When public opinion surveys are conducted by mail, a cover letter explaining the purpose of the survey is usually included. To determine whether the wording of the cover letter influences the response rate, three different cover letters were used in a survey of students at a Midwestern university. Suppose that each of the three cover letters accompanied questionnaires sent to an equal number of randomly selected students. Returned questionnaires were then classified according to the type of cover letter (I, II, or III). Use the accompanying data to test the hypothesis that ?_{1} = 1/3, ?_{2} = 1/3, and ?_{3} = 1/3, where ?_{1}, ?_{2} and ?_{3} are the true proportions of all returned questionnaires accompanied by cover letters I, II, and III, respectively. Use a 0.05 significance level. (Use 2 decimal places.)
Cover-letter Type | |||
I | II | III | |
Frequency | 51 | 36 | 31 |
?^{2} =
The null and alternative hypothesis is
H0: ?1 = ?2 = ?3 = 1/3
H1: At least one proportion is not equal.
Level of significance = 0.05
Test statistic is
O: Observed frequency
E: Expected frequency.
E = n*pi
O | p | E | (O-E) | (O-E)^2 | (O-E)^2/E | |
51 | 0.333333 | 39.33333 | 11.66667 | 136.1111 | 3.460452 | |
36 | 0.333333 | 39.33333 | -3.33333 | 11.11111 | 0.282486 | |
31 | 0.333333 | 39.33333 | -8.33333 | 69.44444 | 1.765537 | |
Total | 118 | 5.51 |
Degrees of freedom = Number of E's - 1 = 3 - 1 = 2
Critical value = 5.991
Test statistic < critical value we fail to reject null hypothesis.
Conclusion:
?1 = ?2 = ?3 = 1/3
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