The image displays a handwritten statistical analysis in a notebook, lying on an orange-colored surface. The notes are written in blue ink on lined paper. A circular watermark, partially visible, is imprinted on the page, centrally located across the mathematical calculations. The content of the page details a Chi-squared test, likely for homogeneity, as indicated by the comparison of frequency distributions across different samples. At the top of the page, the header "AIR MUSLIM STUDENT'S SOCIETY OF NIGERIA FEDERAL UNIVERSITY OF TECHNOLOGY MINNA." is visible, suggesting the context of the work. A faint number "0-419-" is also present in the top right corner. The problem starts with a "Calculate" section presenting a contingency table titled "Samples" with five categories (1, 2, 3, 4, 5) and two main rows of data: "Above" and "Median". A "Total" row for each sample and a "Total" column for each category are also included. The observed frequencies are: - Above: 20, 30, 25, 40, 30 (Row Total: 145) - Median: 25, 35, 30, 45, 32 (Row Total: 167) - Sample Totals (Total row): 46, 65, 55, 85, 62 (Grand Total: 312) Following the observed frequencies, a "Calc" section details the calculation of expected frequencies (denoted as C_ij or E_ij). These are calculated for each cell using the formula: (Row Total * Column Total) / Grand Total. - C_11 = 145 × (46/312) = 20.91 - C_12 = 145 × (65/312) = 30.21 - C_13 = 145 × (55/312) = 25.56 - C_14 = 145 × (85/312) = 39.50 - C_15 = 145 × (62/312) = 28.80 - C_21 = 167 × (46/312) = 24.08 - C_22 = 167 × (65/312) = 34.79 - C_23 = 167 × (55/312) = 29.44 - C_24 = 167 × (85/312) = 45.50 - C_25 = 167 × (62/312) = 33.18 The next section, "Test Statistics", presents the formula for the Chi-squared (χ²) test: χ² = ΣΣ ((a_ij - e_ij)² / e_ij), where a_ij are observed frequencies and e_ij are expected frequencies. The individual components of the Chi-squared statistic are summed: χ² = 0.0396 + 0.00145 + 0.0123 + 0.0063 + 0.05 + 0.0351 + 0.0013 + 0.0106 + 0.0055 + 0.0419. The calculated Chi-squared value is determined as χ² = 0.2081. Finally, the "Conclusion" section states the outcome of the hypothesis test. It compares the calculated χ² value with a critical value from the Chi-squared distribution table. "Since χ² = 0.2081 < χ²_0.05,4 = 9.49." The degrees of freedom (df) are 4 (calculated as (rows-1) * (columns-1) = (2-1) * (5-1) = 1 * 4 = 4), and the significance level is 0.05. Based on this comparison, the conclusion is to "Accept H_o and Conclude that the Pops from which the five Sample drawn have the Same freq. distn." This implies that the null hypothesis (H_o), which states that the populations from which the samples were drawn have the same frequency distribution, is accepted.

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