Answer to Question #258935 in Statistics and Probability for Restarbar

Question #258935

Solve. Given a population of the scores of BSHS honor students in Mathematics as 1, 3, 4, 6, and 8. Suppose a sample size of 3 have to be drawn from it for the first, second and third rank;

1
Expert's answer
2021-11-01T19:52:06-0400

a.

 We have population values 1,3,4,6,8,1,3,4,6,8, population size N=5N=5

μ=1+3+4+6+85=4.4\mu=\dfrac{1+3+4+6+8}{5}=4.4

b.


σ2=15((14.4)2+(34.4)2+(44.4)2\sigma^2=\dfrac{1}{5}((1-4.4)^2+(3-4.4)^2+(4-4.4)^2

+(64.4)2)+(84.4)2)=5.84+(6-4.4)^2)+(8-4.4)^2)=5.84


c.


σ=σ2=5.842.4166\sigma=\sqrt{\sigma^2}=\sqrt{5.84}\approx2.4166


d. We have population values 1,3,4,6,81,3,4,6,8 population size N=5N=5 and sample size n=3.n=3. Thus, the number of possible samples which can be drawn without replacement is



(Nn)=(53)=10\dbinom{N}{n}=\dbinom{5}{3}=10SampleSampleSample meanNo.values(Xˉ)11,3,48/321,3,610/331,3,8441,4,611/351,4,813/361,6,8573,4,613/383,4,8593,6,817/3104,6,86\def\arraystretch{1.5} \begin{array}{c:c:c} Sample & Sample & Sample \ mean \\ No. & values & (\bar{X}) \\ \hline 1 & 1,3,4 & 8/3 \\ \hdashline 2 & 1,3,6 & 10/3 \\ \hdashline 3 & 1,3,8 & 4 \\ \hdashline 4 & 1,4,6 & 11/3 \\ \hdashline 5 & 1,4,8 & 13/3\\ \hdashline 6 & 1,6,8 & 5 \\ \hdashline 7 & 3,4,6 & 13/3 \\ \hdashline 8 & 3,4,8 & 5 \\ \hdashline 9 & 3,6,8 & 17/3 \\ \hdashline 10 & 4,6,8 & 6 \\ \hline \end{array}

The sampling distribution of the sample means.



Xˉff(Xˉ)Xˉf(Xˉ)Xˉ2f(Xˉ)8/311/108/3064/9010/311/1010/30100/9011/311/1011/10121/90411/1012/30144/9013/322/1026/30338/90522/1030/30450/90611/1018/30324/9017/311/1017/30289/90Total1014.461/3\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c} & \bar{X} & f & f(\bar{X}) & \bar{X}f(\bar{X})& \bar{X}^2f(\bar{X}) \\ \hline & 8/3 & 1 & 1/10 & 8/30 & 64/90 \\ \hdashline & 10/3 & 1 & 1/10 & 10/30 & 100/90 \\ \hdashline & 11/3 & 1 & 1/10 & 11/10 & 121/90 \\ \hdashline & 4 & 1& 1/10 & 12/30 & 144/90 \\ \hdashline & 13/3 & 2 & 2/10 & 26/30 & 338/90 \\ \hdashline & 5 & 2 & 2/10 & 30/30 & 450/90 \\ \hdashline & 6 & 1 & 1/10 & 18/30 & 324/90 \\ \hdashline & 17/3 & 1 & 1/10 & 17/30 & 289/90 \\ \hdashline Total & & 10 & 1 & 4.4 & 61/3 \\ \hline \end{array}




E(Xˉ)=Xˉf(Xˉ)=4.4E(\bar{X})=\sum\bar{X}f(\bar{X})=4.4


The mean of the sampling distribution of the sample means is equal to the the mean of the population.


E(Xˉ)=μXˉ=4.4=μE(\bar{X})=\mu_{\bar{X}}=4.4=\mu



e.


Var(Xˉ)=Xˉ2f(Xˉ)(Xˉf(Xˉ))2Var(\bar{X})=\sum\bar{X}^2f(\bar{X})-(\sum\bar{X}f(\bar{X}))^2




=613(4.4)2=2.923=\dfrac{61}{3}-(4.4)^2=\dfrac{2.92}{3}




σXˉ=Var(Xˉ)=2.9230.9866\sigma_{\bar{X}}=\sqrt{Var(\bar{X})}=\sqrt{\dfrac{2.92}{3}}\approx0.9866

Verification:


Var(Xˉ)=σ2n(NnN1)=5.843(5351)Var(\bar{X})=\dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})=\dfrac{5.84}{3}(\dfrac{5-3}{5-1})




=2.923,True=\dfrac{2.92}{3}, True




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