TU BCA Applied Statistics Master Guide: Chi-Square Test, One-Way ANOVA & Time Series (Solved Numericals)
Author: Bhuban Subedi | Subject: Applied Statistics (CAST353) | Semester: Sixth Semester
In data science, machine learning feature selection, clinical trials, and business forecasting, advanced statistical hypothesis tests validate whether observed differences are statistically significant or merely random variations. In the Tribhuvan University BCA sixth semester, Applied Statistics (CAST353) tests students on non-parametric tests, analysis of variance, multiple regression, and econometric time series modeling.
In the final 60-mark TU board examination, computational numericals account for over 75% of the paper. Scoring a distinction requires mastering the exact formula calculations for the Chi-Square ($\chi^2$) Contingency Test, One-Way ANOVA $F$-table constructions, and Least Squares Time Series trend fitting.
In this guide, I will solve standard TU board numericals step-by-step.
1. Chi-Square ($\chi^2$) Test of Independence
The Chi-Square test tests whether two categorical variables are independent:
$$\chi^2 = \sum \frac{(O_{ij} – E_{ij})^2}{E_{ij}}$$
Where $O_{ij}$ is the Observed frequency, and Expected frequency $E_{ij} = \frac{\text{Row Total} \times \text{Column Total}}{\text{Grand Total } N}$.
Degrees of Freedom ($df$) = $(r – 1)(c – 1)$.
Solved TU Board Problem 1: Chi-Square Contingency Test
Problem: A survey of 200 BCA graduates analyzed whether employment status is independent of their specialization stream. Test at $\alpha = 0.05$ significance level:
Observed Frequencies (O):
+----------------+---------------+---------------+------------+
| Stream | Employed | Unemployed | Row Total |
+----------------+---------------+---------------+------------+
| Web Tech | 70 | 30 | **100** |
| AI / Data Sci | 80 | 20 | **100** |
+----------------+---------------+---------------+------------+
| **Col Total** | **150** | **50** | **N = 200**|
+----------------+---------------+---------------+------------+
Step 1: Calculate Expected Frequencies ($E = \frac{\text{Row} \times \text{Col}}{N}$):
– $E_{11} = \frac{100 \times 150}{200} = 75$
– $E_{12} = \frac{100 \times 50}{200} = 25$
– $E_{21} = \frac{100 \times 150}{200} = 75$
– $E_{22} = \frac{100 \times 50}{200} = 25$
Step 2: Chi-Square Calculation Table:
+---+----+----+---------+-------------+---------------------+
| | O | E | (O - E) | (O - E)^2 | (O - E)^2 / E |
+---+----+----+---------+-------------+---------------------+
| 1 | 70 | 75 | -5 | 25 | 25 / 75 = 0.3333 |
| 2 | 30 | 25 | +5 | 25 | 25 / 25 = 1.0000 |
| 3 | 80 | 75 | +5 | 25 | 25 / 75 = 0.3333 |
| 4 | 20 | 25 | -5 | 25 | 25 / 25 = 1.0000 |
+---+----+----+---------+-------------+---------------------+
| | | | | **Calculated \chi^2** | **2.6666**|
+---+----+----+---------+-------------+---------------------+
Step 3: Decision Rule:
– Degrees of Freedom $df = (2 – 1)(2 – 1) = 1$.
– Critical value from $\chi^2$ table at $\alpha = 0.05, df = 1 \implies \chi^2_{crit} = \mathbf{3.841}$.
– Since $\chi^2_{cal} (2.667) < \chi^2_{crit} (3.841)$, we Accept the Null Hypothesis ($H_0$).
– Conclusion: Employment status is independent of academic specialization stream.
2. One-Way Analysis of Variance (ANOVA)
ANOVA tests whether the means of 3 or more independent groups are equal:
$$H_0: \mu_1 = \mu_2 = \mu_3 \quad \text{vs.} \quad H_1: \text{At least one mean is different}$$
+-------------------------------------------------------------------------------+
| ONE-WAY ANOVA SUMMARY TABLE |
+-------------------+---------------+-------+---------------+-------------------+
| Source of Var | Sum of Squares| df | Mean Square | F-Ratio (F_cal) |
+-------------------+---------------+-------+---------------+-------------------+
| **Between Groups**| SSB | k - 1 | MSB = SSB/(k-1)| F = MSB / MSW |
| **Within Groups** | SSW | N - k | MSW = SSW/(N-k)| |
+-------------------+---------------+-------+---------------+-------------------+
| **Total** | SST | N - 1 | | |
+-------------------+---------------+-------+---------------+-------------------+
3. Time Series Analysis: Least Squares Linear Trend
The linear trend equation is given by:
$$Y = a + bX$$
Where:
$$b = \frac{N \sum XY – \sum X \sum Y}{N \sum X^2 – (\sum X)^2}, \quad a = \bar{Y} – b\bar{X}$$
When origin is shifted to the middle year such that $\sum X = 0$:
$$a = \frac{\sum Y}{N}, \quad b = \frac{\sum XY}{\sum X^2}$$
Frequently Asked Questions (FAQ)
Q1: When is a non-parametric test used instead of a parametric test?
Non-parametric tests (such as Chi-Square, Mann-Whitney U, and Kruskal-Wallis) are used when data does not follow a normal distribution or when variables are measured on nominal/ordinal scales.
Q2: What does the Coefficient of Determination ($R^2$) represent in regression?
$R^2$ represents the proportion of variance in the dependent variable that is predictable from the independent variables ($0 \le R^2 \le 1$).
