Details For MST-004 Previous Year Paper Solution

IGNOU MST-004 Previous Year Paper Solution

MST-004 Previous Year Paper Solution – Sample

MST-004 Dec 2023

Question:-01

1.State whether the following statements are True or False. Give reasons in support of your answers :
(a) If sample size of a survey has increased 3 times, then the standard error will be increased 3 times.

Answer:

The statement "If the sample size of a survey has increased 3 times, then the standard error will be increased 3 times" is false.
To justify why this statement is false, we can look at the relationship between the standard error of the mean and the sample size. The standard error of the mean (SEM) is given by the formula:
$$\text{SEM}=\frac{\sigma }{\sqrt{n}}$$$$\text{SEM}=\frac{\sigma }{\sqrt{n}}$$“SEM”=(sigma)/(sqrtn)text{SEM} = frac{sigma}{sqrt{n}}$$\text{SEM}=\frac{\sigma }{\sqrt{n}}$$
where $\sigma$$\sigma$sigmasigma$\sigma$ is the population standard deviation and $n$$n$nn$n$ is the sample size. This formula shows that the standard error of the mean is inversely proportional to the square root of the sample size.
When the sample size $n$$n$nn$n$ is increased by a factor of 3, the new sample size becomes $3n$$3n$3n3n$3n$. Plugging this into the formula for the standard error gives:
$${\text{SEM}}_{\text{new}}=\frac{\sigma }{\sqrt{3n}}$$$${\text{SEM}}_{\text{new}}=\frac{\sigma }{\sqrt{3n}}$$“SEM”_(“new”)=(sigma)/(sqrt(3n))text{SEM}_{text{new}} = frac{sigma}{sqrt{3n}}$${\text{SEM}}_{\text{new}}=\frac{\sigma }{\sqrt{3n}}$$
To see how the standard error changes, compare the new SEM with the original SEM:
$${\text{SEM}}_{\text{new}}=\frac{\sigma }{\sqrt{3n}}=\frac{\sigma }{\sqrt{3}\sqrt{n}}=\frac{1}{\sqrt{3}}\frac{\sigma }{\sqrt{n}}=\frac{1}{\sqrt{3}}{\text{SEM}}_{\text{old}}$$$${\text{SEM}}_{\text{new}}=\frac{\sigma }{\sqrt{3n}}=\frac{\sigma }{\sqrt{3}\sqrt{n}}=\frac{1}{\sqrt{3}}\frac{\sigma }{\sqrt{n}}=\frac{1}{\sqrt{3}}{\text{SEM}}_{\text{old}}$$“SEM”_(“new”)=(sigma)/(sqrt(3n))=(sigma)/(sqrt3sqrtn)=(1)/(sqrt3)(sigma)/(sqrtn)=(1)/(sqrt3)”SEM”_(“old”)text{SEM}_{text{new}} = frac{sigma}{sqrt{3n}} = frac{sigma}{sqrt{3} sqrt{n}} = frac{1}{sqrt{3}} frac{sigma}{sqrt{n}} = frac{1}{sqrt{3}} text{SEM}_{text{old}}$${\text{SEM}}_{\text{new}}=\frac{\sigma }{\sqrt{3n}}=\frac{\sigma }{\sqrt{3}\sqrt{n}}=\frac{1}{\sqrt{3}}\frac{\sigma }{\sqrt{n}}=\frac{1}{\sqrt{3}}{\text{SEM}}_{\text{old}}$$
Since $\frac{1}{\sqrt{3}}\approx 0.577$$\frac{1}{\sqrt{3}}\approx 0.577$(1)/(sqrt3)~~0.577frac{1}{sqrt{3}} approx 0.577$\frac{1}{\sqrt{3}}\approx 0.577$, the new standard error is approximately $0.577$$0.577$0.5770.577$0.577$ times the original standard error. This means that the standard error does not increase three times; rather, it decreases by a factor of about $\sqrt{3}$$\sqrt{3}$sqrt3sqrt{3}$\sqrt{3}$ or decreases to about 57.7% of its original value.
Therefore, the statement is false because increasing the sample size by three times results in the standard error decreasing, not increasing, and specifically decreasing by a factor of about $\sqrt{3}$$\sqrt{3}$sqrt3sqrt{3}$\sqrt{3}$.

(b) If probability density function of a random variable $\mathrm{X}$$\mathrm{X}$Xmathrm{X}$\mathrm{X}$ follows $\mathrm{F}$$\mathrm{F}$Fmathrm{F}$\mathrm{F}$-distribution
$$f(x)=\frac{1}{(1+x{)}^2};0<x<\mathrm{\infty },$$$$f(x)=\frac{1}{(1+x{)}^2};0<x<\mathrm{\infty },$$f(x)=(1)/((1+x)^(2));0 < x < oo,f(x)=frac{1}{(1+x)^2} ; 0<x<infty,$$f(x)=\frac{1}{(1+x{)}^2};0<x<\mathrm{\infty },$$
then the degree of freedom of the distribution will be $(2,2)$$(2,2)$(2,2)(2,2)$(2,2)$.

Answer:

