Years ago a meme went around Facebook and other parts of the internet about the
ID: 3206503 • Letter: Y
Question
Years ago a meme went around Facebook and other parts of the internet about the BBC's "Top 100 Books", with the statement that most people have read only six of the books listed. Your friend at Carleton suggests that Carleton students have read more of those books than either uOttawa students or U of T students. Determined to prove that uOttawa students are well read, you collect some data. Using appropriate sampling techniques, you poll uOttawa, Carleton, and U of T students to see how many of these books they've read. Here are the results of your poll: uOttawa students (148 total): 84 have read 0 to 9 books on the list 43 have read 10 to 19 21 have read 20 or more Carleton students (137 total): 91 have read 0 to 9 books on the list 24 have read 10 to 19 22 have read 20 or more U of T students (119 total):70 have read 0 to 9 books on the list 35 have read 10 to 19 14 have read 20 or more Put the data into a two-way table with the number of books categories in the row s and the university categories in the columns. The data in this table should be observed counts. Create another table showing the corresponding expected counts. Show no more than three decimal places in the second table and make sure your two tables show the totals for the rows, columns, and overall. Perform a hypothesis test to check if the distributions for number of books read are the same across the three universities at the 0.01 significance level. That is, test the independence of the two categorical variables, number of books read and university attended. Is the chi-squared approach appropriate here? Why or why not?Explanation / Answer
(a)
(b)
level of significance 0.01
null hypothesis H0 :universities and students are independent
degree of freedom = (c-1)(r-1)
=(3-1)(3-1)
=4
tabulated chi square at 4 df and 0.01 level of significance=13.277 (from chi square table)
conclusion: since calculated chi square 6.92 is less than tabulated chi sqaure. we accept null hypothesis.
(c) here chi square approach is appropriate since we can find expected frequencies here
0-9 10-19 20 & more total ottava 84 43 21 148 carlton 91 24 22 137 uoft 70 35 14 119 total 245 102 57 grand total =404Related Questions
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