HW5_sol - 1 . 9‐10 a) α = P( X ≤ 98.5) + P( X >...

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Unformatted text preview: 1 . 9‐10 a) α = P( X ≤ 98.5) + P( X > 101.5) = P⎜ ⎛ X − 100 98.5 − 100 ⎞ ⎛ X − 100 101.5 − 100 ⎞ ⎜ 2 / 9 ≤ 2 / 9 ⎟ + P⎜ 2 / 9 > 2 / 9 ⎟ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ = P(Z ≤ −2.25) + P(Z > 2.25) = (P(Z ≤‐ 2.25)) + (1 − P(Z ≤ 2.25)) = 0.01222 + 1 − 0.98778 = 0.01222 + 0.01222 = 0.02444 b) β = P(98.5 ≤ = P⎜ X ≤ 101.5 when μ = 103) ⎛ 98.5 − 103 X − 103 101.5 − 103 ⎞ ⎜ 2/ 9 ≤ 2/ 9 ≤ 2/ 9 ⎟ ⎟ ⎝ ⎠ = P(−6.75 ≤ Z ≤ −2.25) = P(Z ≤ −2.25) − P(Z ≤ −6.75) = 0.01222 − 0 = 0.01222 c) β = P(98.5 ≤ = P⎜ X ≤ 101.5 when μ = 105) ⎛ 98.5 − 105 X − 105 101.5 − 105 ⎞ ⎜ 2/ 9 ≤ 2/ 9 ≤ 2/ 9 ⎟ ⎟ ⎝ ⎠ = P(−9.75≤ Z ≤ −5.25) = P(Z ≤ −5.25) − P(Z ≤ −9.75) = 0 − 0 = 0. The probability of accepting the null hypothesis when it is actually false is smaller in part c since the true mean, μ = 105, is further from the acceptance region. A larger difference exists. 2. 9‐12 μ 0 − zα / 2 ⎜ ⎜ a) b) c) ⎛σ ⎞ ⎛σ ⎞ ⎟ ≤ X ≤ μ 0 + zα / 2 ⎜ ⎟ , where σ =2 ⎟ ⎜ ⎟ ⎝ n⎠ ⎝ n⎠ α=0.01, n=9, then zα / 2 =2,57, then 98.29,101.71 α=0.05, n=9, then zα / 2 =1.96, then 98.69,101.31 α=0.01, n=5, then zα / 2 =2.57, then 97.70,102 .30 d) α=0.05, n=5, then zα / 2 =1.96, then 98.25,101 .75 3. 9‐13 δ =103‐100=3 ⎛ δ n⎞ ⎟ , where σ =2 δ >0 then β = Φ⎜ zα / 2 − ⎜ σ⎟ ⎝ ⎠ a) β= P(98.69< X <101.31|µ=103)=P(‐6.47<Z<‐2.54)=0.0055 b) β= P(98.25< X <101.75|µ=103)=P(‐5.31<Z<‐1.40)=0.0808 a) As n increases, β decreases 4. 9‐14 a) p‐value=2(1‐ Φ( Z 0 ) )=2(1‐ Φ ( 98 − 100 ) )=2(1‐ Φ (3) )=2(1‐.99865)=0.0027 2/ 9 b) p‐value=2(1‐ Φ( Z 0 ) )=2(1‐ Φ ( 101 − 100 ) )=2(1‐ Φ (1.5) )=2(1‐.93319)=0.13362 2/ 9 102 − 100 ) )=2(1‐ Φ (3) )=2(1‐.99865)=0.0027 2/ 9 c) p‐value=2(1‐ Φ( Z 0 ) )=2(1‐ Φ ( 5. 9‐38 a) SE. Mean = σ σ = 1.015 16 N = 1 .1 16 = 0.2750 Or N = 0.2539 (Note: either one will be fine for SE Mean.) z= 15.016 − 14.5 = 1.8764 0.2750 ( since the Z statistic is based on the true standard deviation.) p‐value=1‐ Φ( z ) = 0.0303 b) One‐sided ⎛σ ⎞ x − Z 0.05 ⎜ ⎟≤μ ⎝ n⎠ ⎛ 1.1 ⎞ c) 15.016 − 1.6449⎜ ⎟≤μ ⎝ 16 ⎠ 14.5637 ≤ μ d) p‐value=2(1‐ Φ( z ) ) = 0.0606 6. 