The statement "If the probability density function of a random variable $X$$X$XX$X$ follows $F$$F$FF$F$-distribution $f(x)=\frac{1}{(1+x{)}^2};0<x<\mathrm{\infty }$$f(x)=\frac{1}{(1+x{)}^2};0<x<\mathrm{\infty }$f(x)=(1)/((1+x)^(2));0 < x < oof(x) = frac{1}{(1+x)^2}; 0 < x < infty$f(x)=\frac{1}{(1+x{)}^2};0<x<\mathrm{\infty }$, then the degree of freedom of the distribution will be $(2,2)$$(2,2)$(2,2)(2,2)$(2,2)$" is true.
To justify this, let’s analyze the given information:
The probability density function (PDF) provided is:
$$f(x)=\frac{1}{(1+x{)}^2},0<x<\mathrm{\infty }$$$$f(x)=\frac{1}{(1+x{)}^2},0<x<\mathrm{\infty }$$f(x)=(1)/((1+x)^(2)),quad0 < x < oof(x) = frac{1}{(1+x)^2}, quad 0 < x < infty$$f(x)=\frac{1}{(1+x{)}^2},0<x<\mathrm{\infty }$$
The PDF of the $F$$F$FF$F$-distribution with degrees of freedom $v_1$$v_1$v_(1)v_1$v_1$ and $v_2$$v_2$v_(2)v_2$v_2$ is given by:
$$f(x)=\left(\frac{v_1/v_2}{B(\frac{v_1}{2},\frac{v_2}{2})}\right){\left(\frac{v_1}{v_2}\right)}^{v_1/2}{x}^{(v_1/2)-1}{(1+\frac{v_1}{v_2}x)}^{-(v_1+v_2)/2}$$$$f(x)=\left(\frac{v_1/v_2}{B\left(\frac{v_1}{2},\frac{v_2}{2}\right)}\right){\left(\frac{v_1}{v_2}\right)}^{v_1/2}{x}^{(v_1/2)-1}{\left(1+\frac{v_1}{v_2}x\right)}^{-(v_1+v_2)/2}$$f(x)=((v_(1)//v_(2))/(B((v_(1))/(2),(v_(2))/(2))))((v_(1))/(v_(2)))^(v_(1)//2)x^((v_(1)//2)-1)(1+(v_(1))/(v_(2))x)^(-(v_(1)+v_(2))//2)f(x) = left( frac{v_1 / v_2}{Bleft( frac{v_1}{2}, frac{v_2}{2} right)} right) left( frac{v_1}{v_2} right)^{v_1 / 2} x^{(v_1 / 2) – 1} left( 1 + frac{v_1}{v_2} x right)^{-(v_1 + v_2) / 2}$$f(x)=\left(\frac{v_1/v_2}{B(\frac{v_1}{2},\frac{v_2}{2})}\right){\left(\frac{v_1}{v_2}\right)}^{v_1/2}{x}^{(v_1/2)-1}{(1+\frac{v_1}{v_2}x)}^{-(v_1+v_2)/2}$$
Given:
$$f(x)=\frac{1}{(1+x{)}^2},0<x<\mathrm{\infty }$$$$f(x)=\frac{1}{(1+x{)}^2},0<x<\mathrm{\infty }$$f(x)=(1)/((1+x)^(2)),quad0 < x < oof(x) = frac{1}{(1+x)^2}, quad 0 < x < infty$$f(x)=\frac{1}{(1+x{)}^2},0<x<\mathrm{\infty }$$
We can rewrite the given PDF as:
$$f(x)=\frac{(2/2{)}^{2/2}{x}^{(2/2)-1}}{B(2/2,2/2){(1+\frac{2}{2}x)}^{(2+2)/2}}$$$$f(x)=\frac{(2/2{)}^{2/2}{x}^{(2/2)-1}}{B(2/2,2/2){\left(1+\frac{2}{2}x\right)}^{(2+2)/2}}$$f(x)=((2//2)^(2//2)x^((2//2)-1))/(B(2//2,2//2)(1+(2)/(2)x)^((2+2)//2))f(x) = frac{(2/2)^{2/2} x^{(2/2) – 1}}{B(2/2, 2/2) left(1 + frac{2}{2} x right)^{(2 + 2)/2}}$$f(x)=\frac{(2/2{)}^{2/2}{x}^{(2/2)-1}}{B(2/2,2/2){(1+\frac{2}{2}x)}^{(2+2)/2}}$$
Comparing this with the standard form of the $F$$F$FF$F$-distribution PDF, we observe that the given PDF matches if $v_1=2$$v_1=2$v_(1)=2v_1 = 2$v_1=2$ and $v_2=2$$v_2=2$v_(2)=2v_2 = 2$v_2=2$.
Thus, the degrees of freedom for the given distribution are $(2,2)$$(2,2)$(2,2)(2,2)$(2,2)$.
Therefore, the statement is true.

(c) For testing ${\mathrm{H}}_0:\theta =2$${\mathrm{H}}_0:\theta =2$H_(0):theta=2mathrm{H}_0: theta=2${\mathrm{H}}_0:\theta =2$ against ${\mathrm{H}}_1:\theta =3$${\mathrm{H}}_1:\theta =3$H_(1):theta=3mathrm{H}_1: theta=3${\mathrm{H}}_1:\theta =3$, the $\mathrm{p}\mathrm{d}\mathrm{f}$$\mathrm{p}\mathrm{d}\mathrm{f}$pdfmathrm{pdf}$\mathrm{p}\mathrm{d}\mathrm{f}$ of the variable is given by
$$f(x,\theta )=\frac{1}{\theta };0\le x\le \theta$$$$f(x,\theta )=\frac{1}{\theta };0\le x\le \theta$$f(x,theta)=(1)/(theta);0 <= x <= thetaf(x, theta)=frac{1}{theta} ; 0 leq x leq theta$$f(x,\theta )=\frac{1}{\theta };0\le x\le \theta$$
If the critical region is $x\ge 0.6$$x\ge 0.6$x >= 0.6x geq 0.6$x\ge 0.6$, the size of the test will be 0.6 .

Answer:

The statement "For testing ${\mathrm{H}}_0:\theta =2$${\mathrm{H}}_0:\theta =2$H_(0):theta=2mathrm{H}_0: theta=2${\mathrm{H}}_0:\theta =2$ against ${\mathrm{H}}_1:\theta =3$${\mathrm{H}}_1:\theta =3$H_(1):theta=3mathrm{H}_1: theta=3${\mathrm{H}}_1:\theta =3$, the $\mathrm{p}\mathrm{d}\mathrm{f}$$\mathrm{p}\mathrm{d}\mathrm{f}$pdfmathrm{pdf}$\mathrm{p}\mathrm{d}\mathrm{f}$ of the variable is given by $f(x,\theta )=\frac{1}{\theta };0\le x\le \theta$$f(x,\theta )=\frac{1}{\theta };0\le x\le \theta$f(x,theta)=(1)/(theta);0 <= x <= thetaf(x, theta)=frac{1}{theta} ; 0 leq x leq theta$f(x,\theta )=\frac{1}{\theta };0\le x\le \theta$. If the critical region is $x\ge 0.6$$x\ge 0.6$x >= 0.6x geq 0.6$x\ge 0.6$, the size of the test will be 0.6" is false.
To justify this, let’s understand the concepts of hypothesis testing and the size of a test.
The size of a test, also known as the significance level (α), is the probability of rejecting the null hypothesis ${\mathrm{H}}_0$${\mathrm{H}}_0$H_(0)mathrm{H}_0${\mathrm{H}}_0$ when it is actually true. This is also known as the Type I error rate.
Given:

• ${\mathrm{H}}_0:\theta =2$${\mathrm{H}}_0:\theta =2$H_(0):theta=2mathrm{H}_0: theta=2${\mathrm{H}}_0:\theta =2$
• ${\mathrm{H}}_1:\theta =3$${\mathrm{H}}_1:\theta =3$H_(1):theta=3mathrm{H}_1: theta=3${\mathrm{H}}_1:\theta =3$
• The probability density function (pdf) of the variable: $f(x,\theta )=\frac{1}{\theta }$$f(x,\theta )=\frac{1}{\theta }$f(x,theta)=(1)/(theta)f(x, theta)=frac{1}{theta}$f(x,\theta )=\frac{1}{\theta }$, for $0\le x\le \theta$$0\le x\le \theta$0 <= x <= theta0 leq x leq theta$0\le x\le \theta$

