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
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A researcher is conducting a study about the effectiveness of a new product to the households in a certain Barangay. His population consists of the numbers 8, 4, 2, 1, 13, and 10. Perform the following tasks: 1. Determine the number of sets of all possible random samples using the combination formula 2. List all the possible samples and compute the mean of each sample 3. Construct the sampling distribution. 4. Construct the histogram