Answer
Yes, see explanations.
Work Step by Step
Given $p=0.593, n=300, \hat p=156/300=0.52$
a. State the hypotheses and identify the claim.
$H_o: \mu\geq 0.593$
$H_a: \mu\lt 0.593$ (claim, left tail test)
b. Find the critical value(s).
$\alpha=0.01, z_c=-2.33$
c. Compute the test value.
$z=\frac{0.52-0.593}{\sqrt {0.593\times0.407/300}}=-2.57$
d. Make the decision.
Since $z\lt z_c$, we have enough evidence to reject the null hypothesis and support the claim.
e. Summarize the results.
At the 0.01 level of significance, there is sufficient evidence to conclude that the proportion is less than 59.3%?
