A well-known brokerage firm executive claimed that $70 \%$ of investors are currently confident of meeting their investment goals. An XYZ Investor Optimism Survey, conducted over a two week period, found that in a sample of 400 people, $66 \%$ of them said they are confident of meeting their goals.
Test the claim that the proportion of people who are confident is smaller than $70 \%$ at the 0.10 significance level.
The null and alternative hypothesis would be:
\[
\begin{array}{l}
H_{0}: \mu=0.7 \quad H_{0}: p=0.7 \quad H_{0}: p \leq 0.7 \quad H_{0}: \mu \geq 0.7 \quad H_{0}: p \geq 0.7 \quad H_{0}: \mu \leq 0.7 \text { o6 } \\
H_{1}: \mu \neq 0.7 \quad H_{1}: p \neq 0.7 \quad H_{1}: p> 0.7 \quad H_{1}: \mu< 0.7 \quad H_{1}: p< 0.7 \quad H_{1}: \mu> 0.7 \\
\end{array}
\]
The test ic:
left-tailed two-tailed right-tailed $\phi^{s}$
The test statistic is:
(to 3 decimals)
The $p$-value is:
(to 4 decimals)
Answer
Final Answer: The test statistic is approximately \(\boxed{-1.746}\) and the p-value is approximately \(\boxed{0.0404}\). Since the p-value is less than the significance level of 0.10, we reject the null hypothesis. Therefore, we have enough evidence to support the claim that the proportion of investors who are confident of meeting their goals is less than 70%.
Step 1 ：We are given a claim that the proportion of investors who are confident of meeting their goals is 70% (p=0.7). We are asked to test this claim against the alternative that the proportion is less than 70% (p<0.7). This is a one-tailed (left-tailed) test.
Step 2 ：We are given a sample size of 400 and a sample proportion of 66% (p_hat=0.66).
Step 3 ：We calculate the test statistic (z-score) using the formula: z = (p_hat - p) / sqrt((p*(1-p))/n), where p is the hypothesized population proportion, p_hat is the sample proportion, and n is the sample size.
Step 4 ：Substituting the given values into the formula, we get: z = (0.66 - 0.7) / sqrt((0.7*(1-0.7))/400) = -1.746 (rounded to three decimal places).
Step 5 ：We then calculate the p-value, which is the probability of observing a test statistic as extreme as the one we calculated (or more extreme) under the null hypothesis. For a left-tailed test, the p-value is the probability that a standard normal random variable is less than the test statistic. The p-value is approximately 0.0404.
Step 6 ：Since the p-value is less than the significance level of 0.10, we reject the null hypothesis.
Step 7 ：Therefore, we have enough evidence to support the claim that the proportion of investors who are confident of meeting their goals is less than 70%.
Step 8 ：Final Answer: The test statistic is approximately \(\boxed{-1.746}\) and the p-value is approximately \(\boxed{0.0404}\). Since the p-value is less than the significance level of 0.10, we reject the null hypothesis. Therefore, we have enough evidence to support the claim that the proportion of investors who are confident of meeting their goals is less than 70%.
