Question

A well-known brokerage firm executive claimed that 80% 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 800 people, 86% of them said they are confident of meeting their goals.

Test the claim that the proportion of people who are confident is larger than 80% at the 0.005 significance level.

What would the null and alternative hypothesis would be: ____

The test is: right-tailed

left-tailed

two-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 that the proportion of people who are confident is larger than 80%, we can set up our hypothesis testing as follows:

### Null and Alternative Hypotheses:
- **Null Hypothesis (H0)**: \( p \leq 0.80 \) (The proportion of confident investors is less than or equal to 80%)
- **Alternative Hypothesis (H1)**: \( p > 0.80 \) (The proportion of confident investors is greater than 80%)

### Type of Test:
Since we are checking if the proportion of confident investors is greater than 80%, this is a **right-tailed test**.

### Test Statistic Calculation:
To calculate the test statistic for a proportion, we use the formula for the z-test statistic:

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

Where:
- \( \hat{p} \): Sample proportion (86% or 0.86)
- \( p_0 \): Hypothesized population proportion (80% or 0.80)
- \( n \): Sample size (800)

Substituting the values into the equation:

\[
z = \frac{0.86 - 0.80}{\sqrt{\frac{0.80(1 - 0.80)}{800}}}
\]

Calculating the standard error:

\[
\text{Standard Error} = \sqrt{\frac{0.80 \times 0.20}{800}} = \sqrt{\frac{0.16}{800}} = \sqrt{0.0002} \approx 0.0141421
\]

Now, substituting back to find \( z \):

\[
z \approx \frac{0.06}{0.0141421} \approx 4.243
\]

So, the test statistic is: **4.243** (to three decimals).

### P-value Calculation:
To find the p-value for a right-tailed test with \( z = 4.243 \), we can use a z-table or a statistical software. Looking up \( z = 4.243 \):

The p-value is extremely small; for practical purposes, it can be considered close to **0.0000** (to four decimals).

### Conclusion Based on P-value:
Given that the p-value is significantly smaller than the significance level of \( 0.005 \):

- We **reject the null hypothesis**.

### Summary of Results:
1. **Null and Alternative Hypotheses**:
- \( H_0: p \leq 0.80 \)
- \( H_1: p > 0.80 \)

2. **Test Type**: Right-tailed

3. **Test Statistic**: \( z \approx 4.243 \)

4. **P-value**: \( \approx 0.0000 \)

5. **Conclusion**: Reject the null hypothesis.

Thus, the claim that the proportion of people who are confident is larger than 80% is supported by the data.