A well-known brokerage firm executive claimed that [tex]$60 \%$[/tex] 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 200 people, [tex]$57 \%$[/tex] of them said they are confident of meeting their goals.

Test the claim that the proportion of people who are confident is smaller than [tex][tex]$60 \%$[/tex][/tex] at the 0.01 significance level.

The null and alternative hypotheses would be:
- [tex]H_0: p = 0.6[/tex]
- [tex]H_a: p \ \textless \ 0.6[/tex]

The test is:
- right-tailed
- left-tailed
- two-tailed

The test statistic is: [tex]\square[/tex] (to 3 decimals)

The [tex]p[/tex]-value is: [tex]\square[/tex] (to 4 decimals)

Based on this we:
- Reject the null hypothesis



Answer :

To test the claim that the proportion of people who are confident of meeting their investment goals is smaller than 60% at the 0.01 significance level, we start by setting up our hypotheses and conducting the appropriate statistical test.

### Hypotheses

The null and alternative hypotheses are:

[tex]\[ H_0: p = 0.6 \][/tex]
[tex]\[ H_a: p < 0.6 \][/tex]

This is a left-tailed test because we are testing if the sample proportion is significantly less than the null hypothesis proportion.

### Test Statistic Calculation

The sample proportion ([tex]\( p_{\text{sample}} \)[/tex]) is 0.57 from a sample size of 200. The null hypothesis proportion ([tex]\( p_{\text{null}} \)[/tex]) is 0.6.

The standard error of the sample proportion is calculated using the formula:

[tex]\[ \text{Standard Error} = \sqrt{\frac{p_{\text{null}} (1 - p_{\text{null}})}{n}} \][/tex]

Plugging in the given values:

[tex]\[ \text{Standard Error} = \sqrt{\frac{0.6 \times (1 - 0.6)}{200}} = 0.034641 \][/tex]

The test statistic (z-score) is calculated using the formula:

[tex]\[ z = \frac{p_{\text{sample}} - p_{\text{null}}}{\text{Standard Error}} \][/tex]

Again, using the given values:

[tex]\[ z = \frac{0.57 - 0.6}{0.034641} = -0.866 \][/tex]

### P-Value Calculation

The p-value is the probability of obtaining a test statistic as extreme as the one calculated, under the null hypothesis. Given that this is a left-tailed test, we find the p-value associated with our z-score from standard normal distribution tables or using statistical software.

[tex]\[ p\text{-value} = 0.1932 \][/tex]

### Decision

To conclude the test, we compare the p-value with the significance level ([tex]\(\alpha\)[/tex]).

[tex]\[ \alpha = 0.01 \][/tex]

Since the p-value (0.1932) is greater than [tex]\(\alpha\)[/tex] (0.01), we do not have sufficient evidence to reject the null hypothesis.

### Conclusion

Based on the calculated statistics:

- The test statistic is: [tex]\(-0.866\)[/tex]
- The p-value is: [tex]\(0.1932\)[/tex]

Thus, we do not reject the null hypothesis at the 0.01 significance level. This means that we do not have enough evidence to support the claim that the proportion of people confident in meeting their investment goals is less than 60%.