$$H_{0}: \mu = 4.8$$
$$H_{1}: \mu < 4.8$$
Given $n=200$, sample mean $\bar{x}=4.7$ and sample std $S=0.5$ how do I calculate the P-value of this left-tailed test?
My work:
$$\text{left-side confidence limit for $\mu$ is}\\ \mu < \bar{x} -t_{\alpha}\frac{S}{\sqrt{n}} \implies t_{\alpha} < \frac{\bar{x}-\mu}{S/\sqrt{n}}$$
We want to find:
$$T = \frac{\bar{x}-\mu}{S/\sqrt{n}}$$
$$P(T < t_{a}) = P\left(\frac{\bar{x}-\mu}{S/\sqrt{n}} < t_{\alpha}\right) = P\left(\frac{\bar{x}-\mu}{S/\sqrt{n}} < \frac{\bar{x}-\mu}{S/\sqrt{n}} \right) = P\left( t < \frac{\bar{x}-\mu}{S/\sqrt{n}} \right) = p$$
Substituting in the given numbers we end up with:
$$P(T < -2.828) = p$$
Now I have two questions:
In the second step I have used what I have understood is some form of definition for the p-value, I do not understand it. What does that definition say? $P(t < t_{c})=p$
How do I find $p$ given the final expression? As I have understood it I need to get it from the T-distribution table, but I don't know how to read it properly.
The p-value is the area of the involved tails; in your case it is the area of the left tail, given the observed result of $\overline{X}_{200}=4.7$
Given that $n=200\approx \infty$ you can avoid to use Student T table but you can use the gaussian one, finding that
$$P(Z<-2.83)\approx0.0023$$
the p-value is the white area (you need the specular graphics with the with the white area on the left tail but it does not matter for your calculations) thus it is
$$\mathbb{P}[Z<-2.83]=\mathbb{P}[Z>2.83]=1-0.9977=0.0023$$
note that your exercise is bad written because if you want to use the Test Statistic you used you must assume that your population is gaussian.