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.fn thETIlF-of Uontana straff Ue focated tn the City-of Bllllngs, County ...

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I. Probabilitycalculations using MINITAB.
Background and ButtonPushing Recallthat thecumulative distribution function, or cdf , for a random variable X is
 definedto beF( x ) = P ( X#x ) .There are two common sorts of calculations one
 makes: given avalue xone findsthe correspondingvalue ofF ; or , given a particular
 valueof F, that is, given a particularprobability , one findsthe corresponding value of x.
 ( Interms of functions, in the latter case one is evaluating theinversecumulative distribution
 function. )
 Forthe case of a standard normal random variable, there is a MINITAB function that provides
 analternative to using tables to do calculations. To get to this function one uses the tab
 CALC( and in the two successive dropdown boxes ) PROBABILITY DISTRIBUTIONS ,  andNORMAL. (A mean of zero and a standard deviation of one are the default settings
 forNORMAL ) .Then one chooses‘cumulative probability’ or ‘inverse
 cumulativeprobability’ as appropriate.
Required Calculations
 Headcolumn one in the MINITAB worksheetPROB ;head column three in the  MINITABworksheet ZVALUE. Incolumn one enter the values 0.02 , 0.20
 0.40,0.50, 0.60 , 0.80 and 0.98 . In column three enter the values3.5 , 2.5 ,
 1.5,0.5 , 1.5 , 2.5 and 3.5 .Use MINITAB to find the ‘zvalues’ corresponding
 tothe probabilities in column one, and store the results in column two. Use MINITAB
 tofind the probabilities corresponding to ‘zvalues’ in column three, and store the
 resultsin column four.(Friendly advice: I’d use a good oldfashionedtableto  checkmy work ifI had to do this exercise ! )
The T  Random Variable
 Onemay do the same sorts of calculations for aT random variable as for a standard
 normalrandom variable.One uses the tabs CALC( and in the two successive
 dropdownboxes ) PROBABILITY DISTRIBUTIONS , and t.Notice that , of
 course,in this case one must specify the number of degrees of freedom.
 Thisis a handy function : typical ttables don’t tabulate many values, which makes
 somecalculations using ttables awkward.
II .Calculating a Confidence Interval
 Backgroundand Review
 Recallthat ifX is a normal random variable with mean:, and with known standard deviation,Fis the sample mean based on a simple random sample of size n,, and if then a twosided confidence interval for:of size1 " is:
 Recallwhat this means : the random interval
 covers:1  with probability".
 Recallalso that when one says that the standard deviation is ‘known’ this usually
 justmeans that one has a large sample size .For the same set of hypotheses, but
 forFone has a small sample size, so one can’t safely assume unknown,i.e. when
 thats is close toFa twosided confidence interval for, then:of size1 " is:
The height of male USundergraduates is normally distributed.A random
 sampleof 10 male undergraduates produced the following measurements ( in inches ):
 73.25, 69.5 , 69.5 , 68 , 68, 70.5 , 68 , 69 , 71 , 68.
 Calculatea 96% twosided confidence interval for the mean height ( in inches ) of
 USmale undergraduates.
Remarks: I have a neighbor who gets into her car every morning and backs
 herautomobile out to her mailbox to pickup the newspaper from the paper box, and
 thendrives back into the garage with the paper .....Puzzling, since she appears in
 quiteadequate physical condition to make the arduous roundtrip odyssey of
 about, oh , maybe, .... 60 feet !!!Please don’t approach this problem with
 hermentality :don’t try to get MINITAB to do the entire problem for you !!!
 Rather,use MINITAB to do the nasty bits , — computing the sample mean, the
 samplestandard deviation, and calculating the necessary tvalue, — andthen do the
 remaining10 cents worth of calculation yourself .
III. Understandingwhat a Confidence Interval Means WhetherFis known or unknown, the meaning of a confidence interval is still the  same.To say one has , for instance, a 96% confidence interval for:means that
 hascomputed one particular member of a family of random intervals that cover
:with probability 0.96.The following computer exercise illustrates this idea.
 Firstgenerate 300 simple random samples of size 10 tencorresponding to observations
 ofa random variable with mean 70 and standarddeviation of 2.4 ( i.e. these are simulated
 observationsof male undergraduateheight in inches ) .Use the tabs CALC, RANDOM
 DATA,NORMAL .Enter c1c2 c3 c4 c5 c6 c7 c8 c9 c10in the ‘store columns’
 box, andremember to specify the mean to be 70 and the standard deviation to be 2.4
 Headcolumn 11xbar andcolumn 12s .In column 11 store the
 samplemean of each sample of size 10by using CALC, ROW STATISTICS , etc.
 similarly,in column 12 store the standard deviation of each sample of size 10.
 Incolumn 13 store thelowerendpoint of a 96% twosided confidence interval for
: based on each sample of size 10 , which is
, using the
 CALCULATORtab .... ( note you previously calculated the necessary t value ) .
 Incolumn 14 store theupper96% twosided confidence interval forendpoint of a
:based on each sample of size 10 , which is
. (You may
 wantto head columns 13 and 14 ‘lower’ and ‘upper’ , or some such titles...)
 Inthis artificial, but hopefully instructive exercise, weknow the value of the
 populationmean : it’s 70 , because we chose it that way !So, in the longrun
 96%of all confidence intervals for:A set of, will cover the value 70.
 300random samples isn’t as good an approximation to the ‘longrun’ as
 3,000,000would be – but, nevertheless, the proportion of these300 intervals
 thatcontain 70 shouldn’t be too far from 0.96.Compute it ....!!!!
 Atechnical note :checking whethera number x is in an interval [ a , b ] is
 thesame as checking whether ( x  a )( x  b ) is negative.... ( To see this
 sketcha graph of y = ( x  a )( x  b ) ...)There is a function in the MINITAB
 calculatorfunction menu called‘Signs’ that returns 1 if the argument of the function
 isnegative, and 1 if the argument of the function is positive.Using these ideas, one
 canbuild an indicator variable , maybe headed‘sign’ ? , in column 15 that is 1 if the
 intervalfor the corresponding simple random sample contains 70 , and 1 ifthe interval
 doesn’tcontain 70.
 Thenone can count the number of intervals which contain 70 by either doing a
 sort, or else by forming a subset, if one prefers that approach.
 (By way of example, formy data , and , of course your data will probably be different ,
 287of the300 intervals contained the population mean, 70 , for a sample proportion of
 287/300or about 95.7% ....)