Question #182438

Consider all samples of size 5 from this population. 

1,3,5,9,11,12,15,26

1. Compute the mean and the standard deviation of the population. 

2. List all samples of size 5 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. 

Compare this to mean of the population. 

5. Calculate the standard deviation of the sampling distribution of the 

sample means . Compare this to the standard deviation of the 

population. 


Expert's answer

The given population is-





1. Mean of population μ=∑N=828=10.25\mu=\dfrac{\sum}{N}=\dfrac{82}{8}=10.25


Variance=∑(x−μ)2N=441.58=55.1875\dfrac{\sum(x-\mu)^2}{N}=\dfrac{441.5}{8}=55.1875


Population standard deviation σ=55.1875=7.428\sigma=\sqrt{55.1875}=7.428


2.The samples are listed in the table below:

sample means are also listed in the table




3.Sampling Distribution of sample means is shown in the above table-


4.Mean of sampling distribution xˉ=679=7.44\bar{x}=\dfrac{67}{9}=7.44

Sample mean xˉ<μ\bar{ x}<\mu i.e population mean


5.standard deviation of sampling distribution of sample means-


s=variance=∑(x−xˉ)29=18.12879=2.04=1.42s=\sqrt{variance}=\sqrt{\dfrac{\sum(x-\bar{x})^2}{9}}=\sqrt{\dfrac{18.1287}{9}}=\sqrt{2.04}=1.42


sample mean is also less than the population mean.


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