To calculate the p-value, we will conduct a hypothesis test to determine if there is an association between age groups and promotion likelihood. This requires a chi-square test for independence.

Here's the plan:

1. Set up the hypotheses:

   • Null hypothesis ($H_0$): There is no association between age group and promotion likelihood.
   • Alternative hypothesis ($H_1$): There is an association between age group and promotion likelihood.

2. Create a contingency table and expected frequencies:

   • Observed values are already provided.
   • Calculate the expected values using the formula: $E_{ij} = \frac{(R_i)(C_j)}{N}$ where $R_i$ is the row total, $C_j$ is the column total, and $N$ is the grand total.

3. Calculate the test statistic:

   • Use the chi-square formula: $\chi^2 = \sum \frac{(O_{ij} - E_{ij})^2}{E_{ij}}$

4. Find the p-value:

   • Degrees of freedom = $(\text{Number of rows} - 1) \times (\text{Number of columns} - 1)$.
   • Use a chi-square distribution to find the p-value.

Let me calculate the expected values, the test statistic, and the p-value. numpy as np from scipy.stats import chi2

Observed data (from the table in the image)

observed = np.array([ [11, 23, 23, 6], # Promoted [39, 29, 47, 37] # Not Promoted ])

Total rows, columns, and grand total

row_totals = observed.sum(axis=1) col_totals = observed.sum(axis=0) grand_total = observed.sum()

Calculate expected frequencies

expected = np.outer(row_totals, col_totals) / grand_total

Chi-square test statistic calculation

chi_square_stat = ((observed - expected) ** 2 / expected).sum()

Degrees of freedom

df = (observed.shape[0] - 1) * (observed.shape[1] - 1)

Calculate p-value using the chi-square distribution

p_value = 1 - chi2.cdf(chi_square_stat, df)

chi_square_stat, df, p_valueThe results of the chi-square test are as follows:

• Chi-square statistic ($\chi^2$): 12.198
• Degrees of freedom (df): 3
• P-value: 0.00673

Interpretation:

Since the p-value (0.00673) is less than the significance level ($\alpha = 0.01$), we reject the null hypothesis. This indicates that there is a statistically significant association between age group and promotion likelihood.

Would you like a detailed breakdown of the expected values or any clarification?

Related Questions:

1. How is the chi-square statistic calculated step-by-step?
2. What are the expected frequencies, and how do they compare to the observed frequencies?
3. What does the degrees of freedom represent in this context?
4. How does the p-value change with larger sample sizes?
5. Can we visualize the association between age group and promotion using a chart?

Tip:

When performing a chi-square test, ensure the expected frequencies in each cell are sufficiently large (typically at least 5) to meet test assumptions.