Question

A well-known brokerage firm executive claimed that 10% 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 500 people, 4% of them said they are confident of meeting their goals.

Test the claim that the proportion of people who are confident is smaller than 10% at the 0.05 significance level.


The test is:

right-tailed

two-tailed

left-tailed



The test statistic is:
(to 3 decimals)

The p-value is:
(to 4 decimals)

Based on this we:
Reject the null hypothesis
Fail to reject the null hypothesis

Answer 1

To test the claim, we will conduct a hypothesis test for a proportion.

### Step 1: Define the Hypotheses

- **Null Hypothesis \( H_0 \)**: The proportion of investors who are confident in meeting their goals is equal to 10%, \( p = 0.10 \).
- **Alternative Hypothesis \( H_a \)**: The proportion of investors who are confident in meeting their goals is smaller than 10%, \( p < 0.10 \).

This is a left-tailed test.

### Step 2: Determine the Test Statistic

We use the following formula for the test statistic \( z \):

\[
z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}
\]

Where:
- \( \hat{p} \) = sample proportion
- \( p_0 \) = hypothesized population proportion
- \( n \) = sample size

Given data:
- \( \hat{p} = \frac{4}{100} = 0.04 \)
- \( p_0 = 0.10 \)
- \( n = 500 \)

Now, we can calculate the standard error:

\[
\text{Standard Error (SE)} = \sqrt{\frac{p_0(1 - p_0)}{n}} = \sqrt{\frac{0.10(1 - 0.10)}{500}} = \sqrt{\frac{0.10 \cdot 0.90}{500}} = \sqrt{\frac{0.09}{500}} = \sqrt{0.00018} \approx 0.013416
\]

Now, substituting the values into the formula for \( z \):

\[
z = \frac{0.04 - 0.10}{0.013416} \approx \frac{-0.06}{0.013416} \approx -4.466
\]

### Step 3: Calculate the p-value

To find the p-value for a left-tailed test, we can use the standard normal (Z) distribution.

We can look up the value of \( z = -4.466 \) in the Z-table or use statistical software/calculator. The p-value is the probability of observing a value as extreme or more extreme than the test statistic under the null hypothesis:

\[
\text{p-value} \approx 0.00000425 \text{ (about 0.0000043)}
\]

### Step 4: Make a Decision

Now, we compare the p-value with the significance level \( \alpha = 0.05 \):

- \( \text{p-value} \approx 0.0000043 < 0.05 \)

This indicates that we can reject the null hypothesis.

### Summary of Results

- The test is: **left-tailed**
- The test statistic is: **-4.466**
- The p-value is: **0.0000** (to 4 decimal places: 0.0000)
- Based on this we: **Reject the null hypothesis**.

Thus, there is sufficient evidence to suggest that the proportion of investors who are confident in meeting their goals is smaller than 10%.