Let p be the proportion of plants of a certain kind that can be attacked by late blight. In an experiment with 160 plants 50 of them were attacked. Test the following hypotheses with a significance level 5%
$H_0 : p = 0.4$ against $H_1 : p < 0.4.$
$N = 160, n = 50, \mu = 64.$
So I notice that I have to use normal approximation since
$Np$ and $N(1-p) \ge 5.$
$\sigma = \sqrt{ Np(1-p) } = \sqrt{ 160*0.4*0.6 } = 6.1968.$
$Z = (x - \mu)/\sigma$
According to the answer x is 50.5, what rule in the continuity for correction table have they used? I am having trouble interpreting it. My guess would be that they have used $P(X\le n)$ and $P(X < n + 0.5).$
Is it because $p \le 0.4?$ What variables in the question do I need to use when navigating the table?
According to your null hypothesis the number $X$ attacked by blight is distributed as $\mathsf{Binom}(160, .4).$ The exact P-value of your test is $P(X \le 50) = 0.0138,$ which is smaller than 0.05. Accordingly, you would reject $H_0: p = 0.4$ at the 5% level, concluding that the proportion attacked by blight must be smaller than 0.4.$ (The exact binomial computation from R statistical software is shown below.)
You are correct that you could get a reasonable approximation to the P-value using the normal approximation:
$$P(X \le 50) = P(X < 50.5) = P\left(\frac{X-\mu}{\sigma} < \frac{50.5 - 64}{ 6.1968}\right) \approx P(Z < -2.18) = 0.0146,$$
which is smaller than 0.05. Here $Z$ is standard normal and the value 0.0146 can be obtained from printed normal tables. The normal approximation is accurate to about two decimal places, which is as good as you can expect. It is close enough to the correct value to let you know that you should reject $H_0.$
The normal approximation uses 50.5 (instead of 50) in order to more nearly match the discrete binomial distribution with the continuous normal distribution. In the figure below, you are interested in the probability to the left of the vertical dotted line.