Co-browsing refers to the ability to have a contact center agent and customer jointly navigate an application on a real time basis through the web. A study of businesses indicates that 82 of 128 co-browsing organizations use skills-based routing to match the caller with the right agent, whereas 66 of 173 non-co-browsing organizations use skills-based routing to match the caller with the right agent. Complete parts (a) through (c) below. a. Construct a 2 x2 contingency table. Company Uses Skill-based routing Co-browsing Non-co-browsing Yes No Total b. At the 0.025 level of significance, is there evidence of a difference between co-browsing organizations and non-co-browsing organizations in the proportion that use skills-based routing to match the caller with the right agent? Let it, represent the proportion of co-browsing organizations, and let it2 represent the proportion of non-co-browsing organizations. What are the null and alternative hypotheses to test? Ho : 1 1 2 H 1 : 51 7 2 Calculate the test statistic. X STAT =(Round to three decimal places as needed.) What is the critical value for a = 0.025? X0.025 = (Round to three decimal places as needed.) State the conclusion. Since the test statistic is the critical value, there is evidence to conclude there is a significant difference between the proportions.c. Determine the p-value in (a) and interpret its meaning. The p-value is . (Round to three decimal places as needed.) Interpret the p-value. The p-value is the probability that in a random sample of 301 organizations, the difference between the proportion of co-browsing organizations and non-co-browsing organizations that use skills-based routing to match the caller with the right agent will be the difference in this sample, assuming that the proportion of co-browsing organizations is the proportion of non-co-browsing organizations that use skills-based routing to match the caller with the right agent. d. An earlier Z-test for the difference between two proportions in parts (b) and (c) resulted in a test statistic of ZSTAT = 4.45 against critical values of - 1.96 and 1.96, with a p-value of 0.000. Compare the results of (b) and (c) to the results of the Z-test. O A. The chi-square test concluded that the null hypothesis should be rejected, while the Z-test concluded that the null hypothesis should not be rejected. O B. Both tests agree that the null hypothesis should not be rejected. O C. Both tests agree that the null hypothesis should be rejected. O D. The chi-square test concluded that the null hypothesis should not be rejected, while the Z-test concluded that the null hypothesis should be rejected. O E. The tests cannot be compared since the Z-test was a one-tail test, while the chi-square test was a two-tail test