Solution — 2018 Question 11 Sampling| Hypothesis Testing

Let $X$ denote the random variable of the length (in cm) of a randomly chosen adult fish
Let $\mu$ denote the population mean length (in cm) of a randomly chosen adult fish

${\mathrm{H}}_0:\mu =30$
${\mathrm{H}}_1:\mu >30$

Under $H_0,$ at $0.025\%$ level of significance, $$\bar{X}\sim \mathrm{N}(30,\frac{\frac{29}{36}}{100})$$ $$Z=\frac{\bar{X}-30}{\sqrt{\frac{\frac{29}{36}}{100}}}\sim \mathrm{N}(\mathrm{0,1})$$ approximately by the Central Limit Theorem since $n=100$ is large

From GC,
$p$-value = $0.047335>0.00025\Rightarrow \text{Do not reject }{\mathrm{H}}_0$

Hence there is insufficient evidence at the $0.025\%$ level of significance to conclude whether the mean length of the fish is greater than 30 cm $∎$

Since $n=100$ is large, by the Central Limit Theorem, $\bar{X}$ will be approximately normally distributed. Hence it is not necessary to assume that the lengths of the fish are normally distributed. $∎$

Let $X$ denote the random variable of the length (in cm) of a randomly chosen fish from the second lake
Let $\mu$ denote the population mean length (in cm) of a randomly chosen fish from the second lake

${\mathrm{H}}_0:\mu =30$
${\mathrm{H}}_1:\mu >30$

Under $H_0,$ at $0.1\%$ level of significance, $$\bar{X}\sim \mathrm{N}(30,\frac{0.9}{100})$$ $$Z=\frac{\bar{X}-30}{\sqrt{\frac{0.9}{100}}}\sim \mathrm{N}(\mathrm{0,1})$$ approximately by the Central Limit Theorem since $n=100$ is large

Since the null hypothesis is rejected,