(Show all Work 6.46 Full body scan, Part II. The table below summarizes a data s
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6.46 Full body scan, Part II. The table below summarizes a data set we first encountered in Exercise 6.32 regarding views on full-body scans and political affiliation. The differences in each political group may be due to chance. Complete the following computations under the null hypothesis of independence between an individual's party affiliation and his support of full-body scans. It may be useful to first add on an extra column for row totals before proceeding with the Party Affliation Republican Democrat Independent Should 264 38 16 318 299 351 Answer Should not Don't know/No answer Total 15 369 450 (a) How many Republicans would you expect to not support the use of full-body scans? (b) How many Democrats would you expect to support the use of full-body scans? (c) How many Independents would you expect to not know or not answer? 6.47 Offshore drilling. Part TIL Th tblalaExplanation / Answer
party affiliation
republican
democrat
independent
total
should
264
299
351
914
should not
38
55
77
170
don't know
16
15
22
53
total
318
369
450
1137
Expected frequency is given by:
E = Tr * Tc / T
Where Tr is the total of row of particular class
And Tc is the total of column of particular class
And T is the grand total
a)Republicans expected to not support the use of full-body scans= 170* 318/1137 ~ 48
b)Democrats expected to support the use of full-body scans = 914*369/1137 ~ 297
c) Independents expected to not know or not answer = 53*450/1137 ~ 21
party affiliation
republican
democrat
independent
total
should
264
299
351
914
should not
38
55
77
170
don't know
16
15
22
53
total
318
369
450
1137
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