Problem
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 below Hypothesis Test for a Population Proportion A well-known brokerage firm executive claimed that \( 90 \% \) of investors are currently confiden their investment goals. An XYZ Investor Optimism Survey, conducted over a two week perioc 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 the 0.00 level. The null and alternative hypothêsis would be: \[ \begin{array}{l} H_{0}: \mu \geq 0.9 \quad H_{0}: \mu \leq 0.9 \quad H_{0}: p \leq 0.9 \quad H_{0}: p=0.9 \quad H_{0}: p \geq 0.9 \quad H_{0}: \mu=0.9 \quad \sigma \\ H_{1}: \mu<0.9 H_{1}: \mu>0.9 H_{1}: p>0.9 H_{1}: p \neq 0.9 \quad H_{1}: p<0.9 \quad H_{1}: \mu \neq 0.9 \\ \end{array} \] The test is: two-tailed right-tailed left-tailed \( \sigma^{\sigma} \checkmark \) Type here to search
Solution
Step 1 ：Step 1: Set the null and alternative hypotheses: \(H_0: p = 0.9\) and \(H_1: p > 0.9\)(right-tailed test).
Step 2 ：Step 2: Calculate the test statistic: \(z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} = \frac{0.92 - 0.9}{\sqrt{\frac{0.9(1-0.9)}{700}}}\).
Step 3 ：Step 3: Find the p-value and compare it to the significance level \(\alpha = 0.01\). If p-value \(< \alpha\), reject \(H_0\); otherwise, do not reject \(H_0\).
