Questions: Determine the p-value in (a) and interpret its meaning. The p-value is 0.000. (Round to three decimal places as needed.) Interpret the p-value. The p-value is the probability that in a random sample of 298 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 greater than or less than the difference in this sample, assuming that the proportion of co-browsing organizations is equal to the proportion of non-co-browsing organizations that use skills-based routing to match the caller with the right agent. An earlier Z-test for the difference between two proportions in parts (b) and (c) resulted in a test statistic of ZSTAT=4.92 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. A. Both tests agree that the null hypothesis should not be rejected. B. 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. C. Both tests agree that the null hypothesis should be rejected. D. 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. 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.

Determine the p-value in (a) and interpret its meaning. The p-value is 0.000. (Round to three decimal places as needed.) Interpret the p-value. The p-value is the probability that in a random sample of 298 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 greater than or less than the difference in this sample, assuming that the proportion of co-browsing organizations is equal to the proportion of non-co-browsing organizations that use skills-based routing to match the caller with the right agent. An earlier Z-test for the difference between two proportions in parts (b) and (c) resulted in a test statistic of ZSTAT=4.92 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. A. Both tests agree that the null hypothesis should not be rejected. B. 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. C. Both tests agree that the null hypothesis should be rejected. D. 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. 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.
Transcript text: Determine the $p$-value in (a) and interpret its meaning. The $p$-value is 0.000 . (Round to three decimal places as needed.) Interpret the $p$-value. The $p$-value is the probability that in a random sample of 298 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 $\square$ the difference in this sample, assuming that the proportion of co-browsing organizations is $\square$ the proportion of non-co-browsing organizations that use skills-based routing to match the caller with the right agent. An earlier $Z$-test for the difference between two proportions in parts (b) and (c) resulted in a test statistic of $Z_{\text {STAT }}=4.92$ 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. A. Both tests agree that the null hypothesis should not be rejected. B. 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. C. Both tests agree that the null hypothesis should be rejected. D. 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. 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.
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Solution

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Solution Steps

To solve the given problem, we need to determine the $p$-value and interpret its meaning. Additionally, we need to compare the results of the $Z$-test with the results of the chi-square test.

Part (c)
  1. Determine the $p$-value: The $p$-value is given as 0.000. This indicates the probability of observing a test statistic as extreme as, or more extreme than, the observed value under the null hypothesis.
  2. Interpret the $p$-value: A $p$-value of 0.000 (rounded to three decimal places) suggests that the observed difference is highly significant, and the null hypothesis can be rejected.
Part (d)
  1. Compare the results of the $Z$-test and chi-square test: Given the $Z$-test statistic of 4.92 and critical values of -1.96 and 1.96, the $p$-value is 0.000, indicating that the null hypothesis should be rejected. We need to compare this with the results of the chi-square test.
Step 1: Determine the \( p \)-value

The \( p \)-value is given as 0.000. This indicates the probability of observing a test statistic as extreme as, or more extreme than, the observed value under the null hypothesis.

Step 2: Interpret the \( p \)-value

A \( p \)-value of 0.000 (rounded to three decimal places) suggests that the observed difference is highly significant, and the null hypothesis can be rejected.

Step 3: Compare the results of the \( Z \)-test and chi-square test

Given the \( Z \)-test statistic of \( Z_{\text{STAT}} = 4.92 \) and critical values of \(-1.96\) and \(1.96\), the \( p \)-value is 0.000, indicating that the null hypothesis should be rejected. Assuming the chi-square test also resulted in a \( p \)-value of 0.000, both tests agree that the null hypothesis should be rejected.

Final Answer

  • Interpretation of \( p \)-value: The \( p \)-value of 0.000 indicates that the observed difference is highly significant, and the null hypothesis can be rejected.
  • Comparison of \( Z \)-test and chi-square test results: Both tests agree that the null hypothesis should be rejected.

\[ \boxed{\text{Both tests agree that the null hypothesis should be rejected.}} \]

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