Confidence Intervals for Population Parameters
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Estimation part III
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Confidence Intervals for Population Parameters
- 2. CONFIDENCE INTERVAL FOR THE DIFFERENCE OF TWO MEANS : d = 1 - 2 (For paired observation) 𝑑 ± 𝑡∝ 2 𝑛 − 1 𝑆 𝑑 𝑛 𝑑 = Σ𝑑 𝑛 𝑆 𝑑 = 𝑛Σ𝑑2 − Σ𝑑 2 𝑛 𝑛 − 1
- 3. Example No.1. The comparison of two methods for training managers to be more assertive, suppose that 10 pairs of managerial assertiveness test scores were as shown in the table find a 95% confidence interval for the difference in mean levels of assertiveness, d .
- 4. Method 1 (x) Method 2 (y) d = x – y d2 78 71 7 49 63 44 19 361 72 61 11 121 89 84 5 25 91 74 17 289 49 51 – 2 04 68 55 13 169 76 60 16 256 85 77 8 64 55 39 16 256 d = 110 d2 = 1594 Solution:
- 5. 𝑑 = Σ𝑑 𝑛 = 110 10 = 11 𝑆 𝑑 = 𝑛Σ𝑑2− Σ𝑑 2 𝑛 𝑛−1 = 10 1594 − 110 2 10 10−1 𝑆 𝑑 = 6.53 = 1 – 0.95 = 0.05 𝑡∝ 2 (𝑛 − 1) = 𝑡 0.05 2 (10 − 1) = 𝑡0.025(9) = 2.262 𝑑 ± 𝑡∝ 2 𝑛 − 1 𝑆 𝑑 𝑛 11 ± 2.262 6.53 10 11 ±4.7 6.3d15.17
- 6. CONFIDENCE INTERVAL FOR POPULATION PROPORTION 𝑝 = 𝑥 𝑛 q = 1 – p 𝑝 ± 𝑍∝ 2 𝑝𝑞 𝑛
- 7. Example No.2 A manufacturer wants to assess the proportion of defective items in a large batch produced by a particular machine. He tests 60 random sample of 300 items and find that 45 items are defective. Calculate 95% confidence interval for that of defective items in the batch.
- 8. Solution: n = 300 x = 45 𝒑 = 𝒙 𝒏 = 𝟒𝟓 𝟑𝟎𝟎 = 𝟎. 𝟏𝟓 q = 1 – p = 1 – 0.15 = 0.85 = 1 – 0.95 = 0.05 𝒁 𝜶 𝟐 = 𝒁 𝟎.𝟎𝟓 𝟐 = 𝒁 𝟎.𝟎𝟐𝟓 = 𝟏. 𝟗𝟔 𝒑 ± 𝒁∝ 𝟐 𝒑𝒒 𝒏 𝟎. 𝟏𝟓 ± 𝟏. 𝟗𝟔 𝟎.𝟏𝟓 ×𝟎.𝟖𝟓 𝟑𝟎𝟎 𝟎. 𝟏𝟓 ± 𝟎. 𝟎𝟒𝟎𝟒 0.1096 p 0.1904
- 9. Example No.3 What size sample would have to be taken order to estimate the percentage to within 12% with 90% confidence, in a random sample of 400 carpet shops, it was discovered that 136 of them sold carpets at below the list prices.
- 10. Solution: n = 400 x = 136 𝒑 = 𝒙 𝒏 = 𝟏𝟑𝟔 𝟒𝟎𝟎 = 𝟎. 𝟑𝟒 q = 1 – p = 1 – 0.34 = 0.66 = 1 – 0.90 = 0.10 𝒁 𝜶 𝟐 = 𝒁 𝟎.𝟏𝟎 𝟐 = 𝒁 𝟎.𝟎𝟓 = 𝟏. 𝟔𝟒𝟓 𝒏 = 𝒁∝ 𝟐 𝟐 𝒑𝒁 𝒆 𝟐 = 𝟏.𝟔𝟒𝟓 𝟐 𝟎.𝟑𝟒 𝟎.𝟔𝟔 𝟎.𝟎𝟐 𝟐 n = 1520 1520 shops would have to be sampled.
- 11. CONFIDENCE INTERVAL FOR THE DIFFERENCE OF TWO PROPORTIONS 𝑝1 = 𝑥1 𝑛1 , 𝑞1 = 1 − 𝑝1 𝑝2 = 𝑥2 2 , 𝑞2 = 1 − 𝑝2 𝑝1 − 𝑝2 ± 𝑍∝ 2 𝑝1 𝑞1 𝑛1 + 𝑝2 𝑞2 𝑛2
- 12. Example No.4 A poll is taken among the residents of a city and the surrounding county to determine the feasibility of proposal to construct a civic center. If 2400 of 5000 city residents favor the proposal and 1200 of 2000 county residents favor it, find a 90% confidence interval for the true difference in the fractions favoring the proposal to construct the civic center.
- 13. Solution: 𝑝1 = 𝑥1 𝑛1 = 2400 5000 = 0.48 𝑝2 = 𝑥2 2 = 1200 2000 = 0.60 𝑞1 = 1 − 𝑝1 = 1 − 0.48 = 0.52 𝑞2 = 1 − 𝑝2 = 1 − 0.60 = 0.40 = 1 – 0.90 = 0.10 𝑍 𝛼 2 = 𝑍0.10 2 = 𝑍0.05 = 1.645
- 14. 𝑝1 − 𝑝2 ± 𝑍∝ 2 𝑝1 𝑞1 𝑛1 + 𝑝2 𝑞2 𝑛2 0.48 − 0.60 ± 1.645 0.48 (0.52) 5000 + 0.60 (0.40) 2000 −𝟎. 𝟏𝟐 ± 𝟎. 𝟎𝟐𝟏𝟒 – 0.1414 p1 – p2 – 0.0986
- 15. CONFIDENCE INTERVAL FOR POPULATION VARIANCE 𝑛 − 1 𝑆2 𝑥2 𝛼 2 𝑛 − 1 < 2 < 𝑛 − 1 𝑆2 𝑥2 1− 𝛼 2 𝑛 − 1
- 16. Example No.5 A random sample of 8 cigarettes of a certain brand has an average nicotine content of 3.6 milligrams and standard deviation of 0.9 milligrams. Construct 99% confidence interval for variance.
- 17. Solution: S = 0.9 S2 = 0.81 n = 8 = 1 – 0.99 = 0.01 𝒙 𝟐 𝜶 𝟐 (𝒏 − 𝟏) = 𝒙 𝟐 𝟎.𝟎𝟏 𝟐 (𝟖 − 𝟏) = 𝒙 𝟐 𝟎.𝟎𝟎𝟓(𝟕) = 𝟐𝟎. 𝟐𝟕𝟖 𝒙 𝟐 𝟏− 𝜶 𝟐 (𝒏 − 𝟏) = 𝒙 𝟐 𝟏−𝟎.𝟎𝟎𝟓(𝟖 − 𝟏) = 𝒙 𝟐 𝟎.𝟗𝟗𝟓(𝟕) = 𝟎. 𝟗𝟖𝟗
- 18. 𝑛 − 1 𝑆2 𝑥2 𝛼 2 𝑛 − 1 < 2 < 𝑛 − 1 𝑆2 𝑥2 1− 𝛼 2 𝑛 − 1 𝟕(𝟎.𝟖𝟏) 𝟐𝟎.𝟐𝟕𝟖 < 𝟐 < 𝟕(𝟎.𝟖𝟏) 𝟎.𝟗𝟖𝟗 0.2796 < 𝟐 < 𝟓. 𝟕𝟑𝟑𝟏