Problem
Employ Mail - CHAPT | BG My Gra \( (436 \mathrm{ur} \) \( \leftarrow \rightarrow \mathrm{C} \) a \( 1(1) \) Docum (5 OEC Re 5 OEC Re ХUIE: \( \angle J .00 / J U \) Question 26 Score on last try: 0.6 of 1 pts. See Details for more. > Next question \( \rightleftarrows \) Get a similar question You can retry this question be Hypothesis Test for a Population Proportion A well-known brokerage firm executive claimed that \( 90 \% \) of investors are currently c their investment goals. An XYZ Investor Optimism Survey, conducted over a two week a sample of 700 people, \( 92 \% \) of them said they are confident of meeting their goals. Test the claim that the proportion of people who are confident is larger than \( 90 \% \) at level. The null and alternative hypothesis would be: \[ \begin{array}{cccccc} H_{0}: \mu \geq 0.9 & H_{0}: \mu \leq 0.9 & H_{0}: p<0.9 & H_{0}: p=0.9 & H_{0}: p \geq 0.9 & H_{0}: \mu= \\ H_{1}: \mu<0.9 & H_{1}: \mu>0.9 & H_{1}: p>0.9 & H_{1}: p \neq 0.9 & H_{1}: p<0.9 & H_{1}: \mu \neq \\ \bigcirc & \bigcirc & 0 & \bigcirc & \bigcirc & \bigcirc \end{array} \] The test is: Type here to search (sese \( \iint^{n} ? \) \( \left(^{12}+\frac{13}{13}+\frac{1}{1}\right. \) (a) 2 8 0
Solution
Step 1 ：H_0: p = 0.9
Step 2 ：H_1: p > 0.9
Step 3 ：Use the test statistic: z = \(\frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}\), where \(\hat{p}\) is the sample proportion, \(p_0\) is the null hypothesis proportion, and n is the sample size.
