Question 2)
Given: (i) Test scores of Bilingual and monolingual students
(ii) Type of test: \(T\) test at significance level \((\alpha)=0.05\).
(iii) samples have equal variances and are normally distributed.
To find: (i) Null and Alternate Hypothesis
(ii) p value of the test.
(iii) concision of the test and Justification
▶️Answer/Explanation
Ans:
(i) claim: \(\mu_b>\mu_m\)
So, Null hypothesis: \(\mu_b=\mu_m\)
Alternate Hypothesis: \(\mu_b>\mu_m\).
(ii)
\(\rm {\quad\quad\quad Bilingual \quad\quad\quad Monolingual}\)
Sample mean \(\bar{x}_1=97 \quad\quad\quad \bar{x}_2=94.625\)
sample std deviation \(s_1=3.300 \quad\quad s_2=4.926\)
sample size \(n_1=10 \quad\quad\quad\quad n_2=8\)
\[
\begin{aligned}
\text { Test statistic } & =t=\frac{\bar{x}_1-\bar{x}_2}{\sqrt{\frac{\left(n_1-1\right) s_1^2+\left(n_2-1\right) s_2^2}{n_1+n_2-2}\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}} \\
t & =\frac{97-94.625}{\sqrt{\frac{(10-1)(3.3)^2+(8-1)(4.920)^2}{10+8-2}}\left(\frac{1}{10}+\frac{1}{8}\right)}=1.224 .
\end{aligned}
\]
From \(t\) value to \(p\) value chart we get : \(p_{t=1.224}=0.1194\).
(iii) Since \(P\) value \((0.1194) \geq\) significance level \((0.05)\)
\(\therefore\) we cannot reject the null hypothesis.
There is insufficient evidence that \(\mu_b>\mu_m\).
Question 3)
Given: (i) coordinates of three towns – \(A, B , C\)
(ii) coordinate of Dononique’s farm \(D\).
To find: (i) Distance between town \(A\) and farm \(D\).
(ii) using interpolation to find temperature of farm D.
▶️Answer/Explanation
Answer
(i) For finding the distance between two coordinate paints \(\left(x_1, y_1\right)\) and \(\left(x_3, y_2\right)\)
$
\begin{gathered}
d=\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2} . \\
A(35,26) \text { and } D(24,19) \\
\therefore \quad A D=\sqrt{(35-24)^2+(26-19)^2}=\sqrt{(11)^2+(7)^2} \\
=13.0384 \simeq 13 \text { miles. }
\end{gathered}
$
(ii). similarly,
$$
\begin{aligned}
& B D=\sqrt{(11-24)^2+(24-13)^2}=\sqrt{(-13)^2+(5)^2}=13.928 \simeq 13.9 \text { miles. } \\
& C D=\sqrt{(28-24)^2+(7-19)^2}=\sqrt{(4)^2+(-12)^2}=12.6491 \simeq 12.6 \text { miles. }
\end{aligned}
$$
Since \(C D\) is the lowest and distance between \(C\) and \(D\) is minimum
\(\therefore\) best approximate tumparcture in farm \(D=\) Tempareture in city \(C=30^{\circ} \mathrm{C}\)