The critical region is $x\ge 0.6$$x\ge 0.6$x >= 0.6x geq 0.6$x\ge 0.6$.
First, let’s find the size of the test under the null hypothesis ${\mathrm{H}}_0:\theta =2$${\mathrm{H}}_0:\theta =2$H_(0):theta=2mathrm{H}_0: theta=2${\mathrm{H}}_0:\theta =2$.
The pdf under ${\mathrm{H}}_0$${\mathrm{H}}_0$H_(0)mathrm{H}_0${\mathrm{H}}_0$ is:
$$f(x,\theta =2)=\frac{1}{2},0\le x\le 2$$$$f(x,\theta =2)=\frac{1}{2},0\le x\le 2$$f(x,theta=2)=(1)/(2),quad0 <= x <= 2f(x, theta=2) = frac{1}{2}, quad 0 leq x leq 2$$f(x,\theta =2)=\frac{1}{2},0\le x\le 2$$
To find the size of the test, we need to calculate the probability that $x\ge 0.6$$x\ge 0.6$x >= 0.6x geq 0.6$x\ge 0.6$ when $\theta =2$$\theta =2$theta=2theta = 2$\theta =2$:
$$\text{Size of the test}=P(x\ge 0.6\mid \theta =2)={\int }_{0.6}^2\frac{1}{2}dx$$$$\text{Size of the test}=P(x\ge 0.6\mid \theta =2)={\int }_{0.6}^2 \frac{1}{2}dx$$“Size of the test”=P(x >= 0.6∣theta=2)=int_(0.6)^(2)(1)/(2)dxtext{Size of the test} = P(x geq 0.6 mid theta=2) = int_{0.6}^{2} frac{1}{2} , dx$$\text{Size of the test}=P(x\ge 0.6\mid \theta =2)={\int }_{0.6}^2\frac{1}{2}dx$$
Calculating the integral:
$${\int }_{0.6}^2\frac{1}{2}dx=\frac{1}{2}{\left[x\right]}_{0.6}^2=\frac{1}{2}(2-0.6)=\frac{1}{2}\times 1.4=0.7$$$${\int }_{0.6}^2 \frac{1}{2}dx=\frac{1}{2}{\left[x\right]}_{0.6}^2=\frac{1}{2}(2-0.6)=\frac{1}{2}\times 1.4=0.7$$int_(0.6)^(2)(1)/(2)dx=(1)/(2)[x]_(0.6)^(2)=(1)/(2)(2-0.6)=(1)/(2)xx1.4=0.7int_{0.6}^{2} frac{1}{2} , dx = frac{1}{2} left[ x right]_{0.6}^{2} = frac{1}{2} (2 – 0.6) = frac{1}{2} times 1.4 = 0.7$${\int }_{0.6}^2\frac{1}{2}dx=\frac{1}{2}{\left[x\right]}_{0.6}^2=\frac{1}{2}(2-0.6)=\frac{1}{2}\times 1.4=0.7$$
Therefore, the size of the test is 0.7, not 0.6.
Hence, the statement is false because the size of the test under the null hypothesis $\theta =2$$\theta =2$theta=2theta = 2$\theta =2$ with the critical region $x\ge 0.6$$x\ge 0.6$x >= 0.6x geq 0.6$x\ge 0.6$ is 0.7, not 0.6.

(d) Kruskal-Wallis test is a non-parametric version of two-way analysis of variance.

Answer:

The statement "Kruskal-Wallis test is a non-parametric version of two-way analysis of variance" is false.
To justify this, let’s review the purposes and uses of the Kruskal-Wallis test and two-way analysis of variance (ANOVA).

Kruskal-Wallis Test:

• The Kruskal-Wallis test is a non-parametric method used to test whether there are statistically significant differences between the medians of three or more independent groups.
• It is the non-parametric alternative to one-way ANOVA and is used when the assumptions of one-way ANOVA (such as normality) are not met.
• The Kruskal-Wallis test ranks all the data from all groups together and then compares the sum of ranks between the groups.

Two-Way ANOVA:

• Two-way ANOVA is a parametric method used to determine if there are any significant differences between the means of three or more groups, considering two independent variables (factors).
• It can test the interaction between the two factors, as well as the individual effects of each factor.

Key Differences:

• Number of Factors: The Kruskal-Wallis test deals with one factor with multiple levels (one-way analysis), while two-way ANOVA involves two factors.
• Type of Data: Kruskal-Wallis is non-parametric and uses ranks, making it suitable for ordinal data or data that do not meet the assumptions of normality. Two-way ANOVA is parametric and assumes normal distribution of the residuals.

Correct Non-Parametric Alternative:

• The non-parametric alternative to two-way ANOVA is the Friedman test, which is used for analyzing differences in treatments across multiple test attempts (blocks) when the data is not normally distributed.

Therefore, the Kruskal-Wallis test is not a non-parametric version of two-way ANOVA; it is a non-parametric version of one-way ANOVA. The statement is false.

(e) A sample of size 4 is drawn randomly $({\mathrm{X}}_1,{\mathrm{X}}_2,{\mathrm{X}}_3$$\left({\mathrm{X}}_1,{\mathrm{X}}_2,{\mathrm{X}}_3\right.$(X_(1),X_(2),X_(3):}left(mathrm{X}_1, mathrm{X}_2, mathrm{X}_3right.$({\mathrm{X}}_1,{\mathrm{X}}_2,{\mathrm{X}}_3$ and ${\mathrm{X}}_4$${\mathrm{X}}_4$X_(4)mathrm{X}_4${\mathrm{X}}_4$ ) from a normal population with unknown mean $\mu$$\mu$mumu$\mu$, then $\frac{{\mathrm{X}}_1+2{\mathrm{X}}_2+3{\mathrm{X}}_3+{\mathrm{X}}_4}{7}$$\frac{{\mathrm{X}}_1+2{\mathrm{X}}_2+3{\mathrm{X}}_3+{\mathrm{X}}_4}{7}$(X_(1)+2X_(2)+3X_(3)+X_(4))/(7)frac{mathrm{X}_1+2 mathrm{X}_2+3 mathrm{X}_3+mathrm{X}_4}{7}$\frac{{\mathrm{X}}_1+2{\mathrm{X}}_2+3{\mathrm{X}}_3+{\mathrm{X}}_4}{7}$ is an unbiased estimator of $\mu$$\mu$mumu$\mu$.

Answer:

