In both cases, $\bar{x}=4.40, s=2.90$, and the sample effect as measured by $\bar{x}-\mu_{0}=4.40-4.50=-0.10$ is the same. However, the standard error is smaller with a larger sample size. The se values are
(i) $s e=s / \sqrt{n}=2.90 / \sqrt{100}=0.29$,
(ii) $s e=s / \sqrt{n}=2.90 / \sqrt{10,000}=0.029$.
The test statistics then are also quite different:
(i) $t=\frac{\left(\bar{x}-\mu_{0}\right)}{s e}=\frac{4.40-4.50}{0.29}=-0.345$,
(ii) $t=\frac{\left(\bar{x}-\mu_{0}\right)}{s e}=\frac{4.40-4.50}{0.029}=-3.45$.
PHAS00039 COURSE NOTES ：
a. The slope of the regression equation is $b=1.49$. Since $s_{x}=7.1$ and $s_{y}=13.3$,
$$
r=b\left(\frac{s_{x}}{s_{y}}\right)=1.49\left(\frac{7.1}{13.3}\right)=0.80 .
$$
The variables have a strong, positive association.
b. If $y$ had been measured in kilograms, the $y$ values would have been divided by $2.2$, since $1 \mathrm{~kg}=2.2$ pounds. For instance, Subject 1 had $y=80$ pounds, which is $80 / 2.2=36.4 \mathrm{~kg}$. Likewise, the standard deviation $s_{y}$ of $13.3$ in pounds would have been divided by $2.2$ to get $13.3 / 2.2=6.05 \mathrm{in} \mathrm{kg}$. The slope of $1.49$ would have been divided by $2.2$, giving $0.68$, since $1.49$ pounds $=0.68 \mathrm{~kg}$. Then
$$
r=b\left(s_{x} / s_{y}\right)=0.68(7.1 / 6.05)=0.80 .
$$
The correlation is the same $(0.80)$ if we measure $y$ in pounds or in kilograms.