Multiple Regression Analysis - Verified Model Answer

Given Data

• Regression equation: $\hat{Y} = 52.7 - 2.87X_1 + 0.85X_2$
• Total Sum of Squares: $SST = 1452$
• Sum of Squares due to Regression: $SSR = 1143.3$
• Standard error of $b_2$: $S_{b_2} = 0.55$
• Sample size: $n = 7$; number of predictors: $k = 2$
• Significance level: $\alpha = 0.05$
• Data table (Y, X1, X2) as given.

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a) Interpretation of $b_1$ and $b_2$

$b_1 = -2.87$: Holding number of tables ($X_2$) constant, each additional gigabyte of data size decreases the processed requests by about 2.87 per hour on average.

$b_2 = +0.85$: Holding data size ($X_1$) constant, each additional table used increases the processed requests by about 0.85 per hour on average.

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b) Significance of the Overall Model (F-test)

$$SSE = SST - SSR = 1452 - 1143.3 = 308.7$$

$$MSR = \frac{SSR}{k} = \frac{1143.3}{2} = 571.65$$

$$MSE = \frac{SSE}{n-k-1} = \frac{308.7}{4} = 77.175$$

$$F_{cal} = \frac{MSR}{MSE} = \frac{571.65}{77.175} = 7.41$$

Critical value at $\alpha=0.05$, $df_1=2$, $df_2=4$: $$F_{tab}(2,4) = 6.94$$

Since $F_{cal}=7.41 > F_{tab}=6.94$, reject $H_0$.

Conclusion: The overall regression model is statistically significant at 0.05.

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c) Significance of $X_2$ (Number of Tables)

Hypotheses: $H_0: \beta_2 = 0$ vs $H_1: \beta_2 \neq 0$

$$t_{cal} = \frac{b_2}{S_{b_2}} = \frac{0.85}{0.55} = 1.545$$

Critical value (two-tailed) at $\alpha=0.05$, $df = n-k-1 = 4$: $$t_{tab} = 2.776$$

Since $|t_{cal}| = 1.545 < 2.776$, fail to reject $H_0$.

Conclusion: There is no significant relationship between processed requests and number of tables at the 0.05 level.

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d) Percentage of Variation Explained ($R^2$)

$$R^2 = \frac{SSR}{SST} = \frac{1143.3}{1452} = 0.7874$$

$$R^2 \times 100 = 78.74%$$

About 78.74% of the variation in processed requests is explained by data size and number of tables together.

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e) Standard Error of Estimate

$$S_e = \sqrt{\frac{SSE}{n-k-1}} = \sqrt{\frac{308.7}{4}} = \sqrt{77.175} = 8.785$$

$S_e \approx 8.79$ requests per hour.

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f) Estimate for $X_1 = 9$, $X_2 = 8$

$$\hat{Y} = 52.7 - 2.87(9) + 0.85(8)$$ $$= 52.7 - 25.83 + 6.80 = 33.67$$

Estimated processed requests $\approx 33.67$ per hour.

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Summary

Kruskal-Wallis Test for Missile System Propellant Burning Rates

Step 1: Given Data

Total $N = 5 + 6 + 7 = 18$

Step 2: Hypotheses

• $H_0$: The propellant burning rates are the same for all three missile systems.
• $H_1$: At least one system differs.

Step 3: Rank All 18 Observations (ascending)

Tie at 19.5: average rank $= (10+11)/2 = 10.5$

Step 4: Rank Sums

System I: $R_1 = 2 + 8 + 9 + 15 + 16 = 50$

System II: $R_2 = 1 + 4 + 10.5 + 13 + 14 + 18 = 60.5$

System III: $R_3 = 3 + 5 + 6 + 7 + 10.5 + 12 + 17 = 60.5$

Check: $50 + 60.5 + 60.5 = 171 = \dfrac{N(N+1)}{2} = \dfrac{18(19)}{2} = 171$ ✓

Step 5: Test Statistic

$$H = \frac{12}{N(N+1)} \sum \frac{R_i^2}{n_i} - 3(N+1)$$

$$\frac{50^2}{5} = 500, \quad \frac{60.5^2}{6} = \frac{3660.25}{6} = 610.042, \quad \frac{60.5^2}{7} = \frac{3660.25}{7} = 522.893$$

