Applied Maths 5 Marks Questions

For first data set:
Number of terms in first set i.e. $n_1=6$
Calculate mean value for first data set using formula:
$\begin{aligned} \bar{x} & =\frac{\sum x}{n_1} \\ & =\frac{63+65+68+69+71+72}{6} \\ & =\frac{408}{6}=68\end{aligned}$
For second data set:
Number of terms in second set i.e. $n_2=10$
Calculate mean value for second data set using formula:
$\begin{aligned} \bar{y} & =\frac{\sum y}{n_2} \\ & =\frac{62+61+65+66+69+69+70+71+72+73}{10} \\ & =\frac{678}{10}=67.8\end{aligned}$
Construct the following table for standard error s:

Now, compute the standard error, s using formula:
$\begin{aligned} s & =\sqrt{\frac{\sum(x-\bar{x})^2+\sum(y-\bar{y})^2}{n_1+n_2-2}} \\ s & =\sqrt{\frac{60+153.60}{6+10-2}} \\ & =\sqrt{\frac{213.60}{14}} \\ & =\sqrt{15.257} \\ & =3.906\end{aligned}$
Now using t-test formula:
$t=\frac{\bar{x}-\bar{y}}{s} \sqrt{\frac{n_1 n_2}{n_1+n_2}}$
$\begin{array}{l}=\frac{68-67.8}{3.906} \sqrt{\frac{6 \times 10}{6+10}} \\ =\frac{0.2 \sqrt{3.75}}{3.906} \\ =0.0512 \times 1.936 \\ =0.099\end{array}$
Hence, t-test value for the two data sets is 0.099.

For first data set:
Number of terms in first set i.e. $n_1=7$
Mean of $A$ is
$\begin{aligned} \bar{x} & =\frac{\sum x}{n_1} \\ & =\frac{28+30+32+33+33+29+34}{7} \\ & =\frac{219}{7} \\ & =31.29\end{aligned}$
For second data set:
Number of terms in second set i.e., $n_2=6$ Mean of $B$ is
$\begin{aligned} \bar{y} & =\frac{\sum y}{n_2} \\ & =\frac{29+30+30+24+27+29}{6} \\ & =\frac{169}{6} \\ & =28.16\end{aligned}$
Construct the following table for standard error $s$ :

Now, compute the standard error, $s$ using formula:
$\begin{aligned} s & =\sqrt{\frac{\sum(x-\bar{x})^2+\sum(y-\bar{y})^2}{n_1+n_2-2}} \\ s & =\sqrt{\frac{31.4+26.86}{7+6-2}} \\ & =\sqrt{\frac{58.26}{11}} \\ & =\sqrt{5.29} \\ & =2.3\end{aligned}$
Now using t-test formula:
$\begin{aligned} t & =\frac{\bar{x}-\bar{y}}{s} \sqrt{\frac{n_1 n_2}{n_1+n_2}} \\ t & =\frac{31.29-28.16}{2.3} \sqrt{\frac{7 \times 6}{7+6}} \\ & =\frac{3.13}{2.3} \sqrt{3.23} \\ & =1.36 \times 1.79 \\ & =2.43\end{aligned}$
Hence, $t$-test value for the two data sets is 2.43 .
The degree of freedom $=n_1+n_2-2=7+6-2=11$
Given, the value of $t$ for 11 degrees of freedom at $5 \%$ level of significance $=2.20$, which is less than the calculated value of $t$. Hence the difference is significant i.e., we can discriminate between two horses.

For first data set:
Number of terms in first set i.e. $n_1=6$
Calculate mean value for first data set using formula:
$\begin{aligned} \bar{x} & =\frac{\sum x}{n_1} \\ & =\frac{63+65+68+69+71+72}{6} \\ & =\frac{408}{6}=68\end{aligned}$
For second data set:
Number of terms in second set i.e. $n_2=10$
Calculate mean value for second data set using formula:
$\begin{aligned} \bar{y} & =\frac{\sum y}{n_2} \\ & =\frac{62+61+65+66+69+69+70+71+72+73}{10} \\ & =\frac{678}{10}=67.8\end{aligned}$
Construct the following table for standard error s:

Now, compute the standard error, s using formula:
$\begin{aligned} s & =\sqrt{\frac{\sum(x-\bar{x})^2+\sum(y-\bar{y})^2}{n_1+n_2-2}} \\ s & =\sqrt{\frac{60+153.60}{6+10-2}} \\ & =\sqrt{\frac{213.60}{14}} \\ & =\sqrt{15.257} \\ & =3.906\end{aligned}$
Now using t-test formula:
$t=\frac{\bar{x}-\bar{y}}{s} \sqrt{\frac{n_1 n_2}{n_1+n_2}}$
$\begin{array}{l}=\frac{68-67.8}{3.906} \sqrt{\frac{6 \times 10}{6+10}} \\ =\frac{0.2 \sqrt{3.75}}{3.906} \\ =0.0512 \times 1.936 \\ =0.099\end{array}$
Hence, t-test value for the two data sets is 0.099.