Questions: Each year, Kerri adds to her book collection a number of new books. She has categorized each of her 13 new books as hardcover, paperback, fiction, or nonfiction. The information is displayed in the table below. Fiction (F) Nonfiction (N) Totals ------------ Hardcover (H) 2 4 6 Paperback (P) 1 6 7 Totals 3 10 13 If she randomly chooses one of these 13 books, find the probability that it will be NONFICTION given it is PAPERBACK. 6 7

Solution Steps

The total number of paperback books is calculated as follows: \[ \text{Total Paperback} = 1 + 6 = 7 \]

The number of nonfiction paperback books is given directly from the table: \[ \text{Nonfiction Paperback} = 6 \]

The probability of selecting a nonfiction book given that it is a paperback is calculated using the formula: \[ P(\text{Nonfiction} | \text{Paperback}) = \frac{P(\text{Nonfiction and Paperback})}{P(\text{Paperback})} = \frac{6}{7} \]

The expected frequencies for each cell in the contingency table are calculated as follows:

• For cell (1, 1): \[ E_{1,1} = \frac{R_1 \times C_1}{N} = \frac{6 \times 3}{13} = 1.3846 \]

• For cell (1, 2): \[ E_{1,2} = \frac{R_1 \times C_2}{N} = \frac{6 \times 10}{13} = 4.6154 \]

• For cell (2, 1): \[ E_{2,1} = \frac{R_2 \times C_1}{N} = \frac{7 \times 3}{13} = 1.6154 \]

• For cell (2, 2): \[ E_{2,2} = \frac{R_2 \times C_2}{N} = \frac{7 \times 10}{13} = 5.3846 \]

The expected frequencies are: \[ \begin{bmatrix} 1.3846 & 4.6154 \\ 1.6154 & 5.3846 \end{bmatrix} \]

The Chi-Square test statistic (\(\chi^2\)) is calculated for each cell:

• For cell (1, 1): \[ \frac{(O - E)^2}{E} = \frac{(2 - 1.3846)^2}{1.3846} = 0.2735 \]

• For cell (1, 2): \[ \frac{(O - E)^2}{E} = \frac{(4 - 4.6154)^2}{4.6154} = 0.0821 \]

• For cell (2, 1): \[ \frac{(O - E)^2}{E} = \frac{(1 - 1.6154)^2}{1.6154} = 0.2344 \]

• For cell (2, 2): \[ \frac{(O - E)^2}{E} = \frac{(6 - 5.3846)^2}{5.3846} = 0.0703 \]

The total Chi-Square statistic is: \[ \chi^2 = 0.2735 + 0.0821 + 0.2344 + 0.0703 = 0.0232 \]

The critical value at \(\alpha = 0.05\) for a Chi-Square distribution with 1 degree of freedom is: \[ \chi^2_{\alpha, df} = 3.8415 \]

The p-value associated with the Chi-Square statistic is: \[ P = P(\chi^2 > 0.0232) = 0.8789 \]

Final Answer