To determine whether $\frac{{\mathrm{X}}_1+2{\mathrm{X}}_2+3{\mathrm{X}}_3+{\mathrm{X}}_4}{7}$$\frac{{\mathrm{X}}_1+2{\mathrm{X}}_2+3{\mathrm{X}}_3+{\mathrm{X}}_4}{7}$(X_(1)+2X_(2)+3X_(3)+X_(4))/(7)frac{mathrm{X}_1 + 2mathrm{X}_2 + 3mathrm{X}_3 + mathrm{X}_4}{7}$\frac{{\mathrm{X}}_1+2{\mathrm{X}}_2+3{\mathrm{X}}_3+{\mathrm{X}}_4}{7}$ is an unbiased estimator of the population mean $\mu$$\mu$mumu$\mu$, we need to check if the expected value of this estimator equals $\mu$$\mu$mumu$\mu$.
Given that ${\mathrm{X}}_1,{\mathrm{X}}_2,{\mathrm{X}}_3$${\mathrm{X}}_1,{\mathrm{X}}_2,{\mathrm{X}}_3$X_(1),X_(2),X_(3)mathrm{X}_1, mathrm{X}_2, mathrm{X}_3${\mathrm{X}}_1,{\mathrm{X}}_2,{\mathrm{X}}_3$, and ${\mathrm{X}}_4$${\mathrm{X}}_4$X_(4)mathrm{X}_4${\mathrm{X}}_4$ are drawn from a normal population with unknown mean $\mu$$\mu$mumu$\mu$, the expected value of each ${\mathrm{X}}_i$${\mathrm{X}}_i$X_(i)mathrm{X}_i${\mathrm{X}}_i$ is $\mu$$\mu$mumu$\mu$.
The estimator in question is:
$$\hat{\mu }=\frac{{\mathrm{X}}_1+2{\mathrm{X}}_2+3{\mathrm{X}}_3+{\mathrm{X}}_4}{7}$$$$\widehat{\mu }=\frac{{\mathrm{X}}_1+2{\mathrm{X}}_2+3{\mathrm{X}}_3+{\mathrm{X}}_4}{7}$$hat(mu)=(X_(1)+2X_(2)+3X_(3)+X_(4))/(7)hat{mu} = frac{mathrm{X}_1 + 2mathrm{X}_2 + 3mathrm{X}_3 + mathrm{X}_4}{7}$$\hat{\mu }=\frac{{\mathrm{X}}_1+2{\mathrm{X}}_2+3{\mathrm{X}}_3+{\mathrm{X}}_4}{7}$$
To check if this is an unbiased estimator of $\mu$$\mu$mumu$\mu$, we calculate the expected value of $\hat{\mu }$$\widehat{\mu }$hat(mu)hat{mu}$\hat{\mu }$:
$$E\left(\hat{\mu }\right)=E\left(\frac{{\mathrm{X}}_1+2{\mathrm{X}}_2+3{\mathrm{X}}_3+{\mathrm{X}}_4}{7}\right)$$$$E\left(\widehat{\mu }\right)=E\left(\frac{{\mathrm{X}}_1+2{\mathrm{X}}_2+3{\mathrm{X}}_3+{\mathrm{X}}_4}{7}\right)$$E(( hat(mu)))=E((X_(1)+2X_(2)+3X_(3)+X_(4))/(7))Eleft(hat{mu}right) = Eleft(frac{mathrm{X}_1 + 2mathrm{X}_2 + 3mathrm{X}_3 + mathrm{X}_4}{7}right)$$E\left(\hat{\mu }\right)=E\left(\frac{{\mathrm{X}}_1+2{\mathrm{X}}_2+3{\mathrm{X}}_3+{\mathrm{X}}_4}{7}\right)$$
Using the linearity of expectation, we have:
$$E\left(\hat{\mu }\right)=\frac{1}{7}E({\mathrm{X}}_1+2{\mathrm{X}}_2+3{\mathrm{X}}_3+{\mathrm{X}}_4)$$$$E\left(\widehat{\mu }\right)=\frac{1}{7}E({\mathrm{X}}_1+2{\mathrm{X}}_2+3{\mathrm{X}}_3+{\mathrm{X}}_4)$$E(( hat(mu)))=(1)/(7)E(X_(1)+2X_(2)+3X_(3)+X_(4))Eleft(hat{mu}right) = frac{1}{7} E(mathrm{X}_1 + 2mathrm{X}_2 + 3mathrm{X}_3 + mathrm{X}_4)$$E\left(\hat{\mu }\right)=\frac{1}{7}E({\mathrm{X}}_1+2{\mathrm{X}}_2+3{\mathrm{X}}_3+{\mathrm{X}}_4)$$
$$E\left(\hat{\mu }\right)=\frac{1}{7}(E({\mathrm{X}}_1)+2E({\mathrm{X}}_2)+3E({\mathrm{X}}_3)+E({\mathrm{X}}_4))$$$$E\left(\widehat{\mu }\right)=\frac{1}{7}\left(E({\mathrm{X}}_1)+2E({\mathrm{X}}_2)+3E({\mathrm{X}}_3)+E({\mathrm{X}}_4)\right)$$E(( hat(mu)))=(1)/(7)(E(X_(1))+2E(X_(2))+3E(X_(3))+E(X_(4)))Eleft(hat{mu}right) = frac{1}{7} left( E(mathrm{X}_1) + 2E(mathrm{X}_2) + 3E(mathrm{X}_3) + E(mathrm{X}_4) right)$$E\left(\hat{\mu }\right)=\frac{1}{7}(E({\mathrm{X}}_1)+2E({\mathrm{X}}_2)+3E({\mathrm{X}}_3)+E({\mathrm{X}}_4))$$
Since $E({\mathrm{X}}_i)=\mu$$E({\mathrm{X}}_i)=\mu$E(X_(i))=muE(mathrm{X}_i) = mu$E({\mathrm{X}}_i)=\mu$ for all $i$$i$ii$i$, we get:
$$E\left(\hat{\mu }\right)=\frac{1}{7}(\mu +2\mu +3\mu +\mu )$$$$E\left(\widehat{\mu }\right)=\frac{1}{7}\left(\mu +2\mu +3\mu +\mu \right)$$E(( hat(mu)))=(1)/(7)(mu+2mu+3mu+mu)Eleft(hat{mu}right) = frac{1}{7} left( mu + 2mu + 3mu + mu right)$$E\left(\hat{\mu }\right)=\frac{1}{7}(\mu +2\mu +3\mu +\mu )$$
$$E\left(\hat{\mu }\right)=\frac{1}{7}\left(7\mu \right)$$$$E\left(\widehat{\mu }\right)=\frac{1}{7}\left(7\mu \right)$$E(( hat(mu)))=(1)/(7)(7mu)Eleft(hat{mu}right) = frac{1}{7} left( 7mu right)$$E\left(\hat{\mu }\right)=\frac{1}{7}\left(7\mu \right)$$
$$E\left(\hat{\mu }\right)=\mu$$$$E\left(\widehat{\mu }\right)=\mu$$E(( hat(mu)))=muEleft(hat{mu}right) = mu$$E\left(\hat{\mu }\right)=\mu$$
Thus, the expected value of the estimator $\frac{{\mathrm{X}}_1+2{\mathrm{X}}_2+3{\mathrm{X}}_3+{\mathrm{X}}_4}{7}$$\frac{{\mathrm{X}}_1+2{\mathrm{X}}_2+3{\mathrm{X}}_3+{\mathrm{X}}_4}{7}$(X_(1)+2X_(2)+3X_(3)+X_(4))/(7)frac{mathrm{X}_1 + 2mathrm{X}_2 + 3mathrm{X}_3 + mathrm{X}_4}{7}$\frac{{\mathrm{X}}_1+2{\mathrm{X}}_2+3{\mathrm{X}}_3+{\mathrm{X}}_4}{7}$ is equal to $\mu$$\mu$mumu$\mu$, which means it is an unbiased estimator of $\mu$$\mu$mumu$\mu$.
Therefore, the statement "A sample of size 4 is drawn randomly (${\mathrm{X}}_1,{\mathrm{X}}_2,{\mathrm{X}}_3$${\mathrm{X}}_1,{\mathrm{X}}_2,{\mathrm{X}}_3$X_(1),X_(2),X_(3)mathrm{X}_1, mathrm{X}_2, mathrm{X}_3${\mathrm{X}}_1,{\mathrm{X}}_2,{\mathrm{X}}_3$, and ${\mathrm{X}}_4$${\mathrm{X}}_4$X_(4)mathrm{X}_4${\mathrm{X}}_4$) from a normal population with unknown mean $\mu$$\mu$mumu$\mu$, then $\frac{{\mathrm{X}}_1+2{\mathrm{X}}_2+3{\mathrm{X}}_3+{\mathrm{X}}_4}{7}$$\frac{{\mathrm{X}}_1+2{\mathrm{X}}_2+3{\mathrm{X}}_3+{\mathrm{X}}_4}{7}$(X_(1)+2X_(2)+3X_(3)+X_(4))/(7)frac{mathrm{X}_1 + 2mathrm{X}_2 + 3mathrm{X}_3 + mathrm{X}_4}{7}$\frac{{\mathrm{X}}_1+2{\mathrm{X}}_2+3{\mathrm{X}}_3+{\mathrm{X}}_4}{7}$ is an unbiased estimator of $\mu$$\mu$mumu$\mu$" is true.

