Chi-Square Test for Association

Step 1: Given Data

Observed contingency table:

Significance level $\alpha = 0.05$.

Step 2: Hypotheses

• $H_0$: No association between duration of marriage and acquaintanceship period (independent).
• $H_1$: There is an association.

Step 3: Expected Frequencies

$$E_{ij} = \frac{R_i \times C_j}{N}$$

Step 4: Chi-Square Statistic

$$\chi^2 = \sum \frac{(O-E)^2}{E}$$

$$\chi^2_{cal} = 0.24545 + 0.36818 + 0.27222 + 0.40833 + 0.05556 + 0.08333 = 1.4331$$

Step 5: Degrees of Freedom and Critical Value

$$df = (r-1)(c-1) = (3-1)(2-1) = 2$$

$$\chi^2_{0.05,,2} = 5.991$$

Step 6: Decision

$$\chi^2_{cal} = 1.433 < \chi^2_{tab} = 5.991$$

Fail to reject $H_0$.

Step 7: Conclusion

At the 5% level of significance, there is no significant association between the duration of marriage and the period of acquaintanceship before marriage. The data do not substantiate an association.

Design of Experiments and Latin Square Design

Design of Experiments

Design of experiments refers to the systematic planning and arrangement of experimental conditions so that valid, unbiased, and reliable conclusions can be drawn from the experimental data with minimum experimental error.

According to R.A. Fisher, the design of experiments is based on three fundamental principles:

Three Principles of Design of Experiments

Latin Square Design (LSD)

LSD is a design used for non-homogeneous experimental material. It applies local control simultaneously along both rows and columns, making it more efficient than RBD. The experimental layout is always square (equal number of rows and columns).

Mathematical Model

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

Where:

• $Y_{ijk}$ = observation in $i^{th}$ row and $j^{th}$ column receiving $k^{th}$ treatment
• $\mu$ = general mean effect
• $\alpha_i$ = effect due to $i^{th}$ row
• $\beta_j$ = effect due to $j^{th}$ column
• $T_k$ = effect due to $k^{th}$ treatment
• $e_{ijk}$ = random error due to chance

Layout of a Latin Square Design (4x4 Example)

In an LSD with 4 treatments (A, B, C, D), each treatment appears exactly once in each row and exactly once in each column:

$$ \begin{array}{|c|c|c|c|} \hline \textbf{Col 1} & \textbf{Col 2} & \textbf{Col 3} & \textbf{Col 4} \ \hline A & B & C & D \ \hline B & C & D & A \ \hline C & D & A & B \ \hline D & A & B & C \ \hline \end{array} $$

Here, 4 rows, 4 columns, and 4 treatments -- each appearing exactly once per row and per column.

Why the Number of Treatments Should NOT Be Less Than 3

In an LSD with $m$ treatments, the degrees of freedom for error is given by:

$$df_{error} = (m-1)(m-2)$$

If $m = 2$ (only 2 treatments):

$$df_{error} = (2-1)(2-2) = 1 \times 0 = 0$$

• With zero degrees of freedom for error, it is impossible to estimate the experimental error.
• Without an error estimate, no F-test can be performed, and no valid statistical conclusions can be drawn.

If $m = 3$ (minimum acceptable):

$$df_{error} = (3-1)(3-2) = 2 \times 1 = 2$$

• This gives at least 2 degrees of freedom for error, which allows estimation of experimental error and valid hypothesis testing.

Summary Table

Conclusion: The number of treatments in an LSD must be at least 3 so that the degrees of freedom for error is greater than zero, enabling estimation of experimental error and valid F-tests in the ANOVA table.

