Questions: Question 1, 11.2.5 Part 2 of 4 Points: 0 of 1 Save annual meeting of the American Society of Tropical Medicine, by principal investigator Steve Lindsay). The accompanying information shows the hypotheses and TI-83/84 Plus calculator results for the experiment. Assume that the hypothesis test requirements are all satisfied. Complete parts (a) through (c) below. Malaria Was Present Malaria Was Not Present Dog Was Correct 123 131 Dog Was Wrong 52 14 a. Identify the test statistic and the P-value, and then state the conclusion about the null hypothesis. The test statistic is 19.49 . (Round to two decimal places as needed.) Identify the P-value. The P-value is (Round to three decimal places as needed.) Just output the content of the question, DO NOT output additional information or explanations.

Question 1, 11.2.5
Part 2 of 4
Points: 0 of 1
Save annual meeting of the American Society of Tropical Medicine, by principal investigator Steve Lindsay). The accompanying information shows the hypotheses and TI-83/84 Plus calculator results for the experiment. Assume that the hypothesis test requirements are all satisfied. Complete parts (a) through (c) below.

Malaria Was Present Malaria Was Not Present
Dog Was Correct 123 131
Dog Was Wrong 52 14

a. Identify the test statistic and the P-value, and then state the conclusion about the null hypothesis.

The test statistic is 19.49 .
(Round to two decimal places as needed.)
Identify the P-value.
The P-value is
(Round to three decimal places as needed.)
Just output the content of the question, DO NOT output additional information or explanations.

Transcript text: Question 1, 11.2.5 Part 2 of 4 Points: 0 of 1 Save annual meeting of the American Society of Tropical Medicine, by principal investigator Steve Lindsay). The accompanying information shows the hypotheses and TI-83/84 Plus calculator results for the experiment. Assume that the hypothesis test requirements are all satisfied. Complete parts (a) through (c) below. \begin{tabular}{l|c|c} \hline & Malaria Was Present & Malaria Was Not Present \\ \hline Dog Was Correct & 123 & 131 \\ \hline Dog Was Wrong & 52 & 14 \\ \hline \end{tabular} a. Identify the test statistic and the P-value, and then state the conclusion about the null hypothesis. The test statistic is 19.49 . (Round to two decimal places as needed.) Identify the P-value. The $P$-value is $\square$ (Round to three decimal places as needed.)

Solution

Solution Steps

Step 1: Calculate Expected Frequencies

Expected frequencies are calculated as $(\text{row total} * \text{column total}) / \text{grand total}$ for each cell, resulting in: [[138.906 115.094] [ 36.094 29.906]]

Step 2: Compute Test Statistic

Using the chi-square formula $\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}$, we find $\chi^2 = 19.49$

Step 3: Determine Degrees of Freedom (df)

Degrees of freedom (df) are calculated as $(\text{number of rows} - 1) * (\text{number of columns} - 1)$, resulting in: 1

Step 4: Find P-value

Using the chi-square distribution with df=1, the P-value corresponding to $\chi^2 = 19.49$ is 0

Step 5: Make Decision

Comparing the P-value to the significance level (0.05), we reject the null hypothesis of independence.