Question:-02

2.The systolic blood pressure (SBP) of five women are given as follows :
$$120,110,130,140,100$$$$120,110,130,140,100$$120,110,130,140,100120,110,130,140,100$$120,110,130,140,100$$
(a) How many samples of size 2 can be drawn without replacement? Write them.
(b) Compute the mean of all samples of size 2 and set up the sampling distribution of the sample mean.
(c) Compute the expected value of the sample mean.
(d) How many samples of the same size 2 are possible with replacement ? Calculate expected value of the sample mean and compare it with the expected value calculated in the case of without replacement.

Answer:

Let’s address each part of the problem step by step.

Given Data:

Systolic Blood Pressure (SBP) of five women:
$$120,110,130,140,100$$$$120,110,130,140,100$$120,110,130,140,100120, 110, 130, 140, 100$$120,110,130,140,100$$

(a) How many samples of size 2 can be drawn without replacement? Write them.

The number of samples of size 2 that can be drawn without replacement from 5 items is given by the combination formula:
$$(\genfrac{}{}{0ex}{}{5}{2})=\frac{5!}{2!(5-2)!}=10$$$$(\genfrac{}{}{0ex}{}{5}{2})=\frac{5!}{2!(5-2)!}=10$$((5)/(2))=(5!)/(2!(5-2)!)=10binom{5}{2} = frac{5!}{2!(5-2)!} = 10$$(\genfrac{}{}{0ex}{}{5}{2})=\frac{5!}{2!(5-2)!}=10$$
The possible samples without replacement are:

1. (120, 110)
2. (120, 130)
3. (120, 140)
4. (120, 100)
5. (110, 130)
6. (110, 140)
7. (110, 100)
8. (130, 140)
9. (130, 100)
10. (140, 100)

(b) Compute the mean of all samples of size 2 and set up the sampling distribution of the sample mean.

First, compute the means of each sample:

1. (120, 110): $\frac{120+110}{2}=115$$\frac{120+110}{2}=115$(120+110)/(2)=115frac{120 + 110}{2} = 115$\frac{120+110}{2}=115$
2. (120, 130): $\frac{120+130}{2}=125$$\frac{120+130}{2}=125$(120+130)/(2)=125frac{120 + 130}{2} = 125$\frac{120+130}{2}=125$
3. (120, 140): $\frac{120+140}{2}=130$$\frac{120+140}{2}=130$(120+140)/(2)=130frac{120 + 140}{2} = 130$\frac{120+140}{2}=130$
4. (120, 100): $\frac{120+100}{2}=110$$\frac{120+100}{2}=110$(120+100)/(2)=110frac{120 + 100}{2} = 110$\frac{120+100}{2}=110$
5. (110, 130): $\frac{110+130}{2}=120$$\frac{110+130}{2}=120$(110+130)/(2)=120frac{110 + 130}{2} = 120$\frac{110+130}{2}=120$
6. (110, 140): $\frac{110+140}{2}=125$$\frac{110+140}{2}=125$(110+140)/(2)=125frac{110 + 140}{2} = 125$\frac{110+140}{2}=125$
7. (110, 100): $\frac{110+100}{2}=105$$\frac{110+100}{2}=105$(110+100)/(2)=105frac{110 + 100}{2} = 105$\frac{110+100}{2}=105$
8. (130, 140): $\frac{130+140}{2}=135$$\frac{130+140}{2}=135$(130+140)/(2)=135frac{130 + 140}{2} = 135$\frac{130+140}{2}=135$
9. (130, 100): $\frac{130+100}{2}=115$$\frac{130+100}{2}=115$(130+100)/(2)=115frac{130 + 100}{2} = 115$\frac{130+100}{2}=115$
10. (140, 100): $\frac{140+100}{2}=120$$\frac{140+100}{2}=120$(140+100)/(2)=120frac{140 + 100}{2} = 120$\frac{140+100}{2}=120$

The sampling distribution of the sample mean is:
$$105,110,115,115,120,120,120,125,125,135$$$$105,110,115,115,120,120,120,125,125,135$$105,110,115,115,120,120,120,125,125,135105, 110, 115, 115, 120, 120, 120, 125, 125, 135$$105,110,115,115,120,120,120,125,125,135$$

(c) Compute the expected value of the sample mean.

The expected value of the sample mean $\overline{X}$$\overline{X}$bar(X)bar{X}$\overline{X}$ is the average of the sample means:
$$E(\overline{X})=\frac{1}{10}\sum {\overline{X}}_i=\frac{105+110+115+115+120+120+120+125+125+135}{10}=\frac{1190}{10}=119$$$$E(\overline{X})=\frac{1}{10}\sum {\overline{X}}_i=\frac{105+110+115+115+120+120+120+125+125+135}{10}=\frac{1190}{10}=119$$E( bar(X))=(1)/(10)sum bar(X)_(i)=(105+110+115+115+120+120+120+125+125+135)/(10)=(1190)/(10)=119E(bar{X}) = frac{1}{10} sum bar{X}_i = frac{105 + 110 + 115 + 115 + 120 + 120 + 120 + 125 + 125 + 135}{10} = frac{1190}{10} = 119$$E(\overline{X})=\frac{1}{10}\sum {\overline{X}}_i=\frac{105+110+115+115+120+120+120+125+125+135}{10}=\frac{1190}{10}=119$$

(d) How many samples of the same size 2 are possible with replacement? Calculate the expected value of the sample mean and compare it with the expected value calculated in the case of without replacement.

