Let N � 6 and n � 3. For purpose of studying sampling distributions, assume
that all population values are known. Specically, y1 � 98, y2 � 102, y3 � 154,
y4 � 133, y5 � 190, and y6 � 175. We are interested in �yU, the population
mean. Two sampling plan are proposed.
Plan 1. Eight possible samples may be chosen.
Sample
Number Sample,S P(S)
1 (1,3,5) 1/8
2 (1,3,6) 1/8
3 (1,4,5) 1/8
4 (1,4,6) 1/8
5 (2,3,5) 1/8
6 (2,3,6) 1/8
7 (2,4,5) 1/8
8 (2,4,6) 1/8
Plan 2. Three possible samples may be chosen.
Sample
Number Sample, S P(S)
1 (1,4,6) 1/8
2 (2,3,6) 1/8
3 (1,3,5) 1/8
a What is the value of �yU?
b Let �y be the mean of the sample values. For each sampling plan �nd
E(y), Var(y), bias(y), and MSE(�y).
c Which sampling plan do you think is better? Why?
a) To find the value of �yU (the population mean), we can simply calculate the average of all the population values:
�yU = (y1 + y2 + y3 + y4 + y5 + y6) / N
= (98 + 102 + 154 + 133 + 190 + 175) / 6
= 852 / 6
= 142
Therefore, the value of �yU is 142.
b) For each sampling plan, let's calculate the expected value (E(y)), variance (Var(y)), bias(y), and mean squared error (MSE(�y)).
Plan 1:
Sample mean (�y) = (98 + 102 + 154) / 3 = 118
E(y) = 1/8 * (98 + 102 + 154) + 1/8 * (98 + 102 + 175) + ... + 1/8 * (154 + 133 + 190)
= (98 + 102 + 154 + 98 + 102 + 175 + ... + 154 + 133 + 190) / 8
= (118 + 135 + 137 + 147 + 137 + 147 + 152 + 163) / 8
= 104.375
Var(y) = [(98 - 118)^2 + (102 - 118)^2 + (154 - 118)^2 + ... + (175 - 118)^2] / 8
= (20^2 + 16^2 + 36^2 + 15^2 + 72^2 + 57^2 + ... + 57^2) / 8
= 7591 / 8
= 948.875
Bias(y) = E(y) - �yU
= 104.375 - 142
= -37.625
MSE(�y) = Var(y) + (Bias(y))^2
= 948.875 + (-37.625)^2
= 2037.625
Plan 2:
Sample mean (�y) = (98 + 133 + 175) / 3 = 135.333
E(y) = 1/8 * (98 + 133 + 175) + 1/8 * (102 + 154 + 133) + 1/8 * (98 + 102 + 154)
= (98 + 133 + 175 + 102 + 154 + 133 + 98 + 102 + 154) / 8
= 127.25
Var(y) = [(98 - 135.333)^2 + (133 - 135.333)^2 + (175 - 135.333)^2] / 8
= (38.333^2 + 2.333^2 + 39.667^2) / 8
= 591.064
Bias(y) = E(y) - �yU
= 127.25 - 142
= -14.75
MSE(�y) = Var(y) + (Bias(y))^2
= 591.064 + (-14.75)^2
= 605.59
c) To determine which sampling plan is better, we can compare the mean squared errors (MSE(�y)) of both plans. The sampling plan with a smaller MSE(�y) is generally considered better as it indicates less error or variability between the sample mean and the population mean.
Comparing the MSE values:
Plan 1: MSE(�y) = 2037.625
Plan 2: MSE(�y) = 605.59
Since Plan 2 has a smaller MSE, it is considered a better sampling plan as it results in less error or variability in estimating the population mean.
a) To find the value of the population mean �yU, we need to calculate the average of all the population values:
�yU = (y1 + y2 + y3 + y4 + y5 + y6) / N
= (98 + 102 + 154 + 133 + 190 + 175) / 6
= 852 / 6
= 142
Therefore, the value of �yU is 142.
b) For each sampling plan, we need to find E(y), Var(y), bias(y), and MSE(�y).
Plan 1:
E(y) = (1/8) * (98 + 154 + 190) + (1/8) * (98 + 154 + 175) + (1/8) * (98 + 133 + 190) + (1/8) * (98 + 133 + 175) + (1/8) * (102 + 154 + 190) + (1/8) * (102 + 154 + 175) + (1/8) * (102 + 133 + 190) + (1/8) * (102 + 133 + 175)
= (1/8) * (3*442 + 3*427)
= (1/8) * (1326 + 1281)
= (1/8) * 2607
= 325.875
Var(y) = [(98 - E(y))^2 + (154 - E(y))^2 + (190 - E(y))^2 + (102 - E(y))^2 + (133 - E(y))^2 + (175 - E(y))^2] / (8-1)
= [(98 - 325.875)^2 + (154 - 325.875)^2 + (190 - 325.875)^2 + (102 - 325.875)^2 + (133 - 325.875)^2 + (175 - 325.875)^2] / 7
Bias(y) = E(y) - �yU
MSE(�y) = Var(y) + Bias(y)^2
Plan 2:
E(y), Var(y), bias(y), and MSE(�y) can be calculated in a similar manner as in Plan 1.
c) To determine which sampling plan is better, we need to compare the values of MSE(�y) for both plans. The plan with a lower MSE(�y) indicates that the estimate of the population mean is closer to the true population mean on average.
After calculating MSE(�y) for both plans, we can compare the values and choose the plan with the lower MSE(�y) as the better sampling plan.