Television networks frequently run public opinion polls on issues of concern. One network conducted a scientific poll asking a question concerning the approval rating of the way President Obama was handling the Iraq War. At about the same time a second network ran an online poll using a very similar question. The results of the two polls are summarized in the following table. Poll Scientific Online Approve 339 385 Disapprove 780 573 Total 1119 958 We would like to test to see if the two polls are consistent with respect to the proportion who approve of President Obama's handling of the war If the x^2 test is used to test the null hypothesis, the expected cell count in the online poll for those who approve and that cell's contribution to the value of the test statistic are, respectively, Group of answer choices a. 390.1 and 7.70. b. 333.9 and 6.78. c. 390.1 and 6.69. d. 333.9 and 7.81. e. 362.0 and 1.46.

To determine if the two polls are consistent with respect to the proportion who approve of President Obama's handling of the war, we can use a chi-square test of independence. To perform the test, we first need to calculate the expected cell counts for each cell in the contingency table. The formula for calculating the expected count for any cell is:

Expected count for cell = (Row total * Column total) / Grand total

First, let's calculate the expected cell count for those who approve in the online poll.

Total number of people who approve (across both polls) = 339 + 385 = 724 Total responses in the online poll = 958 Grand total responses (across both polls) = 1119 + 958 = 2077

Expected count for online poll approvers = (724 * 958) / 2077

Next, we will calculate this expected count. Using the numbers given:

Expected count for online poll approvers = (724 * 958) / 2077 ≈ 333.9 (rounded to one decimal place)

Once the expected count is calculated, we can compute the contribution of this cell to the chi-square test statistic using the following formula:

Contribution to chi-square = (Observed count - Expected count)^2 / Expected count

Observed count in the online poll for those who approve is 385. Using the expected count we just calculated:

Contribution to chi-square = (385 - 333.9)^2 / 333.9 Contribution to chi-square ≈ (51.1)^2 / 333.9 ≈ 2601.21 / 333.9 ≈ 7.81 (rounded to two decimal places)

The correct answer is therefore option d: 333.9 and 7.81.

Extra: The chi-square test is commonly used in statistics to test the independence of two categorical variables. In this scenario, we are examining whether the proportion of approval ratings in two different polls is independent of the method of polling (scientific vs. online). The null hypothesis for the chi-square test of independence states that there is no association between the categorical variables—in this case, that the proportions of approval are the same regardless of polling method.

The chi-square test statistic is calculated as the sum of the squared differences between the observed and the expected counts, divided by the expected counts, summed across all cells in the table. A high chi-square statistic relative to the degrees of freedom and corresponding p-value suggests that the observed and expected counts are sufficiently different to reject the null hypothesis, while a low chi-square value suggests that we do not have enough evidence to reject the null hypothesis.

Remember that the chi-square test requires all expected cell counts to be greater than 5 to ensure enough sample size for the approximations to follow the chi-square distribution. If any expected count is less than 5, a different statistical test might be necessary, or one might need to pool some categories to meet this condition.