Mid Term Exam 3 (14th Batch) — Full Solution

Two questions, worth 5 marks each — a Markov chain transition probability matrix problem, and a one-sample $z$-test for hypothesis testing.

Question 1 — Markov Chain: TV Watching vs Reading

A person spends his evening either watching TV (T) or reading books (R). If he watches TV today, tomorrow he is twice as likely to watch TV again as to read. If he reads today, then tomorrow he certainly watches TV.

1. Write the transition probability matrix (TPM).
2. If initially he starts with reading, find the probability that he reads after two days.

Concept Needed

A Markov chain is a random process where the next state depends only on the current state, not on the past history. The Transition Probability Matrix (TPM) lists the probability of moving from every state to every other state in one step. Each row of a TPM must sum to $1$, since from a given state the person must go to some state tomorrow.

Given / Required

Two states: $T$ = Watching TV, $R$ = Reading.

• If today is $T$: tomorrow is twice as likely to be $T$ as $R$, so $P(T\to T) : P(T\to R) = 2:1$.
• If today is $R$: tomorrow is certainly $T$, so $P(R\to T) = 1$, $P(R\to R) = 0$.

Required: (i) the TPM $P$, (ii) $P(\text{reads on day 2} \mid \text{reads on day 0})$.

Step 1: Split the Ratio 2:1 into Probabilities

Since $P(T\to T):P(T\to R) = 2:1$ and the two probabilities must add to $1$:

$P(T\to T) = \frac{2}{3}, \qquad P(T\to R) = \frac{1}{3}$

Step 2: Use the Given Certain Transition from R

$P(R\to T) = 1, \qquad P(R\to R) = 0$

Step 3: Write the TPM

$P = \begin{pmatrix} 2/3 & 1/3 \\ 1 & 0 \end{pmatrix} \quad \text{(rows/columns in order } T, R\text{)}$

Check: Row sums $= \frac{2}{3}+\frac{1}{3}=1$ and $1+0=1$. Both rows sum to $1$, so the TPM is valid.

(Note: $R$ has no self-loop, since $P(R\to R)=0$ — every time he reads, he is guaranteed to watch TV the next day.)

Step 4: Set Up the Initial State Vector

He starts (day $0$) with reading, so:

$\pi_0 = \begin{pmatrix} P(T) & P(R) \end{pmatrix} = \begin{pmatrix} 0 & 1 \end{pmatrix}$

We need the state after 2 steps: $\pi_2 = \pi_0 P^2$.

Step 5: Compute $P^2 = P \times P$

$P^2 = \begin{pmatrix} 2/3 & 1/3 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 2/3 & 1/3 \\ 1 & 0 \end{pmatrix}$

Row 1 (starting from $T$):

$\left(\tfrac{2}{3}\cdot\tfrac{2}{3} + \tfrac{1}{3}\cdot 1,\ \ \tfrac{2}{3}\cdot\tfrac{1}{3} + \tfrac{1}{3}\cdot 0\right) = \left(\tfrac{4}{9}+\tfrac{3}{9},\ \ \tfrac{2}{9}\right) = \left(\tfrac{7}{9},\ \tfrac{2}{9}\right)$

Row 2 (starting from $R$):

$\left(1\cdot\tfrac{2}{3} + 0\cdot 1,\ \ 1\cdot\tfrac{1}{3} + 0\cdot 0\right) = \left(\tfrac{2}{3},\ \tfrac{1}{3}\right)$

So:

$P^2 = \begin{pmatrix} 7/9 & 2/9 \\ 2/3 & 1/3 \end{pmatrix}$

Step 6: Apply the Initial State Vector

Since $\pi_0 = (0,\ 1)$, we pick out the second row of $P^2$ (the row corresponding to starting state $R$):

$\pi_2 = \pi_0 P^2 = \begin{pmatrix} 0 & 1 \end{pmatrix}\begin{pmatrix} 7/9 & 2/9 \\ 2/3 & 1/3 \end{pmatrix} = \begin{pmatrix} 2/3 & 1/3 \end{pmatrix}$

So after two days: $P(T)=\frac{2}{3}$, $P(R)=\frac{1}{3}$.

Verification

Cross-check with a direct step-by-step method:

• Day 0: He reads ($R$), for certain.
• Day 1: Since he read on Day 0, tomorrow he certainly watches TV, so Day 1 $= T$ with probability $1$.
• Day 2: Since Day 1 was $T$, on Day 2: $P(T)=\frac{2}{3}$, $P(R)=\frac{1}{3}$.

This matches Step 6 exactly, confirming the matrix method.

Question 2 — Hypothesis Testing: One-Sample z-Test for Bulb Lifetime

A factory claims that the average lifetime of its bulbs is 1200 hours. A random sample of 49 bulbs gives:

• Sample mean: $\bar{x} = 1202.62$ hours
• Population standard deviation (known): $\sigma = 7$ hours
• Significance level: $\alpha = 0.05$

Assume the population standard deviation is known, so use a $z$-test.

Concept Needed

A one-sample $z$-test is used to test a claim about a population mean $\mu$ when the population standard deviation $\sigma$ is known.

$z = \frac{\bar{x} - \mu_0}{\sigma/\sqrt{n}}$

Compare the computed $z$ with the critical value from the standard normal ($z$) table, or compare the $p$-value with $\alpha$.

Given / Required

Claimed (population) mean $\mu_0 = 1200$ hours, sample size $n = 49$, sample mean $\bar{x} = 1202.62$ hours, population standard deviation $\sigma = 7$ hours, significance level $\alpha = 0.05$.

Required: test whether the factory's claim ($\mu = 1200$ hours) is statistically supported by the sample data, at $\alpha = 0.05$.

Step 1: State the Hypotheses

$H_0: \mu = 1200 \text{ hours (the factory's claim is true)}$

$H_1: \mu \neq 1200 \text{ hours (two-tailed test — the true mean differs from } 1200\text{)}$

Step 2: Choose the Test and Significance Level

Since $\sigma$ is known and $n=49$ is reasonably large, use the $z$-test. This is a two-tailed test at $\alpha = 0.05$, so each tail has area $\alpha/2 = 0.025$. The critical value is $z_{0.025} = 1.96$ (standard table value).

Step 3: Compute the Test Statistic

Standard error:

$\frac{\sigma}{\sqrt{n}} = \frac{7}{\sqrt{49}} = \frac{7}{7} = 1$

Test statistic:

$z = \frac{1202.62 - 1200}{1} = \frac{2.62}{1} = 2.62$

Step 4: Decision Rule (Critical Value Method)

Reject $H_0$ if $|z| > z_{0.025} = 1.96$. Here, $|z| = 2.62 > 1.96$, so reject $H_0$.

Verification: Cross-Check with the p-Value Method

From the $z$-table, for $z = 2.62$: $\Phi(2.62) = 0.9956$. So the one-tail area beyond $z=2.62$ is:

$P(Z > 2.62) = 1 - 0.9956 = 0.0044$

For a two-tailed test, double this:

$p\text{-value} = 2 \times 0.0044 = 0.0088$

Since $p\text{-value} = 0.0088 < \alpha = 0.05$, we again reject $H_0$ — this matches Step 4 exactly, confirming the decision.

Step 5: Conclusion

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