0: m = 18.5 H1: m ≠ 18.5 𝑥 = 17.79 𝑠 = 0.4782 𝑇 = 𝑥 − 𝜇 𝑠 𝑛 = 17.79 − 18.5 0.4782 5 = −3.32 At the 1% significance level the data do not provide sufficient evidence to reject H0, a mean urban cycle fuel efficiency of 18.5 km per litre is plausible. 0.5%0.5% -4.6041 4.6041 99% t(4)
- 11. H0: m = 18.5 H1: m< 18.5 𝑥 = 17.79 𝑠 = 0.4782 𝑇 = 𝑥 − 𝜇 𝑠 𝑛 = 17.79 − 18.5 0.4782 5 = −3.32 H0 is not rejected at the 1% significance level, a mean urban cycle fuel efficiency of 18.5 km per litre is plausible. 1.0% -3.7469 99% t(4)
- 12. H0: m = 18.5 H1: m≠ 18.5 𝑥 = 17.79 𝑠 = 0.4782 𝑇 = 𝑥 − 𝜇 𝑠 𝑛 = 17.79 − 18.5 0.4782 5 = −3.32 𝑃 𝑇 > 3.32 = 𝑝 = 0.0294 1.47%1.47% -3.32 3.32 t(4)
- 13. ◦ Motivation To test whether the means of two populations are equal ◦ Assumptions: Both populations essentially normally distributed Independence of populations Sample variances differ by less than a factor of 3
- 15. Under H0 the statistic T 𝑇 = 𝑥1 − 𝑥2 𝑆 𝑝 1 𝑛1 + 1 𝑛2 is distributed as T ~ t(n1+n2- 2), where 𝑆 𝑝 2 = 𝑛1−1 𝑠1 2+ 𝑛2−1 𝑠2 2 𝑛1+𝑛2−2
- 16. ◦ Two machines on a production line are set to produce the same part. Workers down the line are concerned that the two machines are not producing parts of the same size. Test the hypothesis. ◦ H0: m1 – m2 = 0 H1: m1 – m2 ≠ 0 ◦ Sample machine 1: 28.54, 28.27, 25.81, 28.41, 29.09, 28.09, 29.72, 23.56 ◦ Sample machine 2: 31.85, 29.96, 31.97, 33.68, 23.81, 30.79
- 17. 2.5%2.5% -2.1788 2.1788 95% t(12) 5.0% -1.7823 95% t(12) a1 5.00% 2.50% 1.00% 0.50% a2 10.00% 5.00% 2.00% 1.00% g 90.00% 95.00% 98.00% 99.00% n = n1 + n2 - 2 11 1.7959 2.2010 2.7181 3.1058 12 1.7823 2.1788 2.6810 3.0545 13 1.7709 2.1604 2.6503 3.0123
- 18. ◦ Sample machine 1: 28.54, 28.27, 25.81, 28.41, 29.09, 28.09, 29.72, 23.56 ◦ Sample machine 2: 31.85, 29.96, 31.97, 33.68, 23.81, 30.79 ◦ 𝑥1 = 27.68, 𝑠1 = 2.014, 𝑥2 = 30.34, 𝑠2 = 3.44 ◦ 𝑆 𝑝 2 = 𝑛1−1 𝑠1 2+ 𝑛2−1 𝑠2 2 𝑛1+𝑛2−2 = 8 − 1 4.1+ 6 − 1 11.8 8+6−2 = 7.3 ◦ 𝑇 = 𝑥1−𝑥2 𝑆 𝑝 1 𝑛1 + 1 𝑛2 = −2.66 7.3 1 8 + 1 6 = −1.82 ◦ H0: m1 – m2 = 0 H1: m1 – m2 ≠ 0 ◦ H0 cannot be rejected at the 5% significance level
- 19. Actual H0 H1 Decision H0 ☺ Type 2 H1 Type 1 ☺ a = 5% b Actual H0 H1 Decision H0 1 - a b H1 a 1 - b significance power Distribution of Sample Means z = 1.6449
- 20. ◦ Motivation Decide reasonableness of using particular probability model to describe empirical data ◦ Example uses Does Poisson probability distribution fit empirical data? Are two categorical variables independent?
