The problem asks us to determine if there is a statistically significant difference in promotion rates between men and women based on the provided data. We are given a contingency table showing the number of men and women who were promoted in the last two years (Yes) and those who were not (No). We are to assess this at the normal level of significance, which commonly refers to $\alpha = 0.05$. We will perform a Chi-Square test for independence.

1. Problem Description

The problem asks us to determine if there is a statistically significant difference in promotion rates between men and women based on the provided data. We are given a contingency table showing the number of men and women who were promoted in the last two years (Yes) and those who were not (No). We are to assess this at the normal level of significance, which commonly refers to $\alpha = 0.05$. We will perform a Chi-Square test for independence.

2. Solution Steps

Step 1: State the Null and Alternative Hypotheses.

Null Hypothesis ($H_0$): Gender and promotion rate are independent.

Alternative Hypothesis ($H_1$): Gender and promotion rate are dependent.

Step 2: Construct the contingency table.

The contingency table is given as:

| | Promoted (Yes) | Not Promoted (No) |

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

| Women | 8 | 5 |

| Men | 10 | 7 |

Step 3: Calculate the expected frequencies under the null hypothesis.

First, calculate the row and column totals.

Total Women = 8 + 5 = 13

Total Men = 10 + 7 = 17

Total Promoted = 8 + 10 = 18

Total Not Promoted = 5 + 7 = 12

Grand Total = 13 + 17 = 18 + 12 = 30

The expected frequency for each cell is calculated as:

$E_{ij} = \frac{(\text{Row Total}) \times (\text{Column Total})}{\text{Grand Total}}$

$E_{11}$ (Women, Promoted) = $\frac{13 \times 18}{30} = \frac{234}{30} = 7.8$

$E_{12}$ (Women, Not Promoted) = $\frac{13 \times 12}{30} = \frac{156}{30} = 5.2$

$E_{21}$ (Men, Promoted) = $\frac{17 \times 18}{30} = \frac{306}{30} = 10.2$

$E_{22}$ (Men, Not Promoted) = $\frac{17 \times 12}{30} = \frac{204}{30} = 6.8$

Step 4: Calculate the Chi-Square test statistic.

The Chi-Square test statistic is calculated as:

$\chi^2 = \sum \frac{(O_{ij} - E_{ij})^2}{E_{ij}}$

where $O_{ij}$ is the observed frequency and $E_{ij}$ is the expected frequency for each cell.

$\chi^2 = \frac{(8-7.8)^2}{7.8} + \frac{(5-5.2)^2}{5.2} + \frac{(10-10.2)^2}{10.2} + \frac{(7-6.8)^2}{6.8}$

$\chi^2 = \frac{(0.2)^2}{7.8} + \frac{(-0.2)^2}{5.2} + \frac{(-0.2)^2}{10.2} + \frac{(0.2)^2}{6.8}$

$\chi^2 = \frac{0.04}{7.8} + \frac{0.04}{5.2} + \frac{0.04}{10.2} + \frac{0.04}{6.8}$

$\chi^2 \approx 0.0051 + 0.0077 + 0.0039 + 0.0059$

$\chi^2 \approx 0.0226$

Step 5: Determine the degrees of freedom.

The degrees of freedom (df) for a contingency table are calculated as:

$df = (r-1)(c-1)$

where r is the number of rows and c is the number of columns.

In this case, $r = 2$ and $c = 2$, so $df = (2-1)(2-1) = 1 \times 1 = 1$

Step 6: Find the critical value.

Using a Chi-Square distribution table with df = 1 and $\alpha = 0.05$, the critical value is 3.

8

4

1.

Step 7: Compare the test statistic to the critical value.

Our calculated $\chi^2$ value is 0.

0

2

2

6. The critical value is 3.

8

4

1. Since 0.0226 < 3.841, we fail to reject the null hypothesis.

Step 8: Conclusion.

Since we fail to reject the null hypothesis, we conclude that there is not enough evidence to suggest that gender and promotion rate are dependent. In other words, we do not have sufficient evidence to say that there is a statistically significant difference in promotion rates between men and women at the 0.05 significance level.

3. Final Answer

There is not a statistically significant difference in promotion rates between men and women.