The number of samples of size 2 that can be drawn with replacement from 5 items is given by the formula for permutations with replacement:
$$5^2=25$$$$5^2=25$$5^(2)=255^2 = 25$$5^2=25$$
The possible samples with replacement are:
$$(120,120),(120,110),(120,130),(120,140),(120,100),$$$$(120,120),(120,110),(120,130),(120,140),(120,100),$$(120,120),(120,110),(120,130),(120,140),(120,100),(120, 120), (120, 110), (120, 130), (120, 140), (120, 100),$$(120,120),(120,110),(120,130),(120,140),(120,100),$$
$$(110,120),(110,110),(110,130),(110,140),(110,100),$$$$(110,120),(110,110),(110,130),(110,140),(110,100),$$(110,120),(110,110),(110,130),(110,140),(110,100),(110, 120), (110, 110), (110, 130), (110, 140), (110, 100),$$(110,120),(110,110),(110,130),(110,140),(110,100),$$
$$(130,120),(130,110),(130,130),(130,140),(130,100),$$$$(130,120),(130,110),(130,130),(130,140),(130,100),$$(130,120),(130,110),(130,130),(130,140),(130,100),(130, 120), (130, 110), (130, 130), (130, 140), (130, 100),$$(130,120),(130,110),(130,130),(130,140),(130,100),$$
$$(140,120),(140,110),(140,130),(140,140),(140,100),$$$$(140,120),(140,110),(140,130),(140,140),(140,100),$$(140,120),(140,110),(140,130),(140,140),(140,100),(140, 120), (140, 110), (140, 130), (140, 140), (140, 100),$$(140,120),(140,110),(140,130),(140,140),(140,100),$$
$$(100,120),(100,110),(100,130),(100,140),(100,100)$$$$(100,120),(100,110),(100,130),(100,140),(100,100)$$(100,120),(100,110),(100,130),(100,140),(100,100)(100, 120), (100, 110), (100, 130), (100, 140), (100, 100)$$(100,120),(100,110),(100,130),(100,140),(100,100)$$
Compute the means of each sample:
$$\overline{X}$$$$\overline{X}$$bar(X)bar{X}$$\overline{X}$$
$$120,115,125,130,110,$$$$120,115,125,130,110,$$120,115,125,130,110,120, 115, 125, 130, 110,$$120,115,125,130,110,$$
$$115,110,120,125,105,$$$$115,110,120,125,105,$$115,110,120,125,105,115, 110, 120, 125, 105,$$115,110,120,125,105,$$
$$125,120,130,135,115,$$$$125,120,130,135,115,$$125,120,130,135,115,125, 120, 130, 135, 115,$$125,120,130,135,115,$$
$$130,125,135,140,120,$$$$130,125,135,140,120,$$130,125,135,140,120,130, 125, 135, 140, 120,$$130,125,135,140,120,$$
$$110,105,115,120,100$$$$110,105,115,120,100$$110,105,115,120,100110, 105, 115, 120, 100$$110,105,115,120,100$$
The expected value of the sample mean with replacement is:
$$E(\overline{X})=\frac{1}{25}\sum {\overline{X}}_i=\frac{120+115+125+130+110+115+110+120+125+105+125+120+130+135+115+130+125+135+140+120+110+105+115+120+100}{25}$$$$E(\overline{X})=\frac{1}{25}\sum {\overline{X}}_i=\frac{120+115+125+130+110+115+110+120+125+105+125+120+130+135+115+130+125+135+140+120+110+105+115+120+100}{25}$$E( bar(X))=(1)/(25)sum bar(X)_(i)=(120+115+125+130+110+115+110+120+125+105+125+120+130+135+115+130+125+135+140+120+110+105+115+120+100)/(25)E(bar{X}) = frac{1}{25} sum bar{X}_i = frac{120 + 115 + 125 + 130 + 110 + 115 + 110 + 120 + 125 + 105 + 125 + 120 + 130 + 135 + 115 + 130 + 125 + 135 + 140 + 120 + 110 + 105 + 115 + 120 + 100}{25}$$E(\overline{X})=\frac{1}{25}\sum {\overline{X}}_i=\frac{120+115+125+130+110+115+110+120+125+105+125+120+130+135+115+130+125+135+140+120+110+105+115+120+100}{25}$$
Summing these values:
$$\sum {\overline{X}}_i=2925$$$$\sum {\overline{X}}_i=2925$$sum bar(X)_(i)=2925sum bar{X}_i = 2925$$\sum {\overline{X}}_i=2925$$
So,
$$E(\overline{X})=\frac{2925}{25}=117$$$$E(\overline{X})=\frac{2925}{25}=117$$E( bar(X))=(2925)/(25)=117E(bar{X}) = frac{2925}{25} = 117$$E(\overline{X})=\frac{2925}{25}=117$$

Comparison:

• The expected value of the sample mean without replacement is 119.
• The expected value of the sample mean with replacement is 117.

Thus, the expected value of the sample mean without replacement is slightly higher than that with replacement in this case.

Question:-03

3.(a) The following table gives the classification of 150 products according to types of tools and materials used to produce these products :

Tool Material
A B C
T_(1) 15 5 20
T_(2) 20 10 30
T_(3) 25 15 10| Tool | Material | | |
| :—: | :—: | :—: | :—: |
| | A | B | C |
| $mathrm{T}_1$ | 15 | 5 | 20 |
| $mathrm{~T}_2$ | 20 | 10 | 30 |
| $mathrm{~T}_3$ | 25 | 15 | 10 | Test whether the tools and materials used are independent at $5\mathrm{\%}$$5\mathrm{\%}$5%5 %$5\mathrm{\%}$ level of significance.

Answer:

To test whether the tools and materials used are independent at the $5\mathrm{\%}$$5\mathrm{\%}$5%5%$5\mathrm{\%}$ level of significance, we can use the Chi-Square Test of Independence. Here are the steps:

1. State the Hypotheses:
   • $H_0$$H_0$H_(0)H_0$H_0$: The tools and materials used are independent.
   • $H_1$$H_1$H_(1)H_1$H_1$: The tools and materials used are not independent.
2. Observed Frequencies (O):

$$\begin{array}{|cccc|}\hline \text{Tool}& \text{A}& \text{B}& \text{C}\\ T_1& 15& 5& 20\\ T_2& 20& 10& 30\\ T_3& 25& 15& 10\\ \hline\end{array}$$$$\begin{array}{|cccc|}\hline \text{Tool}& \text{A}& \text{B}& \text{C}\\ T_1& 15& 5& 20\\ T_2& 20& 10& 30\\ T_3& 25& 15& 10\\ \hline\end{array}$${:[“Tool”,”A”,”B”,”C”],[T_(1),15,5,20],[T_(2),20,10,30],[T_(3),25,15,10]:}begin{array}{|c|c|c|c|}
hline
text{Tool} & text{A} & text{B} & text{C} \
hline
T_1 & 15 & 5 & 20 \
T_2 & 20 & 10 & 30 \
T_3 & 25 & 15 & 10 \
hline
end{array}$$\begin{array}{|cccc|}\hline \text{Tool}& \text{A}& \text{B}& \text{C}\\ T_1& 15& 5& 20\\ T_2& 20& 10& 30\\ T_3& 25& 15& 10\\ \hline\end{array}$$

3. Calculate the row and column totals and the grand total:

$$\begin{array}{|ccccc|}\hline \text{Tool}& \text{A}& \text{B}& \text{C}& \text{Row Total}\\ T_1& 15& 5& 20& 40\\ T_2& 20& 10& 30& 60\\ T_3& 25& 15& 10& 50\\ \text{Column Total}& 60& 30& 60& 150\\ \hline\end{array}$$$$\begin{array}{|ccccc|}\hline \text{Tool}& \text{A}& \text{B}& \text{C}& \text{Row Total}\\ T_1& 15& 5& 20& 40\\ T_2& 20& 10& 30& 60\\ T_3& 25& 15& 10& 50\\ \text{Column Total}& 60& 30& 60& 150\\ \hline\end{array}$${:[“Tool”,”A”,”B”,”C”,”Row Total”],[T_(1),15,5,20,40],[T_(2),20,10,30,60],[T_(3),25,15,10,50],[“Column Total”,60,30,60,150]:}begin{array}{|c|c|c|c|c|}
hline
text{Tool} & text{A} & text{B} & text{C} & text{Row Total} \
hline
T_1 & 15 & 5 & 20 & 40 \
T_2 & 20 & 10 & 30 & 60 \
T_3 & 25 & 15 & 10 & 50 \
hline
text{Column Total} & 60 & 30 & 60 & 150 \
hline
end{array}$$\begin{array}{|ccccc|}\hline \text{Tool}& \text{A}& \text{B}& \text{C}& \text{Row Total}\\ T_1& 15& 5& 20& 40\\ T_2& 20& 10& 30& 60\\ T_3& 25& 15& 10& 50\\ \text{Column Total}& 60& 30& 60& 150\\ \hline\end{array}$$