- 21. Number Frequency 0 10 1 17 2 42 3 34 4 12 5 5 >5 0 120 Average number of customers per hour = 2.3 𝑃 𝑋 = 𝑥 = e−𝜆 𝜆 𝑥 𝑥! = e−2.3 2.3 𝑥 𝑥!
- 22. Number Frequency observed 0 10 1 17 2 42 3 34 4 12 5 5 >5 0 120 𝜒2 = 𝑖=1 𝑘 𝑂𝑖 − 𝐸𝑖 2 𝐸𝑖 ≈ 𝜒2 𝑘 − 𝑚 − 1 • k is number of categories • m is number of estimated parameters in model • Only valid if Ei ≥ 5 for all i Frequency expected 0.1003 12.0311 0.2306 27.6714 0.2652 31.8222 0.2033 24.3970 0.1169 14.0283 0.0538 6.4530 0.0300 3.5971 1 120
- 23. Number Frequency observed Frequency expected 0 10 0.1003 12.0311 1 17 0.2306 27.6714 2 42 0.2652 31.8222 3 34 0.2033 24.3970 4 12 0.1169 14.0283 >4 5 0.0838 10.0501 120 1 120 𝜒2 = 𝑖=1 𝑘 𝑂𝑖 − 𝐸𝑖 2 𝐸𝑖 ≈ 𝜒2 𝑘 − 𝑚 − 1 • k = 6 • m = 1 • Valid since Ei ≥ 5 for all i
- 24. significance level p = 0.05 p = 0.01 aR = 5.00% aR = 1.00% n = k - m -1 1 3.841 6.635 2 5.991 9.210 3 7.815 11.345 4 9.488 13.277 5 11.070 15.086 etc. aR c2
- 25. Number Frequency observed Frequency expected (O – E)2/E 0 10 0.1003 12.0311 0.34 1 17 0.2306 27.6714 4.12 2 42 0.2652 31.8222 3.26 3 34 0.2033 24.3970 3.78 4 12 0.1169 14.0283 0.29 >4 5 0.0838 10.0501 2.54 120 1 120 14.32 𝜒2 = 𝑖=1 𝑘 𝑂𝑖 − 𝐸𝑖 2 𝐸𝑖 = 14.32 H0 is rejected at the 5% significance level.
- 26. Motivation • Is there an association between two categorical variables? Method of Follow-up email phone visit satisfied 15 22 17 54 neither 6 6 3 15 dissatisfied 9 2 10 21 30 30 30 90
- 27. • Is there an association between two categorical variables? • What are expected counts with no association? Method of Follow-up email phone visit 90
- 28. • Is there an association between two categorical variables? • What are expected counts with no association? Method of Follow-up satisfied neither dissatisfied 90
- 29. • Is there an association between two categorical variables? • What are expected counts with no association? Method of Follow-up email phone visit satisfied neither dissatisfied 90
- 30. • Observed and expected counts. Method of Follow-up email phone visit satisfied 15 (18) 22 (18) 17 (18) 54 neither 6 (5) 6 (5) 3 (5) 15 dissatisfied 9 (7) 2 (7) 10 (7) 21 30 30 30 90 𝜒2 = 𝑖=1 𝑘 𝑂𝑖 − 𝐸𝑖 2 𝐸𝑖 = 8.07 𝜒2 ≈ 𝜒2 𝑟 − 1 𝑐 − 1
- 31. ◦ Simple random sampling ◦ Systematic random sampling ◦ Stratified random sampling ◦ Cluster (area) sampling
- 32. ◦ Quota sampling ◦ Availability / convenience sampling ◦ Purposive sampling
- 33. By the end of this topic students will be able to: ◦ Perform a single sample t-test of the mean ◦ Perform a two sample t-test ◦ Interpret significance probabilities ◦ Perform a c2 goodness of fit test
- 34. • Buglear, J. Quantitative Methods for Business. Elsevier Butterworth Heinemann. • Hinton, PR. Statistics Explained. Routledge. • Thomas, AB. Research Skills for Management Studies. Routledge. • Thomas, AB. Research Concepts for Management Studies. Routledge. • Wisniewski, M. Quantitative Methods for Decision Makers. FT Prentice Hall.
- 35. Any Questions?