from the book, i have following task
i think there is error in formulation of alternative hypothesis , because word at least means minimum,there for instead of $p < 0.77$, it should be $p \ge 0.77$ , so instead of left test, it is right test and $\alpha=0.01$ means that for right test we will have $1-0.01=0.99$ and critical value is
=NORMSINV(0.99)
which is equal to 2.33
am i right?
On
The verbal claim is that "at least 77%" oppose replacing...etc. I would have formulated as follows:
$H_0: p\geq 0.77, H_1: p < 0.77.$
While what is claimed should generally be the alternate hypothesis, the wording of the problem suggests the authors felt that the seriousness of rejecting a true high estimate of public objection was greater than wrongly overestimating public objection.
The test is thus left-tailed as the authors indicate. If the point estimate is low enough (leading to a value less than -2.33) we would reject $H_0$ as formulated with high confidence.
The text states clearly, "to see if this claim is valid...etc." And associates this with "$H_0: p = 0.77$ (claim)." In context I think they meant to put $H_0: p\geq 0.77$ (claim).
If I were to interpret the intent of the question, I would say that the hypothesis should be $$H_0 : p = 0.77 \quad \text{vs.} \quad H_1 : p > 0.77.$$ The reason is that what is claimed is that at least $77\%$ of people oppose replacement, and in order to support this claim, the evidence needs to suggest with a high degree of confidence (in this case, $99\%$) that the true proportion is in fact this high. If we use the test described in the given answer, this could only lend evidence to the contrary--i.e., that the true proportion is less. It cannot support a finding that the claim is true.
In general, what is claimed should be the alternative (research) hypothesis.
It is worth noting that because the point estimate $\hat p = 55/80 = 0.6875 < 0.77$, the automatic conclusion of the test is to fail to reject $H_0$. There is not sufficient evidence to suggest the true proportion is so high.