Multiple Regression Analysis

Purpose of Multiple Regression Analysis

Multiple regression analysis is applied to:

• Establish a functional relationship between one dependent variable $Y$ and two or more independent variables $X_1, X_2, \ldots$
• Predict/estimate the value of the dependent variable from known independent variables
• Measure the separate effect of each independent variable while holding the others constant
• Identify which independent variables are significant contributors to the variation in $Y$
• Provide a more realistic model since real phenomena depend on several factors simultaneously

Given Data

$n = 10$

Part (i): Multiple Regression Equation

Model: $\hat{Y} = b_0 + b_1 X_1 + b_2 X_2$

Step 1: Summations

$$\sum Y = 681,\ \sum X_1 = 67,\ \sum X_2 = 30,\ \sum X_1^2 = 463,$$ $$\sum X_2^2 = 182,\ \sum X_1 X_2 = 185,\ \sum X_1 Y = 4673,\ \sum X_2 Y = 1810$$

Step 2: Means

$$\bar{Y} = 68.1,\quad \bar{X}_1 = 6.7,\quad \bar{X}_2 = 3.0$$

Step 3: Normal Equations

$$681 = 10b_0 + 67b_1 + 30b_2 \quad (1)$$ $$4673 = 67b_0 + 463b_1 + 185b_2 \quad (2)$$ $$1810 = 30b_0 + 185b_1 + 182b_2 \quad (3)$$

Step 4: Solve

Eliminate $b_0$ using (1) and (2): multiply (1) by $6.7$: $$4562.7 = 67b_0 + 448.9b_1 + 201b_2 \quad (4)$$ $(2) - (4)$: $$110.3 = 14.1b_1 - 16b_2 \quad (5)$$

Eliminate $b_0$ using (1) and (3): multiply (1) by $3$: $$2043 = 30b_0 + 201b_1 + 90b_2 \quad (6)$$ $(3) - (6)$: $$-233 = -16b_1 + 92b_2 \quad (7)$$

From (5): $14.1b_1 - 16b_2 = 110.3$ From (7): $-16b_1 + 92b_2 = -233$

Solve. From (5): $b_1 = \dfrac{110.3 + 16b_2}{14.1}$.

Substitute in (7): $$-16\left(\frac{110.3 + 16b_2}{14.1}\right) + 92b_2 = -233$$

$$-16(110.3 + 16b_2) + 92(14.1)b_2 = -233(14.1)$$

$$-1764.8 - 256b_2 + 1297.2b_2 = -3285.3$$

$$1041.2b_2 = -1520.5$$

$$b_2 = -1.4603$$

Then: $$b_1 = \frac{110.3 + 16(-1.4603)}{14.1} = \frac{110.3 - 23.365}{14.1} = \frac{86.935}{14.1} = 6.1656$$

From (1): $$b_0 = \frac{681 - 67(6.1656) - 30(-1.4603)}{10}$$ $$= \frac{681 - 413.10 + 43.81}{10} = \frac{311.71}{10} = 31.171$$

Regression Equation

$$\boxed{\hat{Y} = 31.171 + 6.166,X_1 - 1.460,X_2}$$

Part (ii): Interpretation of $b_1$ and $b_2$

• $b_1 = 6.166$: Holding the number of absent days ($X_2$) constant, a one-unit increase in the aptitude test score ($X_1$) raises the job-satisfaction score ($Y$) by about 6.17 units. The positive sign shows a direct relationship.

• $b_2 = -1.460$: Holding the aptitude test score ($X_1$) constant, each additional day of absence ($X_2$) decreases the job-satisfaction score ($Y$) by about 1.46 units. The negative sign shows an inverse relationship.

Part (iii): Estimate for $X_1 = 9,\ X_2 = 6$

$$\hat{Y} = 31.171 + 6.166(9) - 1.460(6)$$ $$= 31.171 + 55.494 - 8.760$$ $$= \boxed{77.91}$$

The estimated job-satisfaction score is approximately 78.

Markov Chain and Its Characteristics

Definition of Stochastic Process

A stochastic process is a family of random variables indexed by time as a parameter. It is denoted as {X(t), t ∈ T}, where:

• T = index set (time parameter)
• X(t) = random variable at time t
• I = state space (the set of all possible values assumed by X(t))

The values assumed by the random variable X(t) are called states.

Definition of Markov Chain

A Markov Chain is a stochastic process {X(t), t = 0, 1, 2, ...} that satisfies the Markov property (memoryless property):

$$P[X(t+1) = j \mid X(t) = i,\ X(t-1) = i_{t-1},\ \ldots,\ X(0) = i_0] = P[X(t+1) = j \mid X(t) = i]$$

That is, the future state depends only on the present state, not on the past states.

