1. Compute the mean and the standard deviation of the population.
E [ X ] = ( 2 + 5 + 6 + 8 + 10 + 12 + 13 ) / 7 = 8 E[X]=(2+5+6+8+10+12+13)/7=8 E [ X ] = ( 2 + 5 + 6 + 8 + 10 + 12 + 13 ) /7 = 8
E [ X 2 ] = ( 2 2 + 5 2 + 6 2 + 8 2 + 1 0 2 + 1 2 2 + 1 3 2 ) / 7 = 542 / 7 = 77 3 7 E[X^2 ]=(2^2+5^2+6^2+8^2+10^2+12^2+13^2 )/7=542/7=77\frac{3}{7} E [ X 2 ] = ( 2 2 + 5 2 + 6 2 + 8 2 + 1 0 2 + 1 2 2 + 1 3 2 ) /7 = 542/7 = 77 7 3
V a r [ X ] = E [ X 2 ] − E [ X ] 2 = 77 3 7 − 64 = 13 3 7 Var[X]=E[X^2 ]-E[X]^2=77 \frac{3}{7}-64=13\frac{3}{7} Va r [ X ] = E [ X 2 ] − E [ X ] 2 = 77 7 3 − 64 = 13 7 3
σ ( X ) = V a r [ X ] = 94 / 7 = 3.6645 \sigma(X)=\sqrt{Var[X] }=\sqrt{94/7}=3.6645 σ ( X ) = Va r [ X ] = 94/7 = 3.6645
2. List all samples of size 5 and compute the mean for each sample.
3. Construct the sampling distribution of the sample means.
P ( X ˉ 5 = 6.2 ) = P ( X ˉ 5 = 6.6 ) = P ( X ˉ 5 = 6.8 ) = P ( X ˉ 5 = 7.0 ) = P ( X ˉ 5 = 7.2 ) = P ( X ˉ 5 = 7.4 ) = P ( X ˉ 5 = 7.8 ) = P ( X ˉ 5 = 8.0 ) = P ( X ˉ 5 = 8.6 ) = P ( X ˉ 5 = 8.8 ) = P ( X ˉ 5 = 9.0 ) = P ( X ˉ 5 = 9.2 ) = P ( X ˉ 5 = 9.6 ) = P ( X ˉ 5 = 9.8 ) = 1 / 21 P(\bar X_5=6.2)= P(\bar X_5=6.6)=P(\bar X_5=6.8)=P(\bar X_5=7.0)=P(\bar X_5=7.2)=P(\bar X_5=7.4)=P(\bar X_5=7.8)=P(\bar X_5=8.0)=P(\bar X_5=8.6)=P(\bar X_5=8.8)=P(\bar X_5=9.0)=P(\bar X_5=9.2)=P(\bar X_5=9.6)=P(\bar X_5=9.8)=1/21 P ( X ˉ 5 = 6.2 ) = P ( X ˉ 5 = 6.6 ) = P ( X ˉ 5 = 6.8 ) = P ( X ˉ 5 = 7.0 ) = P ( X ˉ 5 = 7.2 ) = P ( X ˉ 5 = 7.4 ) = P ( X ˉ 5 = 7.8 ) = P ( X ˉ 5 = 8.0 ) = P ( X ˉ 5 = 8.6 ) = P ( X ˉ 5 = 8.8 ) = P ( X ˉ 5 = 9.0 ) = P ( X ˉ 5 = 9.2 ) = P ( X ˉ 5 = 9.6 ) = P ( X ˉ 5 = 9.8 ) = 1/21
P ( X ˉ 5 = 8.2 ) = P ( X ˉ 5 = 8.4 ) = 2 / 21 P(\bar X_5=8.2)=P(\bar X_5=8.4)=2/21 P ( X ˉ 5 = 8.2 ) = P ( X ˉ 5 = 8.4 ) = 2/21
P ( X ˉ 5 = 7.6 ) = 3 / 21 P(\bar X_5=7.6)=3/21 P ( X ˉ 5 = 7.6 ) = 3/21
4. Calculate the mean of the sampling distribution of the sample means. Compare this to mean of the population.
E [ X ˉ 5 ] = ( 6.2 + 6.6 + 6.8 + 7.0 + 7.2 + 7.4 + 7.8 + 8.0 + 8.6 + 8.8 + 9.0 + 9.2 + 9.6 + 9.8 + 2 ⋅ 8.2 + 2 ⋅ 8.4 + 3 ⋅ 7.6 ) / 21 = 8.0 = E [ X ] E[\bar X_5 ]=(6.2+6.6+6.8+7.0+7.2+7.4+7.8+8.0+8.6+8.8+9.0+9.2+9.6+9.8+2\cdot 8.2+2\cdot 8.4+3\cdot 7.6)/21=8.0=E[X] E [ X ˉ 5 ] = ( 6.2 + 6.6 + 6.8 + 7.0 + 7.2 + 7.4 + 7.8 + 8.0 + 8.6 + 8.8 + 9.0 + 9.2 + 9.6 + 9.8 + 2 ⋅ 8.2 + 2 ⋅ 8.4 + 3 ⋅ 7.6 ) /21 = 8.0 = E [ X ]
5. Calculate the standard deviation of the sampling distribution of the sample means . Compare this to the standard deviation of the population.
E [ X ˉ 5 2 ] = ( 6. 2 2 + 6. 6 2 + 6. 8 2 + 7. 0 2 + 7. 2 2 + 7. 4 2 + 7. 8 2 + 8. 0 2 + 8. 6 2 + 8. 8 2 + 9. 0 2 + 9. 2 2 + 9. 6 2 + 9. 8 2 + 2 ⋅ 8. 2 2 + 2 ⋅ 8. 4 2 + 3 ⋅ 7. 6 2 ) / 21 = 1362.8 / 21 = 64 94 105 E[\bar X_5^2] =(6.2^2+6.6^2+6.8^2+ 7.0^2+7.2^2+7.4^2+7.8^2+8.0^2+8.6^2+ 8.8^2+9.0^2+9.2^2+9.6^2 +9.8^2+2⋅8.2^2+2⋅8.4^2 +3⋅7.6^2)/21=1362.8/21=64\frac{94}{105} E [ X ˉ 5 2 ] = ( 6. 2 2 + 6. 6 2 + 6. 8 2 + 7. 0 2 + 7. 2 2 + 7. 4 2 + 7. 8 2 + 8. 0 2 + 8. 6 2 + 8. 8 2 + 9. 0 2 + 9. 2 2 + 9. 6 2 + 9. 8 2 + 2 ⋅ 8. 2 2 + 2 ⋅ 8. 4 2 + 3 ⋅ 7. 6 2 ) /21 = 1362.8/21 = 64 105 94
V a r [ X ˉ 5 ] = E [ X ˉ 5 2 ] − E [ X ˉ 5 ] 2 = 64 94 105 − 64 = 94 105 Var[\bar X_5 ]=E[\bar X_5^2 ]-E[\bar X_5 ]^2=64\frac{94}{105}-64=\frac{94}{105} Va r [ X ˉ 5 ] = E [ X ˉ 5 2 ] − E [ X ˉ 5 ] 2 = 64 105 94 − 64 = 105 94
σ ( X ˉ 5 ) = V a r [ X ˉ 5 ] = 94 / 105 = σ ( X ) / 15 = 0.946 \sigma(\bar X_5 )=\sqrt{Var[\bar X_5 ]}=\sqrt{94/105}=\sigma(X)/\sqrt{15}=0.946 σ ( X ˉ 5 ) = Va r [ X ˉ 5 ] = 94/105 = σ ( X ) / 15 = 0.946