9‐45 a) 1) The parameter of interest is the true mean speed, μ. 2) H0 : μ = 100 3) H1 : μ < 100 4) α = 0.05 5) z0 = x−μ σ/ n 6) Reject H0 if z0 < −zα where −z0.05 = ‐1.65 7) x = 102.2 , σ = 4 z0 = 102.2 − 100 4/ 8 = 1.56 8) Since 1.56> −1.65, do not reject the null hypothesis and conclude the there is insufficient evidence to conclude that the true speed strength is less than 100 at α = 0.05. b) z0 = 1.56 , then p‐value= Φ ( z 0 ) ≅ 0.94 c) β = 1 − Φ⎜ − z 0.05 − d) n = ⎛ ⎜ ⎝ (95 − 100) 8 ⎞ ⎟ = 1‐Φ(‐1.65 ‐ −3.54) = 1‐Φ(1.89) = 0.02938 ⎟ 4 ⎠ Power = 1‐β = 1‐0.02938 = 0.97062 (z α + zβ ) σ 2 2 δ2 = (z 0.05 + z 0.15 )2 σ 2 (95 − 100) 2 = (1.65 + 1.03) 2 (4) 2 = 4.597, n ≅ 5 (5) 2 e) μ ≤ x + z 0.05 ⎜ ⎜ ⎛σ ⎞ ⎟ ⎟ ⎝ n⎠ ⎛4⎞ ⎟ ⎟ ⎝ 8⎠ μ ≤ 102.2 + 1.65⎜ ⎜ μ ≤ 104.53 Because 100 is included in the CI then we don’t have enough confidence to reject the null hypothesis. 7. 9‐57 a. 1) H0 : μ = 300 H1 : μ ≠ 300 2) α = 0.05 3) t0 = x−μ s/ n , n = 27 4) Reject H0 if |t0| > tα/2,n‐1 where tα/2,n‐1 = 2.0555 5) x = 325.4963 , s = 198.7855 t0 = 325.4963 − 300 198.7855 / 27 = 0.6665 6) Since 2.0555 > 0.6665, do not reject the null hypothesis and conclude that the there is sufficient evidence to conclude that the mean body weight is 300 grams at α = 0.05. b. p‐value = 2 P(Tn‐1 > 0.6665) = 0.5110 c. The 95% confidence interval is: ⎛s⎞ ⎛s⎞ x − tα / 2,n −1 ⎜ ⎟ ≤ μ ≤ x + tα / 2,n −1 ⎜ ⎟ = 246.8594 ≤ μ ≤ 404.1332 ⎜ ⎟ ⎜ ⎟ ⎝ n⎠ ⎝ n⎠ Because 300 is within the confidence interval, we fail to reject the null hypothesis. 8. 9‐69 a) In order to use t statistics in hypothesis testing, we need to assume that the underlying distribution is normal. 1) The parameter of interest is the true mean distance, μ. 2) H0 : μ = 280 3) H1 : μ > 280 4) α = 0.05 5) t0 = x−μ s/ n 6) Reject H0 if t0 > tα,n‐1 where t0.05,99 =1.6604 7) x = 260.3 s = 13.41 n = 100 t0 = 260.3 − 280 13.41 / 100 = −14.69 8) Since –14.69 < 1.6604, do not reject the null hypothesis and conclude that there is insufficient evidence to indicate that the true mean distance is greater than 280 at α = 0.05. From table V the t0 value in absolute value is greater than the value corresponding to 0.0005. Therefore, the P‐value is greater than 0.9995. b) From the normal probability plot, the normality assumption seems reasonable: Normal Probability Plot 0.997 0.99 0.98 0.95 0.90 0.75 Probability 0.50 0.25 0.10 0.05 0.02 0.01 0.003 230 240 250 Data 260 270 280 c) d = δ | μ − μ 0 | | 290 − 280 | = = = 0.75 13.41 σ σ Using the OC curve, Chart VII g) for α = 0.05, d = 0.75, and n = 100, β ≅ 0 and power of 1−0 = 1. d) d = δ | μ − μ 0 | | 290 − 280 | = = = 0.75 13.41 σ σ Using the OC curve, Chart VII g) for α = 0.05, d = 0.75, and β ≅ 0.20 (Power=0.80), n = 15 . ...
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