chapter 1

Chapter 7: Statistical Intervals Based on a Single Sample7. IfσL 2zα 2nσ ⎡ σ ⎤ 1 LL ′⎛ ⎞= 2zα = 2z⎜ ⎟ =2 ⎢α2 ⎥4n⎣ n ⎦⎝2⎠2= and we increase the sample size by a factor of 4, the new length isincreased fourfold. If′ , thenn = 25nL =. Thus halving the length requires n to beL′ , so the length is decreased by a factor of 5.58.a. With probability −α⎛ σ1 , zα 1≤ ( X − µ ) ⎜ ⎟ ≤ zα2⎝⎞n ⎠. These inequalities can bemanipulated exactly as was done in the text to isolate µ ; the result isσσX − zα ≤ µ ≤ X + z2α, so a 100( 1 α)%1nn⎛ σ σ ⎞⎜ X − zα , X + z ⎟2 α1⎝ n n ⎠− interval isb. The usual 95% interval has length 3 .92 , while this interval will have lengthnσα+α. With z 2. 241 2α= z1 .0125= and z α= z.0375= 1. 782nσ σ2.24 + 1.78 = 4.02 which is longer.n n( z z )( ) ,σ, the length is9.⎛ σ ⎞a. ⎜ x − 1.645 , ∞⎟ . From 5a, x = 4. 85 , σ = . 75 , n = 20;⎝ n ⎠.754 .85 1.645 = 4.5741.5741,∞204 .− , so the interval is ( )⎛ σ ⎞b. ⎜ x − zα, ∞⎟ ⎝ n ⎠⎛ σ ⎞c. ⎜− ∞, x + zα⎟ ; From 4a, x = 58. 3 , = 3. 0⎝n ⎠358.3+2.33 = ( − ∞,59.70)25221σ , n = 25;