Question 1080747
Suppose that a population consists of the following families living on Melbourne Crescent.  The number of children in each of these 6 families is shown in the table below.      10 MARKS

Name of Family Living on Melbourne Crescent	Number of children
Lyu	                                                1
Sylvestri	                                        2
Ali	                                                3
Johnson                                            	4
Mensah	                                                5
Mesic	                                                4

a.	  If all possible samples of size 2 are selected, how many different samples are possible? 6C2 = (6*5)/(1*2) = 15
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b.	  List all possible samples of size 2 and compute the mean of each sample. (2 marks)
Example of what to do to all 15 2-element samples::
Sample:: (1,3)---- mean = 2
Sample:: (1,4)---- mean = 2.5
Etc
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c.	  Organize the sample means into a sampling distribution of the sample mean. (2 marks)
List the 15 means you come up with.
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d.	  The mean number of children in the population equals 
Find the mean of the set of 15 means (add them and divide by 15)
_______________.  
e.	  The mean of the distribution of sample means is equal to _____________.
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f.	  Graph the relative frequency distribution for the population and for the sample means.  (2 marks)
List the 15 means from lowest to highest.
List the # of times you have a given mean value (This should add up to 15)
List these frequencies from lowest to highest.
Convert each of the frequencies to a relative frequence by dividing by 15
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g.	  Compare the spread of the sampling distribution with the spread of the population. Which distribution has more variability?
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The spread of the population is greater than
the spread of the sampling distribution. 
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Cheers,
Stan H.
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