The probability $P[X(t+1) = j \mid X(t) = i] = p_{ij}$ is called the transition probability from state $i$ to state $j$.

Characteristics of a Markov Chain

1. Markov Property (Memoryless Property)

The conditional probability of any future state depends only on the current state, not on the sequence of states that preceded it. This is the defining characteristic.

2. Transition Probability Matrix (TPM)

The one-step transition probabilities are arranged in a matrix $P = [p_{ij}]$, where:

• $p_{ij} \geq 0$ for all $i, j$
• $\sum_{j} p_{ij} = 1$ for each row $i$ (rows sum to 1)

Such a matrix is called a stochastic matrix.

3. Homogeneity (Time Homogeneous)

A Markov chain is said to be homogeneous if the transition probabilities do not depend on time $t$, i.e., $p_{ij}$ remains constant over time.

4. Chapman-Kolmogorov Equation

The $n$-step transition probability satisfies:

$$p_{ij}^{(n)} = \sum_{k} p_{ik}^{(r)} \cdot p_{kj}^{(n-r)}, \quad 0 < r < n$$

In matrix form: $P^{(n)} = P^n$

5. Steady State (Limiting) Distribution

As $n \to \infty$, the $n$-step transition probabilities converge to a steady state distribution $\pi$, such that:

$$\pi_j = \lim_{n \to \infty} P^n(x)$$

The steady state distribution satisfies: $$\pi P = \pi \quad \text{and} \quad \sum_{j} \pi_j = 1$$

6. Irreducibility

A Markov chain is irreducible if every state can be reached from every other state, i.e., all states communicate with each other.

7. Recurrence and Transience

• A state $i$ is recurrent if the process returns to state $i$ with probability 1.
• A state $i$ is transient if there is a positive probability of never returning to state $i$.

Summary Table

Confidence Level in Estimation and Confidence Interval for a Proportion

Definition: Confidence Level

The confidence level $(1-\alpha)$ is the probability that the interval estimate constructed from sample data will contain the true population parameter. A 96% confidence level means that if repeated random samples were taken and an interval built from each, about 96% of those intervals would enclose the true population proportion.

Given Data

Step 1: Sample Proportion

$$p = \frac{X}{n} = \frac{20}{400} = 0.05, \qquad q = 1 - p = 0.95$$

Step 2: Critical Value

$$\frac{\alpha}{2} = 0.02 \Rightarrow Z_{\alpha/2} = 2.054 \approx 2.05$$

Step 3: Standard Error

$$S.E.(p) = \sqrt{\frac{pq}{n}} = \sqrt{\frac{0.05 \times 0.95}{400}} = \sqrt{\frac{0.0475}{400}} = \sqrt{0.00011875} = 0.010897$$

Step 4: Confidence Interval

$$p \pm Z_{\alpha/2}\cdot S.E.(p)$$

Margin of error: $$E = 2.054 \times 0.010897 = 0.02238$$

Lower limit: $$0.05 - 0.02238 = 0.02762$$

Upper limit: $$0.05 + 0.02238 = 0.07238$$

Step 5: Express as Percentage

$$\boxed{2.76% \leq P \leq 7.24%}$$

(Using $Z = 2.05$: $2.77%$ to $7.24%$; essentially the same.)

Interpretation

We are 96% confident that the true percentage of all toothpaste tubes with leakage from the tail end lies between approximately 2.76% and 7.24%. If the sampling procedure were repeated many times, about 96% of the resulting intervals would contain the true population proportion of leaking tubes.

Chi-Square Test for Association

Step 1: Given Data

Observed contingency table:

Significance level $\alpha = 0.05$.

Step 2: Hypotheses

• $H_0$: No association between duration of marriage and acquaintanceship period (independent).
• $H_1$: There is an association.