Sum $= 500 + 610.042 + 522.893 = 1632.935$

$$H = \frac{12}{18 \times 19}(1632.935) - 3(19) = \frac{12}{342}(1632.935) - 57$$

$$H = 0.035088 \times 1632.935 - 57 = 57.296 - 57 = 0.296 \approx 0.30$$

Correction for ties (optional, one tie group of size 2):

$$C = 1 - \frac{\sum(t^3 - t)}{N^3 - N} = 1 - \frac{2^3 - 2}{18^3 - 18} = 1 - \frac{6}{5814} = 1 - 0.001032 = 0.99897$$

$$H_{corrected} = \frac{0.296}{0.99897} \approx 0.296$$

The tie correction is negligible.

Step 6: Critical Value

$H$ follows $\chi^2$ with $df = k - 1 = 2$.

$$\chi^2_{0.01, 2} = 9.210$$

Step 7: Decision

$$H = 0.30 < 9.210$$

Fail to reject $H_0$.

Conclusion

At the 1% significance level, there is insufficient evidence to conclude that the propellant burning rates differ among the three missile systems. The burning rates may be considered the same.

Latin Square Design (LSD)

Definition

Latin Square Design (LSD) is a design of experiment used for non-homogeneous experimental material where local control is applied simultaneously in two directions (row-wise and column-wise), thereby converting non-homogeneous material into homogeneous. It is more efficient than Randomized Block Design (RBD) because it controls experimental error in two directions at the same time.

The shape of LSD is always a square since it contains an equal number of rows, columns, and treatments. If there are m treatments, the layout is an m x m square.

LSD is based on all three principles of experimental design: Randomization, Replication, and Local Control.

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Conditions for Use of LSD

LSD can be used under the following conditions:

1. The experimental material is non-homogeneous in two directions (rows and columns).
2. The number of rows = number of columns = number of treatments (i.e., the layout must be a square).
3. The number of treatments is neither too small (less than 4) nor too large (more than 8), as a very large square becomes difficult to manage.
4. When it is desired to control variation in two directions simultaneously.
5. When local control needs to be applied both row-wise and column-wise.

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Layout of Latin Square Design

For a 4 x 4 LSD with treatments A, B, C, D, a typical layout is:

Key property: Each treatment appears exactly once in each row and exactly once in each column.

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Mathematical Model

The mathematical model for LSD is:

$$Y_{ijk} = \mu + \alpha_i + \beta_j + \tau_k + e_{ijk}$$

Where:

• $Y_{ijk}$ = observation in the $i$-th row and $j$-th column receiving the $k$-th treatment
• $\mu$ = general mean effect
• $\alpha_i$ = effect due to $i$-th row $(i = 1, 2, \ldots, m)$
• $\beta_j$ = effect due to $j$-th column $(j = 1, 2, \ldots, m)$
• $\tau_k$ = effect due to $k$-th treatment $(k = 1, 2, \ldots, m)$
• $e_{ijk}$ = random error (chance variation), assumed $\sim N(0, \sigma^2)$

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Analysis of Latin Square Design

Hypotheses

• For Rows: $H_0$: There is no significant difference among rows vs $H_1$: There is significant difference among rows.
• For Columns: $H_0$: There is no significant difference among columns vs $H_1$: There is significant difference among columns.
• For Treatments: $H_0$: There is no significant difference among treatments vs $H_1$: There is significant difference among treatments.

