Question

# A news article reports that "Americans have differing views on two potentially inconvenient and invasive practices...

A news article reports that "Americans have differing views on two potentially inconvenient and invasive practices that airports could implement to uncover potential terrorist attacks." This news piece was based on a survey conducted among a random sample of 1115 adults nationwide, interviewed by telephone November 7-10, 2010, where one of the questions on the survey was "Some airports are now using 'full-body' digital x-ray machines to electronically screen passengers in airport security lines. Do you think these new x-ray machines should or should not be used at airports?" Below is a summary of responses based on party affiliation.

 Republican Democrat Independent Total Should 258 293 345 896 Should not 37 54 75 166 Don't know/No answer 16 15 22 53 Total 311 362 442 1115

We want to test to determine if party affiliation and opinion are independent using the hypotheses:
?0:H0: Party affiliation and opinion are independent
??:HA: Party affiliation and opinion are not independent

Round all numeric answers to four decimal places.

1. What is the expected value for the number of Republicans who think x-ray machines should be used at airports?

2. Calculate the test statistic for this hypothesis test.
? z t X^2 F  =

3. Calculate the degrees of freedom for this test.

4. Calculate the p-value for this hypothesis test.

5. Based on the p-value, we have:
A. some evidence
B. strong evidence
C. extremely strong evidence
D. very strong evidence
E. little evidence
that the null model is not a good fit for our observed data.

The statistical software output for this problem is :

Contingency table results:
Rows: var6
Columns: None

Cell format
Count
(Expected count)
R D I Total
S 258
(249.92)
293
(290.9)
345
(355.19)
896
S N 37
(46.3)
54
(53.89)
75
(65.8)
166
D K 16
(14.78)
15
(17.21)
22
(21.01)
53
Total 311 362 442 1115

Chi-Square test:

Statistic DF Value P-value
Chi-square 4 4.1524681 0.3858

Expected value = 249.92

Test statistics is   = 4.1525

degrees of freedom = 4

P-value = 0.3858

Conclusion : little evidence

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