Question #312499

Consider all samples of size 5 from this population:6 8 10 12 13



1. Compute the mean and variance of the population



2.list all samples of size 4 and compute the mean for each sample



3.construct the sampling distribution of the sample means



4.Calculate the mean of the sampling distribution of the sample means and compare this to the mean of the population

1
Expert's answer
2022-03-17T07:31:15-0400

Solution

Population size N=5N=5


1.(a) mean

μ=XiN\mu=\dfrac{\sum X_i}N


μ=6+8+10+12+135\mu=\dfrac{6+8+10+12+13}5


μ=9.8\mu=9.8


(b) Variance


XiXiXˉ(XiXˉ)263.814.4481.83.24100.20.04122.24.84133.210.24\def\arraystretch{1.5}\begin{array}{c:c:c}X_i &X_i -\bar X& (X_i -\bar X)^2\\\hline6&-3.8&14.44\\\hdashline8&-1.8&3.24\\\hdashline10&0.2&0.04\\\hdashline12&2.2&4.84\\\hdashline13&3.2&10.24\\\hline\end{array}


σ2=(XiXˉ)N=32.85\sigma^2=\dfrac{\sum( X_i-\bar X)}N=\dfrac{32.8}5


σ2=6.56\sigma^2=6.56


2. Samples of size 4 and their means

Sample size n=4n=4


Possible samples

=C54=(54)=5=C_5^4= \binom{5}{4}=5 samples


Sample meanfrequencyProbability9.0011/59.2511/59.7511/510.2511/510.7511/5\def\arraystretch{1.5}\begin{array}{c:c:c}Sample~mean& frequency & Probability \\\hline9.00&1&1/5\\\hdashline9.25&1&1/5\\\hdashline9.75&1&1/5\\\hdashline10.25&1&1/5\\\hdashline10.75&1&1/5 \\\hline\end{array}no.Samplesmeans16,8,10,129.0026,8,10,139.2536,8,12,139.7546,10,12,1310.2558,10,12,1310.75\def\arraystretch{1.5}\begin{array}{c:c:c}no.&Samples&means\\\hline1&6,8,10,12&9.00\\\hdashline2&6,8,10,13&9.25\\\hdashline3&6,8,12,13&9.75\\\hdashline4&6,10,12,13&10.25\\\hdashline5&8,10,12,13&10.75\\\hline\end{array}


3. Sampling distribution of the sample means

meanfreqProb9.0011/59.2511/59.7511/510.2511/510.7511/5\def\arraystretch{1.5}\begin{array}{c:c:c}mean& freq & Prob\\\hline9.00&1&1/5\\\hdashline9.25&1&1/5\\\hdashline9.75&1&1/5\\\hdashline10.25&1&1/5\\\hdashline10.75&1&1/5 \\\hline\end{array}


4. The mean of the sampling distribution

Xˉ=X1P(1)+X2P(2)+X3P(3)+X4P(4)+X5P(5)\bar X=X_1P(1)+X_2P(2)+X_3P(3)+X_4P(4)+X_5P(5)


Xˉ=9.00(1/5)+9.25(1/5)+9.75(1/5)+10.25(1/5)+10.75(1/5)\bar X=9.00(1/5)+9.25(1/5)+9.75(1/5)+10.25(1/5)+10.75(1/5)


Xˉ=9.8\bar X=9.8


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



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