Step 3: Expected Frequencies

$$E_{ij} = \frac{R_i \times C_j}{N}$$

Step 4: Chi-Square Statistic

$$\chi^2 = \sum \frac{(O-E)^2}{E}$$

$$\chi^2_{cal} = 0.24545 + 0.36818 + 0.27222 + 0.40833 + 0.05556 + 0.08333 = 1.4331$$

Step 5: Degrees of Freedom and Critical Value

$$df = (r-1)(c-1) = (3-1)(2-1) = 2$$

$$\chi^2_{0.05,,2} = 5.991$$

Step 6: Decision

$$\chi^2_{cal} = 1.433 < \chi^2_{tab} = 5.991$$

Fail to reject $H_0$.

Step 7: Conclusion

At the 5% level of significance, there is no significant association between the duration of marriage and the period of acquaintanceship before marriage. The data do not substantiate an association.

Design of Experiments and Latin Square Design

Design of Experiments

Design of experiments refers to the systematic planning and arrangement of experimental conditions so that valid, unbiased, and reliable conclusions can be drawn from the experimental data with minimum experimental error.

According to R.A. Fisher, the design of experiments is based on three fundamental principles:

Three Principles of Design of Experiments

Latin Square Design (LSD)

LSD is a design used for non-homogeneous experimental material. It applies local control simultaneously along both rows and columns, making it more efficient than RBD. The experimental layout is always square (equal number of rows and columns).

Mathematical Model

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

Where:

• $Y_{ijk}$ = observation in $i^{th}$ row and $j^{th}$ column receiving $k^{th}$ treatment
• $\mu$ = general mean effect
• $\alpha_i$ = effect due to $i^{th}$ row
• $\beta_j$ = effect due to $j^{th}$ column
• $T_k$ = effect due to $k^{th}$ treatment
• $e_{ijk}$ = random error due to chance

Layout of a Latin Square Design (4x4 Example)

In an LSD with 4 treatments (A, B, C, D), each treatment appears exactly once in each row and exactly once in each column:

$$ \begin{array}{|c|c|c|c|} \hline \textbf{Col 1} & \textbf{Col 2} & \textbf{Col 3} & \textbf{Col 4} \ \hline A & B & C & D \ \hline B & C & D & A \ \hline C & D & A & B \ \hline D & A & B & C \ \hline \end{array} $$

Here, 4 rows, 4 columns, and 4 treatments -- each appearing exactly once per row and per column.

Why the Number of Treatments Should NOT Be Less Than 3

In an LSD with $m$ treatments, the degrees of freedom for error is given by:

$$df_{error} = (m-1)(m-2)$$

If $m = 2$ (only 2 treatments):

$$df_{error} = (2-1)(2-2) = 1 \times 0 = 0$$

• With zero degrees of freedom for error, it is impossible to estimate the experimental error.
• Without an error estimate, no F-test can be performed, and no valid statistical conclusions can be drawn.

If $m = 3$ (minimum acceptable):

$$df_{error} = (3-1)(3-2) = 2 \times 1 = 2$$

• This gives at least 2 degrees of freedom for error, which allows estimation of experimental error and valid hypothesis testing.

Summary Table

Conclusion: The number of treatments in an LSD must be at least 3 so that the degrees of freedom for error is greater than zero, enabling estimation of experimental error and valid F-tests in the ANOVA table.

Multiple Regression Analysis

Purpose of Multiple Regression Analysis

Multiple regression analysis is applied to:

• Establish a functional relationship between one dependent variable $Y$ and two or more independent variables $X_1, X_2, \ldots$
• Predict/estimate the value of the dependent variable from known independent variables
• Measure the separate effect of each independent variable while holding the others constant
• Identify which independent variables are significant contributors to the variation in $Y$
• Provide a more realistic model since real phenomena depend on several factors simultaneously

Given Data

$n = 10$

Part (i): Multiple Regression Equation

Model: $\hat{Y} = b_0 + b_1 X_1 + b_2 X_2$

Step 1: Summations

$$\sum Y = 681,\ \sum X_1 = 67,\ \sum X_2 = 30,\ \sum X_1^2 = 463,$$ $$\sum X_2^2 = 182,\ \sum X_1 X_2 = 185,\ \sum X_1 Y = 4673,\ \sum X_2 Y = 1810$$

