6. do 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
computations.
Party Affiliation
Republican Democrat Independent
Should
264
299
351
Answer Should not
38
55
77
Don't know/No answer 16
15
22
Total
318
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 do Full body scan Part II The table below summarizes a data set we first encountered in Exercise 632 regarding views on fullbody scans and political affiliati class=

Respuesta :

Using the independence between an individual's party affiliation and his support of full-body  scans, it is found that:

a) 57 Republicans would be expected to not support the use of full-body scans.

b) 305 Democrats would be expected to support the use of full-body scans.

c) 18 Independents would be expected to not know or not answer.

The independence between an individual's party affiliation and his support of full-body  scans means that no matter the political party, the proportions for each option would be the same.

In the sample, there are 318 + 369 + 450 = 1137 people.

Item a:

  • 38 + 55 + 77 = 170 people do not support the use of full-body scans.
  • They should be equally divided between parties, hence:

[tex]\frac{170}{3} = 56.67[/tex]

Rounding, 57 Republicans would be expected to not support the use of full-body scans.

Item b:

264 + 299 + 351 = 914 people support the use of full-body scans, hence:

[tex]\frac{914}{3} = 304.67[/tex]

Rounding, 305 Democrats would be expected to support the use of full-body scans.

Item c:

16 + 15 + 22 = 53 people do not know or do not answer, hence:

[tex]\frac{53}{3} = 17.67[/tex]

Rounding, 18 Independents would be expected to not know or not answer.

A similar problem is given at https://brainly.com/question/24372153