chapter 1

Chapter 7: Statistical Intervals Based on a Single SampleSection 7.242.2a. 22. 307χ (.1 column, 15. 1,15=d.f. row)2b. χ 34. 381. 1,25=2c. χ 44. 313. 01,25=2d. χ 46. 925. 005,25=2e. 11. 523χ (from .99. 99,25=column, 25 d.f. row)2f. χ 10. 519. 995,25=43.22a. χ 18. 307b. χ 3. 940. 05,10=. 95,10=c. Sinced. Since10.987χ14.61χ= and2.975,22= and2.95,2536.78χ37.65χ22 2= , ( χ ≤ χ ≤ χ ) . 952.025,222.05,25P .. 975,22.025,22=22 2= , ( χ ≤ χ ≤ χ ) . 90P .. 95,25.05,25=244. n – 1 = 8 , 17. 5432χ , χ 2. 180. 025,8=. 975,8=⎛ 8(7.90) 8(7.90) ⎞⎜ , ⎟ =⎝ 17.543 2.180 ⎠3 .60 , 28.98 = 1.90,5.38 .( 3.60,28.98)( ) ( ), so the 95% interval for. The 95% interval for σ is2σ is45. n = 22 implies that d.f. = n – 1 = 21, so the .995 and .005 columns of Table A.7 give thenecessary chi-squared critical values as 8.033 and 41.399. x =1701. 3 andΣx 2 i= 132,097.35 , so s2 = 25. 368( 25.368) 21( 25.368)⎛ 21⎜⎝ 41.399,8.033⎞⎟ =⎠. The interval for( 12.868,66.317)Σ i2σ isand that for σ is ( 3 .6,8.1)Validity ofthis interval requires that fracture toughness be (at least approximately) normally distributed.46.a. Using a normal probability plot, we ascertain that it is plausible that this sample wastaken from a normal population distribution.2b. With s = 1.579 , n = 15, and χ 23 . 685 the 95% upper confidence bound for σis( )14 1.57923.6852= 1.214. 05 ,14 =231