Question 1116576
Ho: The proportions are 0.2, 0.1, 0.1, 0.2, 0.4
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H1: At least one of the proportions in Ho: is false
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Degrees of Freedom(DF) = 5 - 1 = 4
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The number of observations in the sample is 14 +7 +44 + 11 +14 = 90
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The expected frequency counts are
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E(1) = 0.2 * 90 = 18
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E(2) = 0.1 * 90 = 9
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E(3) = 0.1 * 90 = 9
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E(4) = 0.2 * 90 = 18
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E(5) = 0.4 * 90 = 36
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chi^2 test statistic = summation i = 1, 5 of [((O(i) - E(i))^2/E(i)] where O(i) is the observed frequency count
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chi^2(X) = ((14-18)^2/18) +((7-9)^2/9) ((44-9)^2/9) +((11-18)^2/18) +((14-36)^2/36) =  77.55
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Use a Chi-Square Distribution Calculator to find P(X > 77.55) = 0
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Since the P-value (0) is less than the significance level (0.10), we cannot accept the null hypothesis.
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