Answer
No. Yes. See explanations.
Work Step by Step
Given $\mu=4, \sigma=0.6, n=20, \bar X=4.2$
a. State the hypotheses and identify the claim.
$H_o: \mu=4$
$H_a: \mu\ne 4$ (claim, two tail test)
b. Find the critical value(s).
$\alpha/2=0.005, |z_c|=2.575$
c. Compute the test value.
$z=\frac{4.2-4}{0.6/\sqrt {20}}=1.49$
d. Make the decision
$z\lt 2.575$ is in the non-reject region and we do not reject the null hypothesis.
e. Summarize the results.
At $\alpha=0.01$, there is no reason to believe that the
new food is responsible for a change in the growth of the leaves.
At $c=0.99$, $z_c=2.575$, the margin of error is
$E=2.575\times\frac{4.2}{\sqrt {20}}=2.4$ so the interval for the mean
is $(4.2-2.4,4.2+2.4)$ which gives $1.8\leq \mu\leq 6.8$ and clearly
the mean $\mu=4$ fall in this range, so the results concur.