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Notation

Let:

• $m$ = number of treatments (also number of rows and columns)
• $N = m^2$ = total number of observations
• $T$ = Grand total of all observations
• $R_i$ = Total of $i$-th row
• $C_j$ = Total of $j$-th column
• $T_k$ = Total of $k$-th treatment

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Correction Factor and Sums of Squares

Correction Factor (CF): $$CF = \frac{T^2}{N} = \frac{T^2}{m^2}$$

Total Sum of Squares (TSS): $$TSS = \sum_{i}\sum_{j} Y_{ij}^2 - CF$$

Sum of Squares for Rows (SSR): $$SSR = \frac{1}{m}\sum_{i=1}^{m} R_i^2 - CF$$

Sum of Squares for Columns (SSC): $$SSC = \frac{1}{m}\sum_{j=1}^{m} C_j^2 - CF$$

Sum of Squares for Treatments (SSTr): $$SSTr = \frac{1}{m}\sum_{k=1}^{m} T_k^2 - CF$$

Error Sum of Squares (SSE): $$SSE = TSS - SSR - SSC - SSTr$$

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Degrees of Freedom

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Mean Squares

$$MSR = \frac{SSR}{m-1}, \quad MSC = \frac{SSC}{m-1}, \quad MSTr = \frac{SSTr}{m-1}, \quad MSE = \frac{SSE}{(m-1)(m-2)}$$

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ANOVA Table for LSD

Multiple Correlation and Partial Correlation

Definition of Multiple Correlation

Multiple correlation is the correlation between one variable (the dependent variable) and the combined linear effect of two or more other variables (independent variables) taken together. For a trivariate distribution $X_1, X_2, X_3$, the multiple correlation coefficient $R_{1.23}$ measures the degree of association between $X_1$ and the joint effect of $X_2$ and $X_3$. Its square gives the proportion of variance in $X_1$ explained by $X_2$ and $X_3$.

Given Data

• $r_{12} = 0.5$
• $r_{23} = 0.6$
• $r_{13} = 0.7$

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Part (i): Partial Correlation $r_{12.3}$

$$r_{12.3} = \frac{r_{12} - r_{13},r_{23}}{\sqrt{(1 - r_{13}^2)(1 - r_{23}^2)}}$$

Numerator: $$0.5 - (0.7)(0.6) = 0.5 - 0.42 = 0.08$$

Denominator: $$\sqrt{(1 - 0.49)(1 - 0.36)} = \sqrt{(0.51)(0.64)} = \sqrt{0.3264} = 0.5713$$

Result: $$r_{12.3} = \frac{0.08}{0.5713} = 0.1400$$

A weak positive partial correlation between $X_1$ and $X_2$ when $X_3$ is held constant.

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Part (ii): Multiple Correlation $R_{1.23}$

$$R_{1.23} = \sqrt{\frac{r_{12}^2 + r_{13}^2 - 2,r_{12},r_{13},r_{23}}{1 - r_{23}^2}}$$

Numerator: $$(0.5)^2 + (0.7)^2 - 2(0.5)(0.7)(0.6) = 0.25 + 0.49 - 0.42 = 0.32$$

Denominator: $$1 - (0.6)^2 = 1 - 0.36 = 0.64$$

Result: $$R_{1.23} = \sqrt{\frac{0.32}{0.64}} = \sqrt{0.5} = 0.7071$$

Interpretation: $R_{1.23}^2 = 0.50$, so about 50% of the variation in $X_1$ is jointly explained by $X_2$ and $X_3$.

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Final Answers

• $r_{12.3} = 0.1400$
• $R_{1.23} = 0.7071$

Type I and Type II Errors + Test of Hypothesis on Mobile Popularity

Given Data

• Sample size: $n = 600$
• Number with Samsung: $X = 350$
• Claim: equally popular, so $p_0 = 0.5$
• Level of significance: $\alpha = 0.05$

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Part 1: Type I and Type II Error

Type I Error ($\alpha$): Rejecting the null hypothesis $H_0$ when it is actually true. Its probability equals the level of significance $\alpha$.

