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Question 4 View Policies Current Attempt in Progress A null hypothesis for a goodness-of-fit test and a frequency table from

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Answer #1

Total Count = 185+786+480+315+191+143 = 2100

(a) The expected count is 2100×0.36 = 756

(b) The contribution for the category labelled B is:

(Observed – Expected)2/(Expected) = (786 – 756)2/756 = 1.19047 ≈ 1.190 [Rounded to three decimal places]

(c) The degrees of freedom for the chi-square distribution for this table = 6 – 1 = 5 degrees of freedom

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