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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