Solution — 2019 Question 12 Sampling| Hypothesis Testing

Let $X$ denote the random variable of the mass (in grams) of a randomly chosen cake
Let $\mu$ denote the population mean mass (in grams) of a randomly chosen cake

${\mathrm{H}}_0:\mu =100$
${\mathrm{H}}_1:\mu <100$

Under $H_0,$ at $0.05\%$ level of significance, $$\bar{X}\sim \mathrm{N}(100,\frac{\frac{6568}{295}}{60})$$ $$Z=\frac{\bar{X}-100}{\sqrt{\frac{\frac{6568}{295}}{60}}}\sim \mathrm{N}(\mathrm{0,1})$$ approximately by the Central Limit Theorem since $n=60$ is large

From GC,
$p$-value = $0.024423>0.0005\Rightarrow \text{Do not reject }{\mathrm{H}}_0$

Hence there is insufficient evidence at the $0.05\%$ level of significance to conclude whether the mean mass of the cakes is less than 100 grams $∎$

${\mathrm{H}}_0:\mu =100$
${\mathrm{H}}_1:\mu \ne 100$

Under $H_0,$ at $0.05\%$ level of significance, $$\bar{X}\sim \mathrm{N}(100,\frac{19}{60})$$ $$Z=\frac{\bar{X}-100}{\sqrt{\frac{19}{60}}}\sim \mathrm{N}(\mathrm{0,1})$$ approximately by the Central Limit Theorem since $n=60$ is large

Since the null hypothesis is rejected,