Question

Green Sea Urchins. From the paper "Effects of Chronic Nitrate Exposure on Gonad Growth in Green Sea Urchin Strongylocen. trouts drocbachiensis" (A quaculture, Vol. 242, No. 1-4, pp. 357-363) by $\mathrm{S}$Siikavuopio et al., the weights, $x$of adult green sea urchins are normally distributed with mean $52.0\mathrm{g}$and standard deviation $17.2\mathrm{g}$For samples of $12$such weights, identify the distribution of each of the following variables.

a. $\frac{\overline{x}-52.0}{17.2/\sqrt{12}}$

b. $\frac{\hat{x}-52.0}{s/\sqrt{12}}$

Short answer

Part (a) The standard normal distribution is used a mean of $0$ and a standard deviation of $1$ and follows a normal distribution.

Part (b) It follows the $t$distribution with $n-1=11$degrees of freedom.

Step-by-step solution

Part (a) Step 1: Given information

Adult green sea urchin weights, $x$are normally distributed with a mean of $\mu =52.0\mathrm{gm}$and a standard deviation of $\sigma =17.2\mathrm{gm}$

Sample size $n=12$

Part (a) Step 2: Concept

The formula used: $z=\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}$

Part (a) Step 3: Calculation

$\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}=\frac{\overline{x}-52}{\frac{17.2}{\sqrt{12}}}$the normal distribution is followed. i.e. follows a normal distribution with a mean of $0$ and a standard deviation of $1$

Part (b) Step 1: Calculation

$\frac{\overline{x}-\mu }{s/\sqrt{n}}=\frac{\overline{x}-52}{s/\sqrt{12}}$ With $n-1=11$ degrees of freedom, it follows the $t$ distribution.