In each of the following situations it is believed that $X - N(μ, 400)$. Find the p - value of the observed sample mean. Hence, decide the result of the test if it is conducted at the 5% significance level.
Part a i) $H_0:\:μ\:=85,\:H_1:\:μ\:\ne \:85\: \\n=16\: \bar{\left[X\right]}=95$
My answer was reject $H_0$ as $P(\bar{X}>95) = 0.0228$, and as $0.0228<0.025$ then there is sufficient evidence to reject $H_0$.
However the textbook states reject $H_0$ as the probability is 0.0455 (nothing else is mentioned). Where did I go wrong?
The alternative hypothesis is double sided, so you also need to account for the case $$ P(\bar{X}<75) $$ which is just as extreme, but in the other direction, and by symmetry of the normal distribution the total probability you need is twice $P(\bar{X}>95)$, so
$$ \begin{align} p &= 2P(\bar{X}>95)\\ &= 2P\left( Z>\frac{95-85}{\sqrt{400/16}} \right)\\ &= 2*0.02275013\\ &=0.0455 \end{align} $$