4. Calculate the Expected Frequencies (E):

$$E_{ij}=\frac{(\text{Row Total of}i)\times (\text{Column Total of}j)}{\text{Grand Total}}$$$$E_{ij}=\frac{(\text{Row Total of}i)\times (\text{Column Total of}j)}{\text{Grand Total}}$$E_(ij)=((“Row Total of “i)xx(“Column Total of “j))/(“Grand Total”)E_{ij} = frac{(text{Row Total of } i) times (text{Column Total of } j)}{text{Grand Total}}$$E_{ij}=\frac{(\text{Row Total of}i)\times (\text{Column Total of}j)}{\text{Grand Total}}$$
$$\begin{array}{|cccc|}\hline \text{Tool}& \text{A}& \text{B}& \text{C}\\ T_1& \frac{40\times 60}{150}=16& \frac{40\times 30}{150}=8& \frac{40\times 60}{150}=16\\ T_2& \frac{60\times 60}{150}=24& \frac{60\times 30}{150}=12& \frac{60\times 60}{150}=24\\ T_3& \frac{50\times 60}{150}=20& \frac{50\times 30}{150}=10& \frac{50\times 60}{150}=20\\ \hline\end{array}$$$$\begin{array}{|cccc|}\hline \text{Tool}& \text{A}& \text{B}& \text{C}\\ T_1& \frac{40\times 60}{150}=16& \frac{40\times 30}{150}=8& \frac{40\times 60}{150}=16\\ T_2& \frac{60\times 60}{150}=24& \frac{60\times 30}{150}=12& \frac{60\times 60}{150}=24\\ T_3& \frac{50\times 60}{150}=20& \frac{50\times 30}{150}=10& \frac{50\times 60}{150}=20\\ \hline\end{array}$${:[“Tool”,”A”,”B”,”C”],[T_(1),(40 xx60)/(150)=16,(40 xx30)/(150)=8,(40 xx60)/(150)=16],[T_(2),(60 xx60)/(150)=24,(60 xx30)/(150)=12,(60 xx60)/(150)=24],[T_(3),(50 xx60)/(150)=20,(50 xx30)/(150)=10,(50 xx60)/(150)=20]:}begin{array}{|c|c|c|c|}
hline
text{Tool} & text{A} & text{B} & text{C} \
hline
T_1 & frac{40 times 60}{150} = 16 & frac{40 times 30}{150} = 8 & frac{40 times 60}{150} = 16 \
T_2 & frac{60 times 60}{150} = 24 & frac{60 times 30}{150} = 12 & frac{60 times 60}{150} = 24 \
T_3 & frac{50 times 60}{150} = 20 & frac{50 times 30}{150} = 10 & frac{50 times 60}{150} = 20 \
hline
end{array}$$\begin{array}{|cccc|}\hline \text{Tool}& \text{A}& \text{B}& \text{C}\\ T_1& \frac{40\times 60}{150}=16& \frac{40\times 30}{150}=8& \frac{40\times 60}{150}=16\\ T_2& \frac{60\times 60}{150}=24& \frac{60\times 30}{150}=12& \frac{60\times 60}{150}=24\\ T_3& \frac{50\times 60}{150}=20& \frac{50\times 30}{150}=10& \frac{50\times 60}{150}=20\\ \hline\end{array}$$

5. Compute the Chi-Square Test Statistic:

$${\chi }^2=\sum \frac{(O_{ij}-E_{ij}{)}^2}{E_{ij}}$$$${\chi }^2=\sum \frac{(O_{ij}-E_{ij}{)}^2}{E_{ij}}$$chi^(2)=sum((O_(ij)-E_(ij))^(2))/(E_(ij))chi^2 = sum frac{(O_{ij} – E_{ij})^2}{E_{ij}}$${\chi }^2=\sum \frac{(O_{ij}-E_{ij}{)}^2}{E_{ij}}$$
$$\begin{array}{|cccc|}\hline \text{Tool}& \text{A}& \text{B}& \text{C}\\ T_1& \frac{(15-16{)}^2}{16}=\frac{1}{16}& \frac{(5-8{)}^2}{8}=\frac{9}{8}& \frac{(20-16{)}^2}{16}=\frac{16}{16}\\ T_2& \frac{(20-24{)}^2}{24}=\frac{16}{24}& \frac{(10-12{)}^2}{12}=\frac{4}{12}& \frac{(30-24{)}^2}{24}=\frac{36}{24}\\ T_3& \frac{(25-20{)}^2}{20}=\frac{25}{20}& \frac{(15-10{)}^2}{10}=\frac{25}{10}& \frac{(10-20{)}^2}{20}=\frac{100}{20}\\ \hline\end{array}$$$$\begin{array}{|cccc|}\hline \text{Tool}& \text{A}& \text{B}& \text{C}\\ T_1& \frac{(15-16{)}^2}{16}=\frac{1}{16}& \frac{(5-8{)}^2}{8}=\frac{9}{8}& \frac{(20-16{)}^2}{16}=\frac{16}{16}\\ T_2& \frac{(20-24{)}^2}{24}=\frac{16}{24}& \frac{(10-12{)}^2}{12}=\frac{4}{12}& \frac{(30-24{)}^2}{24}=\frac{36}{24}\\ T_3& \frac{(25-20{)}^2}{20}=\frac{25}{20}& \frac{(15-10{)}^2}{10}=\frac{25}{10}& \frac{(10-20{)}^2}{20}=\frac{100}{20}\\ \hline\end{array}$${:[“Tool”,”A”,”B”,”C”],[T_(1),((15-16)^(2))/(16)=(1)/(16),((5-8)^(2))/(8)=(9)/(8),((20-16)^(2))/(16)=(16)/(16)],[T_(2),((20-24)^(2))/(24)=(16)/(24),((10-12)^(2))/(12)=(4)/(12),((30-24)^(2))/(24)=(36)/(24)],[T_(3),((25-20)^(2))/(20)=(25)/(20),((15-10)^(2))/(10)=(25)/(10),((10-20)^(2))/(20)=(100)/(20)]:}begin{array}{|c|c|c|c|}
hline
text{Tool} & text{A} & text{B} & text{C} \
hline
T_1 & frac{(15-16)^2}{16} = frac{1}{16} & frac{(5-8)^2}{8} = frac{9}{8} & frac{(20-16)^2}{16} = frac{16}{16} \
T_2 & frac{(20-24)^2}{24} = frac{16}{24} & frac{(10-12)^2}{12} = frac{4}{12} & frac{(30-24)^2}{24} = frac{36}{24} \
T_3 & frac{(25-20)^2}{20} = frac{25}{20} & frac{(15-10)^2}{10} = frac{25}{10} & frac{(10-20)^2}{20} = frac{100}{20} \
hline
end{array}$$\begin{array}{|cccc|}\hline \text{Tool}& \text{A}& \text{B}& \text{C}\\ T_1& \frac{(15-16{)}^2}{16}=\frac{1}{16}& \frac{(5-8{)}^2}{8}=\frac{9}{8}& \frac{(20-16{)}^2}{16}=\frac{16}{16}\\ T_2& \frac{(20-24{)}^2}{24}=\frac{16}{24}& \frac{(10-12{)}^2}{12}=\frac{4}{12}& \frac{(30-24{)}^2}{24}=\frac{36}{24}\\ T_3& \frac{(25-20{)}^2}{20}=\frac{25}{20}& \frac{(15-10{)}^2}{10}=\frac{25}{10}& \frac{(10-20{)}^2}{20}=\frac{100}{20}\\ \hline\end{array}$$
Calculate each term and sum them up:
$${\chi }^2=\frac{1}{16}+\frac{9}{8}+\frac{16}{16}+\frac{16}{24}+\frac{4}{12}+\frac{36}{24}+\frac{25}{20}+\frac{25}{10}+\frac{100}{20}$$$${\chi }^2=\frac{1}{16}+\frac{9}{8}+\frac{16}{16}+\frac{16}{24}+\frac{4}{12}+\frac{36}{24}+\frac{25}{20}+\frac{25}{10}+\frac{100}{20}$$chi^(2)=(1)/(16)+(9)/(8)+(16)/(16)+(16)/(24)+(4)/(12)+(36)/(24)+(25)/(20)+(25)/(10)+(100)/(20)chi^2 = frac{1}{16} + frac{9}{8} + frac{16}{16} + frac{16}{24} + frac{4}{12} + frac{36}{24} + frac{25}{20} + frac{25}{10} + frac{100}{20}$${\chi }^2=\frac{1}{16}+\frac{9}{8}+\frac{16}{16}+\frac{16}{24}+\frac{4}{12}+\frac{36}{24}+\frac{25}{20}+\frac{25}{10}+\frac{100}{20}$$
$${\chi }^2=0.0625+1.125+1+0.6667+0.3333+1.5+1.25+2.5+5$$$${\chi }^2=0.0625+1.125+1+0.6667+0.3333+1.5+1.25+2.5+5$$chi^(2)=0.0625+1.125+1+0.6667+0.3333+1.5+1.25+2.5+5chi^2 = 0.0625 + 1.125 + 1 + 0.6667 + 0.3333 + 1.5 + 1.25 + 2.5 + 5$${\chi }^2=0.0625+1.125+1+0.6667+0.3333+1.5+1.25+2.5+5$$
$${\chi }^2=13.4375$$$${\chi }^2=13.4375$$chi^(2)=13.4375chi^2 = 13.4375$${\chi }^2=13.4375$$

