Question #337153

Solve the given problems.

1. A population consists of the four numbers 1, 2, 4 and 5. List all the possible

samples of size n = 2

which can be drawn with replacement from the population.

Find the following:

a. Population mean

b. Population variance

c. Population standard deviation.

d. Mean of the sampling distribution of sample means

e. Variance of the sampling distribution of sample means

f. Standard deviation of the sampling distribution of sample means


1
Expert's answer
2022-05-06T12:13:56-0400

a. We have population values 1,2,4 and 5 population size N=4 and sample size n=2.

Mean of population (μ)(\mu) = 

1+2+4+54=3\dfrac{1+2+4+5}{4}=3


b. Variance of population 


σ2=Σ(xixˉ)2n\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n}


=4+1+1+44=52=\dfrac{4+1+1+4}{4}=\dfrac{5}{2}

c.

σ=σ2=521.5811\sigma=\sqrt{\sigma^2}=\sqrt{\dfrac{5}{2}}\approx1.5811

d. The number of possible samples which can be drawn with replacement is 24=16.2^4=16.

noSampleSamplemean (xˉ)11,12/221,23/231,45/241,56/252,13/262,24/272,46/282,57/294,15/2104,26/2114,48/2124,59/2135,16/2145,27/2155,49/2165,510/2\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} no & Sample & Sample \\ & & mean\ (\bar{x}) \\ \hline 1 &1,1 & 2/2 \\ \hdashline 2 & 1,2 & 3/2 \\ \hdashline 3 & 1,4 & 5/2 \\ \hdashline 4 & 1,5 & 6/2 \\ \hdashline 5 & 2,1 & 3/2 \\ \hdashline 6 & 2,2 & 4/2 \\ \hdashline 7 & 2,4 & 6/2 \\ \hdashline 8 & 2,5 & 7/2 \\ \hdashline 9 & 4,1 & 5/2 \\ \hdashline 10 & 4,2 & 6/2 \\ \hdashline 11 & 4,4 & 8/2 \\ \hdashline 12 & 4,5 & 9/2 \\ \hdashline 13 & 5,1 & 6/2 \\ \hdashline 14 & 5,2 & 7/2 \\ \hdashline 15 & 5,4 & 9/2 \\ \hdashline 16 & 5,5 & 10/2 \\ \hdashline \end{array}



b.


Xˉf(Xˉ)Xˉf(Xˉ)Xˉ2f(Xˉ)2/21/162/324/643/22/166/3218/644/21/164/3216/645/22/1610/3250/646/24/1624/32144/647/22/1614/3298/648/21/168/3264/649/22/1618/32162/6410/21/1610/32100/64\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) &\bar{X}^2 f(\bar{X})\\ \hline 2/2 & 1/16 & 2/32 & 4/64\\ \hdashline 3/2 & 2/16 & 6/32 & 18/64\\ \hdashline 4/2 & 1/16 & 4/32 & 16/64\\ \hdashline 5/2 & 2/16 & 10/32 & 50/64\\ \hdashline 6/2 & 4/16 & 24/32 & 144/64\\ \hdashline 7/2 & 2/16 & 14/32 & 98/64\\ \hdashline 8/2 & 1/16 & 8/32 & 64/64\\ \hdashline 9/2 & 2/16 & 18/32 & 162/64\\ \hdashline 10/2 & 1/16 & 10/32 & 100/64\\ \hdashline \end{array}


Mean of sampling distribution 

μXˉ=E(Xˉ)=Xˉif(Xˉi)=3=μ\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=3=\mu



e.The variance of sampling distribution 

Var(Xˉ)=σXˉ2=Xˉi2f(Xˉi)[Xˉif(Xˉi)]2Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2=65664(3)2=16240=54=σ2n=\dfrac{656}{64}-(3)^2=\dfrac{162}{40}=\dfrac{5}{4}= \dfrac{\sigma^2}{n}

f.

σXˉ=541.1180\sigma_{\bar{X}}=\sqrt{\dfrac{5}{4}}\approx1.1180


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