Sample Solution

MST-004 Solved Assignment 2025

1. State whether the following statements are True or False. Give reason in support of your answer:
   $(5\times 2=10)$$(5\times 2=10)$(5xx2=10)(5 \times 2=10)$(5\times 2=10)$
   (a) If the probability of non rejection of ${\mathrm{H}}_0$${\mathrm{H}}_0$H_(0)\mathrm{H}_0${\mathrm{H}}_0$ when ${\mathrm{H}}_1$${\mathrm{H}}_1$H_(1)\mathrm{H}_1${\mathrm{H}}_1$ is true is 0.4 then power of the test will be 0.6 .
   (b) If $T_1$$T_1$T_(1)T_1$T_1$ and $T_2$$T_2$T_(2)T_2$T_2$ are two estimators of the parameter $\theta$$\theta$theta\theta$\theta$ such that $\mathrm{Var}\left(T_1\right)=1/\mathrm{n}$$\mathrm{Var}\left(T_1\right)=1/\mathrm{n}$Var(T_(1))=1//n\operatorname{Var}\left(T_1\right)=1 / \mathrm{n}$\mathrm{Var}\left(T_1\right)=1/\mathrm{n}$ and $\mathrm{Var}\left({\mathrm{T}}_2\right)=\mathrm{n}$$\mathrm{Var}\left({\mathrm{T}}_2\right)=\mathrm{n}$Var(T_(2))=n\operatorname{Var}\left(\mathrm{T}_2\right)=\mathrm{n}$\mathrm{Var}\left({\mathrm{T}}_2\right)=\mathrm{n}$ then ${\mathrm{T}}_1$${\mathrm{T}}_1$T_(1)\mathrm{T}_1${\mathrm{T}}_1$ is more efficient than ${\mathrm{T}}_2$${\mathrm{T}}_2$T_(2)\mathrm{T}_2${\mathrm{T}}_2$.
   (c) A $95\mathrm{\%}$$95\mathrm{\%}$95%95 \%$95\mathrm{\%}$ confidence interval is smaller than $99\mathrm{\%}$$99\mathrm{\%}$99%99 \%$99\mathrm{\%}$ confidence interval.
   (d) If the level of significance is the same, the area of the rejection region in a two-tailed test is less than that in a one-tailed test.
   (e) Non parametric tests are more powerful than the parametric tests.
2. If a finite population has four elements: $6,1,3,2$$6,1,3,2$6,1,3,26,1,3,2$6,1,3,2$.
   (a) How many different samples of size $\mathrm{n}=2$$\mathrm{n}=2$n=2\mathrm{n}=2$\mathrm{n}=2$ can be selected from this population if you sample without replacement?
   (b) List all possible samples of size $\mathrm{n}=2$$\mathrm{n}=2$n=2\mathrm{n}=2$\mathrm{n}=2$.
   (c) Compute the sample mean for each of the samples given in part $b$$b$bb$b$.
   (d) Find the sampling distribution of $\bar{\mathrm{x}}$$\overline{\mathrm{x}}$bar(x)\overline{\mathrm{x}}$\bar{\mathrm{x}}$ and draw the histogram.
   (e) Compute standard error.
   (f) If all four population values are equally likely, calculate the value of the population mean ${}^{\mu }$${}^{\mu }$^(mu){ }^\mu${}^{\mu }$. Do any of the samples listed in part (b) produce a value of $\overline{x}$$\overline{x}$bar(x)\bar{x}$\overline{x}$ exactly equal to ${\mu }_{\text{?}}$${\mu }_{\text{?}}$mu_(“? “)\mu_{\text {? }}${\mu }_{\text{?}}$
3. A study was conducted to compare the mean numbers of police emergency calls per 8 -hour shift in two districts of a large city. Samples of 1008 -hour shifts were randomly selected from the police records for each of the two regions and the number of emergency calls was recorded for each shift. The sample statistics are listed here:

4. A bond proposal for school construction will be submitted to the voters at the next municipal election. A major portion of the money derived form this bond issue will be used to build schools in a rapidly developing selection of the city, and the remainder will be used to renovate and update school buildings in the rest of the city. To assess the viability of the bond proposal, a random sample of ${\mathrm{n}}_1=50$${\mathrm{n}}_1=50$n_(1)=50\mathrm{n}_1=50${\mathrm{n}}_1=50$ residents in the developing section and ${\mathrm{n}}_2=100$${\mathrm{n}}_2=100$n_(2)=100\mathrm{n}_2=100${\mathrm{n}}_2=100$ residents from the other parts of the city were asked whether they plan to vote for the proposal. The results are tabulated below

6. If magnitude of earthquakes recorded in a region of a country follows a distribution with parameter $\mu$$\mu$mu\mu$\mu$ whose pdf is given below:

Solution

Question:-1

State whether the following statements are True or False. Give reason in support of your answer:

Answer:

• Evaluation: True.
• Justification: The power of a test is the probability of rejecting $H_0$$H_0$H_(0)H_0$H_0$ when $H_1$$H_1$H_(1)H_1$H_1$ is true, i.e., $\text{Power}=P(\text{Reject}{H}_0\mid H_1\text{true})$$\text{Power}=P(\text{Reject}{H}_0\mid H_1\text{true})$“Power”=P(“Reject “H_(0)∣H_(1)” true”)\text{Power} = P(\text{Reject } H_0 \mid H_1 \text{ true})$\text{Power}=P(\text{Reject}{H}_0\mid H_1\text{true})$. The probability of non-rejection of $H_0$$H_0$H_(0)H_0$H_0$ when $H_1$$H_1$H_(1)H_1$H_1$ is true is $P(\text{Fail to reject}{H}_0\mid H_1\text{true})=1-\text{Power}$$P(\text{Fail to reject}{H}_0\mid H_1\text{true})=1-\text{Power}$P(“Fail to reject “H_(0)∣H_(1)” true”)=1-“Power”P(\text{Fail to reject } H_0 \mid H_1 \text{ true}) = 1 – \text{Power}$P(\text{Fail to reject}{H}_0\mid H_1\text{true})=1-\text{Power}$. Given this probability is 0.4, we have $1-\text{Power}=0.4$$1-\text{Power}=0.4$1-“Power”=0.41 – \text{Power} = 0.4$1-\text{Power}=0.4$, so $\text{Power}=1-0.4=0.6$$\text{Power}=1-0.4=0.6$“Power”=1-0.4=0.6\text{Power} = 1 – 0.4 = 0.6$\text{Power}=1-0.4=0.6$.

• Evaluation: True, assuming both estimators are unbiased.
• Justification: The efficiency of an estimator (for unbiased estimators) is determined by its variance; a lower variance indicates a more efficient estimator. Since $\mathrm{Var}(T_1)=1/n$$\mathrm{Var}(T_1)=1/n$Var(T_(1))=1//n\operatorname{Var}(T_1) = 1/n$\mathrm{Var}(T_1)=1/n$ and $\mathrm{Var}(T_2)=n$$\mathrm{Var}(T_2)=n$Var(T_(2))=n\operatorname{Var}(T_2) = n$\mathrm{Var}(T_2)=n$, and assuming $n>0$$n>0$n > 0n > 0$n>0$, we have $1/n<n$$1/n<n$1//n < n1/n < n$1/n<n$ for $n>1$$n>1$n > 1n > 1$n>1$ (and even for $0<n<1$$0<n<1$0 < n < 10 < n < 1$0<n<1$, $1/n>n$$1/n>n$1//n > n1/n > n$1/n>n$, but variance is typically positive and context suggests $n$$n$nn$n$ is a sample size, so $n\ge 1$$n\ge 1$n >= 1n \geq 1$n\ge 1$). Thus, $\mathrm{Var}(T_1)<\mathrm{Var}(T_2)$$\mathrm{Var}(T_1)<\mathrm{Var}(T_2)$Var(T_(1)) < Var(T_(2))\operatorname{Var}(T_1) < \operatorname{Var}(T_2)$\mathrm{Var}(T_1)<\mathrm{Var}(T_2)$, so $T_1$$T_1$T_(1)T_1$T_1$ is more efficient. If the estimators are biased, efficiency comparisons require mean squared error, but the context suggests unbiasedness is assumed.

• Evaluation: True.
• Justification: The width of a confidence interval is proportional to the critical value from the standard normal or t-distribution. For a 95% confidence interval, the critical value (e.g., $z\approx 1.96$$z\approx 1.96$z~~1.96z \approx 1.96$z\approx 1.96$ for large samples) is smaller than for a 99% confidence interval (e.g., $z\approx 2.576$$z\approx 2.576$z~~2.576z \approx 2.576$z\approx 2.576$). Since the width is typically of the form $\text{estimate}\pm z\cdot \text{standard error}$$\text{estimate}\pm z\cdot \text{standard error}$“estimate”+-z*”standard error”\text{estimate} \pm z \cdot \text{standard error}$\text{estimate}\pm z\cdot \text{standard error}$, a smaller $z$$z$zz$z$ results in a narrower interval, assuming the same standard error and sample size.

• Evaluation: False.
• Justification: The level of significance $\alpha$$\alpha$alpha\alpha$\alpha$ represents the total probability of the rejection region. In a one-tailed test, the entire rejection region is in one tail, so the area is $\alpha$$\alpha$alpha\alpha$\alpha$. In a two-tailed test, the rejection region is split between both tails, with each tail having area $\alpha /2$$\alpha /2$alpha//2\alpha/2$\alpha /2$, so the total area is still $\alpha$$\alpha$alpha\alpha$\alpha$. Thus, the total area of the rejection region is the same in both tests, not less in the two-tailed test. For example, if $\alpha =0.05$$\alpha =0.05$alpha=0.05\alpha = 0.05$\alpha =0.05$, a one-tailed test has a rejection region of 0.05 in one tail, while a two-tailed test has 0.025 in each tail, totaling 0.05.

• Evaluation: False.
• Justification: Non-parametric tests are generally less powerful than parametric tests when the assumptions of the parametric tests (e.g., normality, equal variances) are met. Parametric tests use specific distributional assumptions to model the data, leading to higher power for detecting true effects. Non-parametric tests, which do not assume a specific distribution, are more robust but sacrifice power. For example, the Mann-Whitney U test (non-parametric) is less powerful than the t-test (parametric) for normally distributed data with equal variances. However, non-parametric tests can be more powerful when parametric assumptions are violated, but the statement is absolute and does not account for this condition.