Type II Error ($\beta$): Accepting the null hypothesis $H_0$ when it is actually false.

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Part 2: Test of Hypothesis

Step 1: Hypotheses

$$H_0: p = 0.5 \quad \text{(equally popular)}$$ $$H_1: p \neq 0.5 \quad \text{(not equally popular)} \quad \text{(two-tailed test)}$$

Step 2: Sample Proportion

$$\hat{p} = \frac{X}{n} = \frac{350}{600} = 0.5833$$

Step 3: Test Statistic (Z-test for proportion)

$$Z = \frac{\hat{p} - p_0}{\sqrt{\dfrac{p_0 q_0}{n}}}$$

where $q_0 = 1 - 0.5 = 0.5$.

$$\text{S.E.} = \sqrt{\frac{0.5 \times 0.5}{600}} = \sqrt{\frac{0.25}{600}} = \sqrt{0.0004167} = 0.020412$$

$$Z = \frac{0.5833 - 0.5}{0.020412} = \frac{0.0833}{0.020412} = 4.08$$

On the continuity correction: applying it gives $Z_{cal} = 4.04$; without it (the standard approach at this level) $Z = 4.08$. The two are numerically equivalent for the decision. Using the raw count form:

$$Z = \frac{X - np_0}{\sqrt{np_0 q_0}} = \frac{350 - 300}{\sqrt{150}} = \frac{50}{12.247} = 4.08$$

Step 4: Critical Value

At $\alpha = 0.05$, two-tailed:

$$Z_{\alpha/2} = Z_{0.025} = 1.96$$

Step 5: Decision

$$|Z| = 4.08 > 1.96$$

Since the calculated value exceeds the critical value, we reject $H_0$.

Step 6: Conclusion

At the 5% level of significance, there is sufficient evidence to reject the claim that Samsung and Huawei mobiles are equally popular in Kathmandu. The data indicates Samsung is significantly more popular.

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Final Result: $Z_{cal} = 4.08$ (without continuity correction) vs $Z_{tab} = 1.96$; reject $H_0$. With the continuity correction $Z_{cal} = 4.04$, and the decision and conclusion are identical.

Short Notes on Any Two

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a. Difference Between Parametric and Non-Parametric Test

**Note ** The Mann-Whitney U-test is a non-parametric test used as an alternative to the T-test when T-test fails to satisfy the normality assumptions. This clearly illustrates the key distinction: parametric tests demand distributional assumptions, while non-parametric tests do not.

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b. Required Assumptions for Linear Regression Model

For a linear regression model to produce valid and reliable results, the following assumptions must be satisfied:

1. Linearity The relationship between the dependent variable (Y) and independent variable(s) (X) must be linear in nature.

2. Independence of Errors The error terms (residuals) must be independent of each other. There should be no autocorrelation among residuals.

3. Normality of Errors The error terms should follow a normal distribution with mean zero:

$$E(\varepsilon_i) = 0$$

4. Homoscedasticity (Constant Variance) The variance of error terms should be constant across all levels of the independent variable:

$$Var(\varepsilon_i) = \sigma^2 \text{ (constant)}$$

5. No Multicollinearity (for multiple regression) In multiple regression, the independent variables (X1, X2, ...) should not be highly correlated with each other. In multiple regression with variables X1 and X2, each variable is tested separately for its contribution, implying they should be independently contributing.

6. No Measurement Error The independent variables are assumed to be measured without error.

7. Significance of Regression Coefficients The regression coefficients (b1, b2, ...) should be significantly different from zero, tested using the t-test:

$$t_{cal} = \frac{b_i}{S_{b_i}}$$

**Example ** With n=20, b1=4, Sb1=1.2, the calculated t = 3.33 > t-tab = 2.110, confirming a significant linear relationship between Y and X1, validating the regression model assumption.

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