Step 2: Means

$$\bar{Y} = 68.1,\quad \bar{X}_1 = 6.7,\quad \bar{X}_2 = 3.0$$

Step 3: Normal Equations

$$681 = 10b_0 + 67b_1 + 30b_2 \quad (1)$$ $$4673 = 67b_0 + 463b_1 + 185b_2 \quad (2)$$ $$1810 = 30b_0 + 185b_1 + 182b_2 \quad (3)$$

Step 4: Solve

Eliminate $b_0$ using (1) and (2): multiply (1) by $6.7$: $$4562.7 = 67b_0 + 448.9b_1 + 201b_2 \quad (4)$$ $(2) - (4)$: $$110.3 = 14.1b_1 - 16b_2 \quad (5)$$

Eliminate $b_0$ using (1) and (3): multiply (1) by $3$: $$2043 = 30b_0 + 201b_1 + 90b_2 \quad (6)$$ $(3) - (6)$: $$-233 = -16b_1 + 92b_2 \quad (7)$$

From (5): $14.1b_1 - 16b_2 = 110.3$ From (7): $-16b_1 + 92b_2 = -233$

Solve. From (5): $b_1 = \dfrac{110.3 + 16b_2}{14.1}$.

Substitute in (7): $$-16\left(\frac{110.3 + 16b_2}{14.1}\right) + 92b_2 = -233$$

$$-16(110.3 + 16b_2) + 92(14.1)b_2 = -233(14.1)$$

$$-1764.8 - 256b_2 + 1297.2b_2 = -3285.3$$

$$1041.2b_2 = -1520.5$$

$$b_2 = -1.4603$$

Then: $$b_1 = \frac{110.3 + 16(-1.4603)}{14.1} = \frac{110.3 - 23.365}{14.1} = \frac{86.935}{14.1} = 6.1656$$

From (1): $$b_0 = \frac{681 - 67(6.1656) - 30(-1.4603)}{10}$$ $$= \frac{681 - 413.10 + 43.81}{10} = \frac{311.71}{10} = 31.171$$

Regression Equation

$$\boxed{\hat{Y} = 31.171 + 6.166,X_1 - 1.460,X_2}$$

Part (ii): Interpretation of $b_1$ and $b_2$

• $b_1 = 6.166$: Holding the number of absent days ($X_2$) constant, a one-unit increase in the aptitude test score ($X_1$) raises the job-satisfaction score ($Y$) by about 6.17 units. The positive sign shows a direct relationship.

• $b_2 = -1.460$: Holding the aptitude test score ($X_1$) constant, each additional day of absence ($X_2$) decreases the job-satisfaction score ($Y$) by about 1.46 units. The negative sign shows an inverse relationship.

Part (iii): Estimate for $X_1 = 9,\ X_2 = 6$

$$\hat{Y} = 31.171 + 6.166(9) - 1.460(6)$$ $$= 31.171 + 55.494 - 8.760$$ $$= \boxed{77.91}$$

The estimated job-satisfaction score is approximately 78.

Confidence Level in Estimation and Confidence Interval for a Proportion

Definition: Confidence Level

The confidence level $(1-\alpha)$ is the probability that the interval estimate constructed from sample data will contain the true population parameter. A 96% confidence level means that if repeated random samples were taken and an interval built from each, about 96% of those intervals would enclose the true population proportion.

Given Data

Step 1: Sample Proportion

$$p = \frac{X}{n} = \frac{20}{400} = 0.05, \qquad q = 1 - p = 0.95$$

Step 2: Critical Value

$$\frac{\alpha}{2} = 0.02 \Rightarrow Z_{\alpha/2} = 2.054 \approx 2.05$$

Step 3: Standard Error

$$S.E.(p) = \sqrt{\frac{pq}{n}} = \sqrt{\frac{0.05 \times 0.95}{400}} = \sqrt{\frac{0.0475}{400}} = \sqrt{0.00011875} = 0.010897$$

Step 4: Confidence Interval

$$p \pm Z_{\alpha/2}\cdot S.E.(p)$$

Margin of error: $$E = 2.054 \times 0.010897 = 0.02238$$

Lower limit: $$0.05 - 0.02238 = 0.02762$$

Upper limit: $$0.05 + 0.02238 = 0.07238$$

Step 5: Express as Percentage

$$\boxed{2.76% \leq P \leq 7.24%}$$

(Using $Z = 2.05$: $2.77%$ to $7.24%$; essentially the same.)

