# 第 5 章 假设检验

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• 5 5.1 5.2 5.3 5.4 5.5 5.6

• 5.1.1 5.1.2 5.1

• 5.1.1 1 160.50 1015.43

• HoHa16, Excel,15.43

• 215.43160.570.57160.50,0.575 .XLS

• 101160.5A210A2:J2K1K2A2J2=AVERAGE(A2:J2),

• N1 ,O116L1L2ABS=ABS(K2-\$O\$1)K2O1 10

• 16Excel1000 11000K2L2K3:L1001KL16

• 3. 10000.570.5710000.57 0.5716

• A1L1001

• Excel16

• 1000160.57160.51010000.570.571616

• 15.89160.1110000.110.400.11

• 5.1.2

• 1 HoHa2zt t

• 3a/2

• 4 z tPztPPa Ho

• 5

• 5.2

5.2.1 5.2.2 P 5.2.3

• 5.2.1

150.15014.9820.05

• Ho=15 Ha15z15a=0.05

• 1.961.96P

• 5 .xlsz

• B415B50.1E414.982E650B6=B3/SQRT(E3)0.014142B90.05E10z=(E2-B2)/B4-1.27279 zPz

• 5.2.2 P Pzz

• ExcelNORMSDIST1.2729, NORMSDIST1.27279NORMSDIST1-1.27279

• E9=2*NORMSDIST(-ABS(E10))0.203092 PP0.2030920.05B13

• IFIF

• logical-testE9B9 Value-if-trueValue-if-falseB13(15),(P)14.982:

• 15E414.95z()3.53553,P0.000407P

• 5.2.3

z NORMSINV

• 10 NORMSINV

• NORMSINVProbabilityzzzProbabilityB9/2B10z-1.95996

• B10=NORMSINV(B9/2)ABS=ABS(NORMSINV(B9/2))1.95996 Excel

• B14ExcelIFIFLogical_testE10B10FALSEE101.27279,1.959961Value_if_true

• Value_if_falseB13B14E414.98214.95-3.53553B13B14B4B5B9E4E5Excel

• 1200.01530120.010.01?

• Ho=120, Ha120z3.651484P0.000261z2.575835

• 5.3 241.55024.490.01

• Ho24 Ha24z24 a=0.01

• 2.33,2.0537Pa

• 5 .xlsz

• B4B5E4E5B9241.524.9500.01E9=(NORMSDIST(-ABS(E10)))0.00000105025B10=ABS(NORMSINV(B9))2.33B13B1424245024

• 5.4 St \$9350\$935012\$9323\$800.05\$9350

• Ho=9350 Ha9350t9350a =0.05

• 11t2.201t2.201P,0.025

• (1) 7 .xlst

• B49350B580E49323E512B90.05B6E10t(2)tPtExcel

• tTINVtTINVB10 TINVTINV

• TINVtExcelt

• ProbabilityB90.05Deg_freedomE5-111t2.20986273B10 PE9 TDISTTDIST

• XE101.169 Deg_freedomE5-1 11Tails2 2 P0.267061E9

• (3)B13=IF(E9B10,"","")B14

• 5.5

0.653152140.1

• HO=0.65 Ha0.65z a=0.1

• 1.65,1.65P0.05

• (1)7.xls

• 0.65Pi=214/315=0.6794PB40.65

• B90.1 E4=214/315 E5n315 B5

=SQRT((B4*(1-B4))/E5)B50.026874

• E10

=ABS((E4-B4)/B5) z1.092687(2)NORMSINVzB10=ABS(NORMSINV(B9/2))1.95995

• (3)PE9=2*(NORMSDIST(-ABS(E10)))0.274531 (4)B13=IF(E9B10,"","")B14

• 5.6.1 5.6.2 5.6.3 ()5.6

• 5.6.1

11- a a

• P

• 2 (1)HoHa

(2) ~ (n-1) Ho n-1

• (3)(4) Ha(5)

• 5.6.2

641.6501.90.051.6

• Ho1.6Ha>1.6 S1.91.6

• (1) 7 .xls

• B4()1.6B51.9B650B90.05 (2) P E9=(B5^2*(B6-1))/B4^2 E9 69.09766

• E10CHIDISTCHIDIST

• XE9Deg_freedomB6-1P0.030749

• B10CHIINVCHIINV

• ProbabilityB90.05Deg_freedomB6-1 66.33865B10(3)IFB16

• IFIF

• Logical_testE10
• 66.33865 69.0976666.33865 1.60

• 5.6.3 () 0.05 1.60?

Ho=1.6 Ha1.6.

• 5 .xlsE11=2*E10E10B18IF(E110.05

• 0.05cL2 cR2 0.05CHINV (cR2)0.025 cL21 (1-0.025=0.975)

• B12=CHIINV(B9/2,B6-1)cR2 70.22236B13=CHIINV(1-B9/2,B6-1) cL2 31.5549369.0976