Please help calculate this. I don\'t even know where to start! ume a population

Question

Please help calculate this. I don't even know where to start! ume a population of3, 4, and 11 Assume that samples of szen 2 are randomly seleded replacement form the population. 4,4 4,11 11,4 11,11 Find te value of the population standard deviation Round to three decimal places as needed) h Fed tandard deviaton each ot the nine samples, then summarize the samping d ofte standard inthe format o atable representing the probability distribution distinct standard deviaton Use asoending order of the sample standard deviations integers or tractors e. Find be mean of samping distribution of the sample standard deviaions. The mean Desameling dartution e samele standard deviesions (Reund to tree dedma plates as needed d Dohe sample landard owiationstargr he value ofeepopuaton stardard devasorn n general, do sample devietons make good atmalars of opulaton elandad deviatiana Why or why nor? standard

Explanation / Answer

a. The given data set comprises three numbers 3, 4 and 11. Use the following formula to compute the population standard deviation.

Sigma=sqrt[1/N sigma (X-mu)^2], where, N is population size, X corresponds to each data, and mu is population mean. TObtain the population mean, mu =sigma x/N=18/3=6

=sqrt[1/3 {(3-6)^2+(4-6)^2+(11-6)^2}]

=3.56

b. To obtain the standard deviation of each of th enine samples, use, the following formula.

s=sqrt[1/n-1 sigma (x-xbar)^2], where, n is the sample size, x is each value of the sample data, xbar is sample mean.

For first data (3,3), xbar=3, s1=sqrt[1/2-1 {(3-3)^2+(3-3)^2}]=0

Similarly, for rest of the set of data, the standard deviations are as follows:

0.7071, 5.6569, 0.7071, 0, 4.9497, 5.6569, 4.9497, 0.

The table of sample standard deviation (in ascending order) and corresponding probabilities are shown below. Note, there are three 0's. Thus, probability distribution of 0 will be 3/9. Similarly, find the probability distribution for rest of the sample standard deviations.

0 0.3333

0.7071 0.2222

4.9497 0.2222

5.6569 0.2222

c. The mean of the sampling distribution of sample standard deviation is as follows:

3*0.3333+2*0.2222+2*0.2222+2*0.2222=2.333

d. Since, sample standard deviation target the population standard deviation, therefore, sample standard deviations are unbiased estimator. Option D.

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