Question

Market researchers wonder if people in different countries have different preferred sports to watch during the winter Olympics. They take a random sample of people from each country and survey them about their favorite sport to watch. Here are the responses and partial results of a chi-square test (expected counts appear below observed counts): Chi-square test: Country vs. favorite winter Olympics sport China India Total Bobsleigh 8 10 18 9 9 Curling 28 61 89 44.5 44.5 Hockey 39 29 68 34 34 Speed skating 75 50 125 62.5 62.5 Total 150 150 300 They want to use these results to carry out a X2 test for homogeneity. Assume that all conditions for inference were met. What are the values of the test statistic and P-value for their test? Choose 1 answer: A x² = 6.309; 0.05 < P-value < 0.10 B x2 = 6.309; 0.15 < P-value < 0.20 x2 = 18.928; P-value < 0.0005 x2 = 18.928; 0.0005 < P-value < 0.001 D

Answer 1

The test statistic is 18.928 and the P-value is less than 0.0005.

Step-by-step explanation:

To test if people in different countries have different preferred sports to watch during the winter Olympics, the market researchers conducted a chi-square test of homogeneity. The test statistic for the chi-square test of homogeneity is calculated by summing up the squared differences between the observed and expected values, divided by the expected values. In this case, the test statistic is 18.928. The P-value is the probability of observing a test statistic as extreme as the one calculated, assuming the null hypothesis is true. In this case, the P-value is less than 0.0005.

Answer 2

The values of the test statistic and P-value for the chi-square test of homogeneity are 18.928 and P-value < 0.0005, respectively.

Step-by-step explanation:

To test for homogeneity, we use the chi-square test. The chi-square test compares the observed values with the expected values if the two populations followed the same distribution. The test statistic for the chi-square test of homogeneity is calculated as the sum of the squared differences between the observed and expected values divided by the expected values. In this case, the test statistic is 18.928.

The degrees of freedom for a chi-square test of homogeneity are calculated as (number of rows - 1) multiplied by (number of columns - 1). Here, the degrees of freedom is (2 - 1) * (4 - 1) = 3.

The p-value for the chi-square test of homogeneity is the probability of observing a test statistic as extreme as the one calculated, assuming that the null hypothesis is true. The p-value for this test is less than 0.0005. Therefore, we can conclude that there is strong evidence to suggest that there are differences in preferred sports to watch during the winter Olympics between different countries.