From the uploaded image, it appears that the question involves hypothesis testing for the proportion of investors who are confident in meeting their goals. Here’s a detailed breakdown:

Problem Setup

1. Claim: The proportion of investors who are confident is less than 50%.
2. Sample Proportion: $\hat{p} = 0.41$ (41%).
3. Sample Size: $n = 800$.
4. Significance Level: $\alpha = 0.10$.

Hypotheses

• Null Hypothesis ($H_0$): $p = 0.50$.
• Alternative Hypothesis ($H_a$): $p < 0.50$ (left-tailed test).

Test Statistic

The formula for the test statistic for a proportion is: $z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1 - p_0)}{n}}}$

Plugging in the values: $z = \frac{0.41 - 0.50}{\sqrt{\frac{0.50 \cdot 0.50}{800}}}$ $z = \frac{-0.09}{\sqrt{\frac{0.25}{800}}} = \frac{-0.09}{0.01768} \approx -5.091$

P-Value

Using the standard normal table (Z-table), the p-value corresponding to $z = -5.091$ is approximately $0.0000$, as the value is extremely small.

Conclusion

Since the p-value ($0.0000$) is less than $\alpha = 0.10$, we reject the null hypothesis. This supports the claim that the proportion of confident investors is less than 50%.

Would you like further clarification or additional details about hypothesis testing or this specific question? Here are five related questions to expand your understanding:

1. How do we interpret a p-value in hypothesis testing?
2. Why do we use a Z-test for proportions, and what are the assumptions?
3. What happens if the sample size $n$ is smaller—how does it affect the result?
4. How does the choice of significance level $\alpha$ affect hypothesis test conclusions?
5. What is the difference between left-tailed, right-tailed, and two-tailed tests?

Tip

When conducting hypothesis tests, always check if the sample size is sufficiently large to justify the use of the normal approximation for proportions ($np_0 \geq 10$ and $n(1 - p_0) \geq 10$).