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Question 26
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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:
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(sese \( \iint^{n} ? \)
\( \left(^{12}+\frac{13}{13}+\frac{1}{1}\right. \)
(a) 2
8
0
Answer
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.
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.
