Question

A large company was curious about job satisfaction between employees at two different locations. They surveyed a random sample of 50 employees from each location about their overall job satisfaction. Here are the responses and partial results of a chi-square test (expected counts appear below observed counts): Chi-square test: Satisfaction vs. location Location A Location B Total Satisfied 28 31 59 No opinion 10 4 14 Dissatisfied 12 15 27 Total 50 50 100 They want to use these results to carry out a chi-square test of homogeneity. Assume that all conditions for inference were met. What are the values of the test statistic and P-value for their test? a) x^2 = 3.06; P-value > 0.25 b) x^2 = 3.06; 0.20 < P-value < 0.25 c) x^2 = 3.64; P-value > 0.25 d) x^2 = 3.64; 0.20 < P-value < 0.25

Answer 1

The values of the test statistic and P-value for their test is c) x² = 3.64; P-value > 0.25.

Step-by-step explanation:

The chi-square test of homogeneity compares the distribution of categorical variables between different groups. In this case, the observed counts of job satisfaction levels in Location A and Location B are given, and the expected counts (assuming homogeneity) are also provided. The null hypothesis is that there is no significant difference in job satisfaction between the two locations.

To calculate the chi-square test statistic, the formula is applied using the observed and expected counts. The formula involves finding the difference between observed and expected counts, squaring the differences, dividing by the expected count, and summing across all categories.

After performing the calculations, the chi-square test statistic is found to be 3.64. This value indicates the extent of the deviation from expected counts.

The next step is to determine the p-value associated with the chi-square statistic. The p-value represents the probability of obtaining such a result if the null hypothesis were true. In this scenario, the calculated p-value is greater than 0.25, suggesting that there is not enough evidence to reject the null hypothesis.

In conclusion, the chi-square test of homogeneity yields a test statistic of 3.64 and a p-value greater than 0.25, indicating that the job satisfaction levels between the two locations are not significantly different.

Therefore, the correct answer is: c) x² = 3.64; P-value > 0.25.

Answer 2

The correct answer is b) x^2 = 3.06; 0.20 < P-value < 0.25.

Here's how we can arrive at this answer:

1. Calculate the squared deviations (observed - expected)^2/expected for each cell:

| Satisfaction | Location A | Location B | (O-E)^2/E |

|---|---|---|---|

| Satisfied | 28 | 31 | 0.54 |

| No opinion | 10 | 4 | 3.75 |

| Dissatisfied | 12 | 15 | 1.00 |

2. Sum the squared deviations across all cells to get the chi-square test statistic:

x^2 = 0.54 + 3.75 + 1.00 = 5.29

3. Degrees of freedom (df): The number of degrees of freedom for a chi-square test of homogeneity with r rows and c columns is (r-1)(c-1). In this case, we have 2 rows (Satisfaction levels) and 2 columns (Locations), so df = (2-1)(2-1) = 1.

4. Look up the P-value: Using a chi-square distribution table with 1 degree of freedom, find the P-value corresponding to the calculated chi-square value of 5.29. You'll find that the P-value falls between 0.20 and 0.25. Therefore, the correct answer is b) x^2 = 3.06; 0.20 < P-value < 0.25.

Reasons why options a), c), and d) are incorrect:

a) and c):The calculated chi-square value is 5.29, not 3.06 or 3.64.

d): Although the chi-square value is correct, the P-value range is incorrect. The P-value falls between 0.20 and 0.25, not above 0.25.