Interpretation

We are 96% confident that the true percentage of all toothpaste tubes with leakage from the tail end lies between approximately 2.76% and 7.24%. If the sampling procedure were repeated many times, about 96% of the resulting intervals would contain the true population proportion of leaking tubes.

Markov Chain and Its Characteristics

Definition of Stochastic Process

A stochastic process is a family of random variables indexed by time as a parameter. It is denoted as {X(t), t ∈ T}, where:

• T = index set (time parameter)
• X(t) = random variable at time t
• I = state space (the set of all possible values assumed by X(t))

The values assumed by the random variable X(t) are called states.

Definition of Markov Chain

A Markov Chain is a stochastic process {X(t), t = 0, 1, 2, ...} that satisfies the Markov property (memoryless property):

$$P[X(t+1) = j \mid X(t) = i,\ X(t-1) = i_{t-1},\ \ldots,\ X(0) = i_0] = P[X(t+1) = j \mid X(t) = i]$$

That is, the future state depends only on the present state, not on the past states.

The probability $P[X(t+1) = j \mid X(t) = i] = p_{ij}$ is called the transition probability from state $i$ to state $j$.

Characteristics of a Markov Chain

1. Markov Property (Memoryless Property)

The conditional probability of any future state depends only on the current state, not on the sequence of states that preceded it. This is the defining characteristic.

2. Transition Probability Matrix (TPM)

The one-step transition probabilities are arranged in a matrix $P = [p_{ij}]$, where:

• $p_{ij} \geq 0$ for all $i, j$
• $\sum_{j} p_{ij} = 1$ for each row $i$ (rows sum to 1)

Such a matrix is called a stochastic matrix.

3. Homogeneity (Time Homogeneous)

A Markov chain is said to be homogeneous if the transition probabilities do not depend on time $t$, i.e., $p_{ij}$ remains constant over time.

4. Chapman-Kolmogorov Equation

The $n$-step transition probability satisfies:

$$p_{ij}^{(n)} = \sum_{k} p_{ik}^{(r)} \cdot p_{kj}^{(n-r)}, \quad 0 < r < n$$

In matrix form: $P^{(n)} = P^n$

5. Steady State (Limiting) Distribution

As $n \to \infty$, the $n$-step transition probabilities converge to a steady state distribution $\pi$, such that:

$$\pi_j = \lim_{n \to \infty} P^n(x)$$

The steady state distribution satisfies: $$\pi P = \pi \quad \text{and} \quad \sum_{j} \pi_j = 1$$

6. Irreducibility

A Markov chain is irreducible if every state can be reached from every other state, i.e., all states communicate with each other.

7. Recurrence and Transience

• A state $i$ is recurrent if the process returns to state $i$ with probability 1.
• A state $i$ is transient if there is a positive probability of never returning to state $i$.

Summary Table

Source Sum of Squares (SS) Degrees of Freedom (df) --------------------------------------------------- Regression 740 2 Error 510 17 Total 1250 19 $\alpha = 0.05$ --- $$df{Total} = df{Reg} + df{Error} = 2 + 17 = 19$$ Since $df{Total} = n...

Parameter Value ------------------ Confidence level $(1-\alpha)$ 0.95 Maximum allowable error $(E)$ 2.0 points Population standard deviation $(\sigma)$ 30 Critical value $Z{\alpha/2}$ at 95% 1.96 Estimation is the statistical procedure o...

Male Faculty Female Faculty --------- Sample size $n1 = 60$ $n2 = 30$ Mean working hours $\bar{X}1 = 12$ $\bar{X}2 = 9$ Standard deviation $S1 = 4$ $S2 = 3$ Level of significance: $\alpha = 0.01$ (two-tailed) --- Null Hypothesis $H0: \mu...

High Level Mid Level Low Level :-::-::-: 85 90 55 75 100 75 85 95 80 60 85 75 70 90 75 - $n1 = 5,\ n2 = 6,\ n3 = 4$ - Total $N = 15$, groups $k = 3$ - $\alpha = 0.05$ - $H0: \mu1 = \mu2 = \mu3$ (no significant difference) - $H1:$ at leas...

