chapter 1

Chapter 9: Inferences Based on Two Samples4.a. From Exercise 2, the C.I. iss2s21 2( x − y) ± ( 1.96) + = 2100 ± 1.96( 433.33) = 2100 ± 849. 33m1250.67,2949.33( )n= . In the context of this problem situation, the interval ismoderately wide (a consequence of the standard deviations being large), so theinformation about µ1and µ2is not as precise as might be desirable.b. From Exercise 3, the upper bound is5700 + 1.645 396.93 = 5700 + 652.95 = 6352. .( ) 955.a. H a says that the average calorie output for sufferers is more than 1 cal/cm 2 /min below thatfor nonsufferers.21σ+σm n(.64− 2.05) − ( −1)22=2(.04) (.16)10+102= .1414z == −2.90. At level .01, H o is rejected if ≤ −2. 33.14142.90 < -2.33, reject H o ., soz ; since –b. P = Φ( − 2 .90) = . 0019⎛ − 1.2 + 1⎞⎜⎟⎝ .1414 ⎠c. β = 1 − Φ − 2.33 − = 1 − Φ( − .92) = . 8212d.( 2.33 + 1.28)2( − .2).2m = n == 65.15 , so use 66.2262