"A survey was undertaken by Bruskin/Goldring Research for Quicken to determine how people plan to meet their financial goals in the next year. Respondents were allowed to select more than one way to meet their goals. Thirty-one percent said that they were using a financial planner to help them meet their goals. Twenty-four percent were using family/friends to help them meet their financial goals followed by broker/accountant ($19%$), computer software ($17%$), and books ($14%$). Suppose another researcher takes a similar survey of $575$ people to test these results.
Edit:If 191 people respond that they are going to use a financial planner to help them meet their goals, is this proportion enough evidence to reject the 31% figure generated in the Bruskin/Goldring survey using α = 0.10."
Hypotheses:
$H_0: p=0.31$
$H_a: p \neq 0.31$
$n=575$
$x=191$
$a=0.1$
$\hat p=191/575=0.33$
$\hat q=1-\hat p=0.69$
$$\frac{(\hat p-p)}{\sqrt{ pq/n}} = (0.33-0.31)/\sqrt{(0.31 \times 0.69/575)} = 1.03$$
$z_\text{table}=.15151$
Fail to reject null
Edit2: There is not enough evidence to declare that the proportion is any different than $0.31.$
The numerical answer is wrong so what mistake did I make? What's the correct equation to solve for test equation?
Your problem is that you are not testing enough. They don't want to know whether the proportion of planners is the same. They want to know whether all of the proportions are the same.
The test you need is a $\chi^2$ goodness-of-fit test. Your null hypothesis is that $p_1 = .31, p_2 = .24, p_3 = .19, p_4 = .17, p_5 = .14$
Given $n = 575$, you can estimate how many of each you expect to get. You can calculate the test statistic as $$\sum \frac{(O-E)^2}{E}$$ once you have data to work with.
The degrees of freedom is the number of categories $-1$, so $4$ in this case.
Decide on your level of significance and look up the critical value of the $\chi^2$ for that choice of $\alpha$ and df = $4$.
EDIT: Since the question has changed, my answer must as well. The question now specifies that you are only checking one proportion, not all of them. So a $\chi^2$ test is not necessary, and a test of one proportion is sufficient. Your procedures looks correct, but you appear to be rounding too much.
$\hat p = 191/575 \approx 0.33217$
$\hat q = 0.66783$
Standard error of the proportion $\sigma_p = 0.019642$
$z$ score is $\frac{.33217-.31}{.019642} = 1.1287$
The problem does not specify whether a one-sided difference is being tested for, so we assume a two-tailed test. Since $\alpha = 0.10$, that means that each tail has area $0.05$. Using a table, or invnorm$(0.95)$ on a TI-83, we get $z_{critical} = 1.6448.
Since the test statistic -1.6448 < 1.1287 < 1.16448, we fail to reject the null hypothesis.