Hypothesis Test: Mean Drying Time of Spray Paint

Given Data

• Population (claimed) mean: $\mu_0 = 122$ seconds
• Sample size: $n = 50$
• Sample mean: $\bar{X} = 116$ seconds
• Sample standard deviation: $s = 16.8$ seconds
• Significance level: $\alpha = 0.05$

Step 1: Hypotheses

• $H_0: \mu = 122$ (no reduction)
• $H_1: \mu < 122$ (drying time reduced) - left-tailed test

Step 2: Test Statistic

Since $n = 50 > 30$, use the Z-test: $$Z = \frac{\bar{X} - \mu_0}{s/\sqrt{n}}$$

Step 3: Calculation

$$s/\sqrt{n} = \frac{16.8}{\sqrt{50}} = \frac{16.8}{7.0711} = 2.3759$$

$$Z = \frac{116 - 122}{2.3759} = \frac{-6}{2.3759} = -2.525$$

Step 4: Critical Value

Left-tailed at $\alpha = 0.05$: $Z_{tab} = -1.645$

Step 5: Decision

Since $Z_{cal} = -2.525 < -1.645$, we reject $H_0$.

Step 6: P-value

$$p = P(Z < -2.525) = 1 - P(Z < 2.525) = 1 - 0.9942 = 0.0058$$

Since $p = 0.0058 < \alpha = 0.05$, reject $H_0$.

Conclusion

The data provide strong evidence at the 5% level of significance that the new chemical additive reduces the mean drying time below 122 seconds. The small p-value ($\approx 0.0058$) confirms high significance.

Three variables $X1, X2, X3$ with zero-order correlation coefficients: $$r{12} = 0.40, \quad r{23} = 0.50, \quad r{13} = 0.60$$ Note on notation: The problem statement contains a typo. "$R{123}$" is standardly interpreted as the multiple...

Level of Significance and Hypothesis Testing

Concept of Level of Significance

The level of significance (denoted α) is the probability of rejecting the null hypothesis $H_0$ when it is actually true. It is the maximum acceptable probability of committing a Type I error.

• Fixed in advance of the test (typically $1%$, $5%$, or $10%$).
• At $5%$ significance, we accept a $5%$ chance of wrongly rejecting a true $H_0$.
• It determines the critical (tabulated) value used to bound the acceptance/rejection regions.
• A smaller $\alpha$ makes the test more conservative (harder to reject $H_0$).

Numerical Solution

Given Data

Sample proportion: $$\hat{p} = \frac{x}{n} = \frac{80}{100} = 0.80$$

Step 1: Hypotheses

• $H_0: P = \dfrac{2}{3}$ (two-thirds prefer the first color)
• $H_1: P > \dfrac{2}{3}$ (more than two-thirds prefer the first color)

This is a right-tailed test.

Step 2: Test Statistic

$$Z = \frac{\hat{p} - P}{\sqrt{\dfrac{P(1-P)}{n}}}$$

Numerator: $$\hat{p} - P = 0.80 - 0.6667 = 0.1333$$

Denominator: $$\sqrt{\frac{0.6667 \times 0.3333}{100}} = \sqrt{\frac{0.2222}{100}} = \sqrt{0.002222} = 0.04714$$

Therefore: $$Z = \frac{0.1333}{0.04714} = 2.83$$

Step 3: Critical Value

For a right-tailed test at $\alpha = 0.05$: $$Z_{tab} = 1.645$$

Step 4: Decision

$$Z_{cal} = 2.83 > Z_{tab} = 1.645$$

Since the calculated value falls in the rejection region, reject $H_0$.

Step 5: Conclusion

At the $5%$ level of significance, there is sufficient evidence to conclude that more than two-thirds of all customers prefer the first color of the laptop.

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.

Short Notes on Any Two

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.

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.

When all possible random samples of size n are drawn from a population of size N, and the mean of each sample is computed, the probability distribution formed by these sample means is called the sampling distribution of the sample mean. ...