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HW3_Solutions

# HW3_Solutions - Chapter 6 Interval Estimation EXERCISES 6.1...

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Unformatted text preview: Chapter 6 Interval Estimation EXERCISES 6.1 6.1.1. (a) We are 99% confident that the estimate value of the parameter lies in the confidence interval. (b) 99% confidence interval is wider (c) When μ is known but σ 2 is unknown we use t-distribution for the sample size n ≤ 30. If the distribution is binomial and there are enough number of samples such that np ≥ 5 and np( 1 − p) ≥ 5, then we use normal approximation. (d) More the information higher the confidence interval. So the sample size is inversely proportional to the width of the confidence interval. 6.1.2. For y = U θ as a pivot p ( a < U θ < b ) = 0.98 F y (a) = 0.01, F y (b) = 0.99 This implies a n = 0.01, b n = 0.99 a = n √ 0.01, b = n √ 0.99 p n √ 0.01 < U θ < n √ 0.99 = 0.98 98% confidence interval for θ is U n √ 0.99 , U n √ 0.01 6.1.3. (a) p − 2.81 ≤ x − μ σ/ √ n ≤ 2.75 = k p − 2.81 · σ √ n ≤ x − μ ≤ 2.75 · σ √ n = k p − 2.81 · σ √ n − x ≤ − μ ≤ 2.75 · σ √ n + x = k p x − 2.75 · σ √ n ≤ μ ≤ x + 2.81 · σ √ n = k 99 100 CHAPTER 6 Interval Estimation (b) Confidence interval for μ is given by x − 2.75 · σ √ n , x + 2.81 · σ √ n (c) Confidence level = k 6.1.4. n = 50 x = 15.65 σ = 0.59 95% confidence interval for mean μ is x − z 0.025 σ √ n , x − z 0.025 σ √ n = 15.65 − 1.96 0.59 √ 50 , 15.65 + 1.96 0.59 √ 50 = ( 15.945, 15.80 ) 6.1.5. (a) Here x i ∼ N(μ , σ 2 ) (n − 1 )s 2 σ 2 ∼ χ 2 (n − 1 ) Pivot = (n − 1 )s 2 σ 2 ; where only σ 2 is unknown. p a < (n − 1 )s 2 σ 2 < b = 1 − α p χ 2 1 − α/ 2 < (n − 1 )s 2 σ 2 < χ 2 α/ 2 = 1 − α p 1 χ 2 1 − α/ 2 < σ 2 (n − 1 )s 2 < 1 χ 2 α/ 2 = 1 − α p (n − 1 )s 2 χ 2 α/ 2 < σ 2 < (n − 1 )s 2 χ 2 1 − α/ 2 = 1 − α So ( 1 − α) · 100% confidence interval for σ 2 is given by (n − 1 )s 2 χ 2 α/ 2 < σ 2 < (n − 1 )s 2 χ 2 1 − α/ 2 (b) n = 21, x = 44.3, s = 3.96 α = 0.1, 1 − α = 0.9, α/ 2 = 0.05 χ 2 1 − α/ 2,20 = χ 2 0.95,20 = 10.851 χ 2 α/ 2 = χ 2 0.05,20 = 31.410 90% confidence interval is given by (n − 1 )s 2 χ 2 α/ 2 < σ 2 < (n − 1 )s 2 χ 2 1 − α/ 2 = 20 ( 3.96 ) 2 31.41 < σ 2 < 20 ( 3.96 ) 2 10.851 We are 90% confident that σ 2 lies in the interval (9.985, 28.903). 6.1.9. (a) p a ≤ x − μ σ/ √ n ≤ b = 1 − α p − z α/ 2 < x − μ σ/ √ n < z α/ 2 = 1 − α p − z α/ 2 · σ √ n < x − μ < z α/ 2 · σ √ n = 1 − α Instructor’s Solutions Manual 101 p − z α/ 2 · σ √ n − x < − μ < z α/ 2 · σ √ n − x = 1 − α p x − z α/ 2 · σ √ n < μ < x + z α/ 2 · σ √ n = 1 − α So the confidence interval is x − z α/ 2 · σ √ n , x + z α/ 2 · σ √ n ....
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HW3_Solutions - Chapter 6 Interval Estimation EXERCISES 6.1...

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