hw05_solution - 15.7 Denote p i as the proportion of having...

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Unformatted text preview: 15.7 Denote p i as the proportion of having outcome i (i = 1 to 5) H : p 1 = p 2 = p 3 = p 4 = p 5 (or p 1 = 0.2, p 2 = 0.2, p 3 = 0.2, p 4 = 0.2, and p 5 = 0.2) H 1 : At least one proportion is not equal to its specified value Rejection rule: reject H if p-value < 0.05 (for α = 0.05) Since each expected frequency >= 5, the “rule of five” for using the chi-square approximation is satisfied. Using Excel function, we get p-value = CHIDIST(4.2, 4) = 0.3796 Since p-value = 0.3796 > 0.05, we fail to reject the null hypothesis. Thus, we conclude with 95% confidence that there is insufficient evidence to infer the proportions of the five outcomes differ. 15.11 Denote p i as the proportion of having choice j as the correct answer, (i, j) = (1, a), (2, b), (3, c), (4, d), or (5, e). H : p 1 = p 2 = p 3 = p 4 = p 5 (or p 1 = 0.2, p 2 = 0.2, p 3 = 0.2, p 4 = 0.2, and p 5 = 0.2) H 1 : At least one proportion is not equal to its specified value Rejection rule: reject H if p-value < 0.05 (for α = 0.05) Since each expected frequency >= 5, the “rule of five” for using the chi-square approximation is satisfied. Since p-value = 0.1712 > 0.05, we fail to reject the null hypothesis. Thus, we conclude with 95% confidence that there is insufficient evidence to infer the professor does not randomly distribute the correct answer over the five choices. 15.23 The two variables are “shirt condition” and “daily shift”....
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This note was uploaded on 01/06/2012 for the course MANAGEMENT 000 taught by Professor 游啟璋 during the Fall '11 term at National Taiwan University.

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hw05_solution - 15.7 Denote p i as the proportion of having...

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