**Chi-Square Test for Independence: Customer Service Analysis**The table presents cross-tabulation data showing how many people rated a customer service representative as friendly and polite, based on whether the representative greeted them first. The analysis involves the following steps:### Data Table**Friendliness/Politeness**| Staff Greeting | No | Yes | Total ||----------------|-----|-----|-------|| **No** | 15 | 4 | 19 || **Yes** | 14 | 20 | 34 || **Total** | 29 | 24 | 53 |### Instructions**a. Hypotheses for Chi-Square Test**- \( H_0 \): The greeting by staff and customer rating of friendliness/politeness are independent.- \( H_1 \): The greeting by staff and customer rating of friendliness/politeness are not independent.**b. Expected Frequencies**Calculate expected frequencies for each cell, rounding to two decimal places if needed.| Friendliness/Politeness | No | Yes | Grand Total ||-------------------------|--------|--------|-------------|| Staff Greeting **No** | (Fill) | (Fill) | 19 || Staff Greeting **Yes** | (Fill) | (Fill) | 34 || Grand Total | 29 | 24 | 53 |**c. Compute the Chi-Square Statistic**\[\chi^2 = \text{(calculation here)}\]### d. Conclusion Using a Level of Significance of 0.05- **Critical Value**: (Insert value) - **p-value**: (Insert value) - Determine if the test statistic is greater than or less than critical value. - Compare p-value to 0.05 significance level to decide on rejecting \( H_0 \).### Final DecisionBased on the chi-square test, state whether to reject or fail to reject the null hypothesis and conclude if greeting affects customer service rating.

There are two categorical variables which are Staff Greeting and Friendliness/Politeness. Staff Greeting has 2 types and Friendliness/Politeness has 2 types. We have to test whether the Greeting and Customer Service Rating are independent or not. This is Chi-Square test of independence.The null and alternative hypotheses are H0: Greeting and Customer Service Rating are independent. H1: Greeting and Customer Service Rating are not independent. No Yes No Observed 15 4 Expected 10.40 8.60 O - E 4.60 -4.60 (O - E)² / E 2.04 2.46 Yes Observed 14 20 Expected 18.60 15.40 O - E -4.60 4.60 (O - E)² / E 1.14 1.38 Expected values: Staff Greeting No Yes Grand Total No 10.40 8.60 19 Yes 18.60 15.40 34 Grand Total 29 24 55The value of the test statistic is 7.02.From Chi-Square distribution table, at 1 d.f and 0.05 level, the chi-square critical value is 3.841. EXCEL command is “= CHIDIST (x, deg_freedom)” For p-value, enter in EXCEL as “= CHIDIST (7.02, 1)” P-value = 0.008. Conclusion: The test statistic greater than the critical value, and the p-value is less than the chosen level of significance, 0.05. Therefore, reject the null hypothesis and conclude that Greeting and Customer Service Rating are independent.