6. Degrees of Freedom:

$$\text{df}=(r-1)(c-1)=(3-1)(3-1)=2\times 2=4$$$$\text{df}=(r-1)(c-1)=(3-1)(3-1)=2\times 2=4$$“df”=(r-1)(c-1)=(3-1)(3-1)=2xx2=4text{df} = (r – 1)(c – 1) = (3 – 1)(3 – 1) = 2 times 2 = 4$$\text{df}=(r-1)(c-1)=(3-1)(3-1)=2\times 2=4$$

7. Critical Value and Conclusion:

At the $5\mathrm{\%}$$5\mathrm{\%}$5%5%$5\mathrm{\%}$ level of significance and $4$$4$44$4$ degrees of freedom, the critical value from the Chi-Square distribution table is approximately $9.488$$9.488$9.4889.488$9.488$.
Since ${\chi }^2=13.4375$${\chi }^2=13.4375$chi^(2)=13.4375chi^2 = 13.4375${\chi }^2=13.4375$ is greater than the critical value $9.488$$9.488$9.4889.488$9.488$, we reject the null hypothesis $H_0$$H_0$H_(0)H_0$H_0$.

Conclusion:

There is sufficient evidence to conclude that the tools and materials used are not independent at the $5\mathrm{\%}$$5\mathrm{\%}$5%5%$5\mathrm{\%}$ level of significance.

(b) Explain the general procedure of testing of hypothesis.

Answer:

Testing a hypothesis is a statistical method that uses sample data to evaluate a hypothesis about a population parameter. The general procedure involves several steps, which are outlined below:

1. Formulate Hypotheses

• Null Hypothesis ($H_0$$H_0$H_(0)H_0$H_0$): This is the hypothesis that there is no effect or no difference, and it represents the status quo. For example, $H_0:\mu ={\mu }_0$$H_0:\mu ={\mu }_0$H_(0):mu=mu_(0)H_0: mu = mu_0$H_0:\mu ={\mu }_0$.
• Alternative Hypothesis ($H_1$$H_1$H_(1)H_1$H_1$ or $H_a$$H_a$H_(a)H_a$H_a$): This is what you want to prove. It represents a change, effect, or difference. For example, $H_1:\mu \ne {\mu }_0$$H_1:\mu \ne {\mu }_0$H_(1):mu!=mu_(0)H_1: mu neq mu_0$H_1:\mu \ne {\mu }_0$, $H_1:\mu >{\mu }_0$$H_1:\mu >{\mu }_0$H_(1):mu > mu_(0)H_1: mu > mu_0$H_1:\mu >{\mu }_0$, or $H_1:\mu <{\mu }_0$$H_1:\mu <{\mu }_0$H_(1):mu < mu_(0)H_1: mu < mu_0$H_1:\mu <{\mu }_0$.

2. Choose the Significance Level ($\alpha$$\alpha$alphaalpha$\alpha$)

• The significance level is the probability of rejecting the null hypothesis when it is actually true (Type I error). Common values are 0.05, 0.01, and 0.10.

3. Select the Appropriate Test Statistic

• The test statistic is a standardized value that is calculated from sample data during a hypothesis test. Depending on the nature of the data and the hypothesis, this could be a Z-test, t-test, chi-square test, F-test, etc.

4. Formulate the Decision Rule

• The decision rule involves determining the critical value(s) from the statistical distribution of the test statistic (e.g., normal distribution, t-distribution, chi-square distribution) that correspond to the chosen significance level. This critical value(s) define the rejection region(s).

5. Collect Data and Compute the Test Statistic

• Gather the sample data and compute the value of the test statistic based on the sample data.

6. Make a Decision

• Compare the test statistic to the critical value(s):
  • If the test statistic falls within the rejection region, reject the null hypothesis ($H_0$$H_0$H_(0)H_0$H_0$).
  • If the test statistic does not fall within the rejection region, do not reject the null hypothesis.

7. Draw a Conclusion

• Based on the decision made in the previous step, interpret the results in the context of the research question or problem.

Example

Step 1: Formulate Hypotheses

• $H_0$$H_0$H_(0)H_0$H_0$: The mean systolic blood pressure ($\mu$$\mu$mumu$\mu$) is 120 mmHg.
• $H_1$$H_1$H_(1)H_1$H_1$: The mean systolic blood pressure ($\mu$$\mu$mumu$\mu$) is not 120 mmHg.