Question

A well-known brokerage firm executive claimed that 90% 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, 81% of them said they are confident of meeting their goals.

Test the claim that the proportion of people who are confident is smaller than 90% at the 0.05 significance level.

The null and alternative hypothesis would be:

A. H_0 : p = 0.9
H_1 : p ≠ 0.9

B. H_0 : p ≥ 0.9
H_1 : p < 0.9

C. H_0 : p = 0.9
H_1 : p ≠ 0.9

D. H_0 : μ ≤ 0.9
H_1 : μ > 0.9

E. H_0 : μ ≥ 0.9
H_1 : μ < 0.9

F. H_0 : μ ≥ 0.9
H_1 : μ < 0.9

The test is:

A. right-tailed

B. left-tailed

C. two-tailed


The test statistic is: ___ (to 3 decimals)

The p-value is: ___ (to 4 decimals)

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

Answer 1

To conduct this statistical test, let's start by identifying the correct hypotheses and test type.

### Hypotheses
Since we want to test whether the proportion of people who are confident is smaller than 90%, the appropriate null and alternative hypotheses are:

**B. H_0 : p ≥ 0.9
H_1 : p < 0.9**

### Test Type
Since we are testing if the proportion is less than (i.e., we are evaluating if it has decreased from 90%), the test is:

**B. left-tailed**

### Test Statistic Calculation
We will use the formula for the test statistic for proportions:

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

Where:
- \(\hat{p} = 0.81\) (sample proportion)
- \(p_0 = 0.9\) (hypothesized population proportion)
- \(n = 200\) (sample size)

Plugging in the values:

1. Calculate the standard error:
\[
SE = \sqrt{\frac{p_0(1 - p_0)}{n}} = \sqrt{\frac{0.9(1 - 0.9)}{200}} = \sqrt{\frac{0.9 \times 0.1}{200}} = \sqrt{\frac{0.09}{200}} \approx \sqrt{0.00045} \approx 0.0212
\]

2. Now, calculate the z statistic:
\[
z = \frac{0.81 - 0.9}{0.0212} \approx \frac{-0.09}{0.0212} \approx -4.245
\]

So, the test statistic is approximately:

**Test Statistic: -4.245 (to 3 decimals: -4.245)**

### P-value Calculation
We will find the p-value corresponding to the calculated z-value in a standard normal distribution. The p-value for a left-tailed test can be found using a standard normal distribution table or calculation.

Looking up \(z = -4.245\), or using a calculator:

\[
p\text{-value} \approx 0.000014 (to 4 decimals: 0.0000)
\]

### Conclusion
Since the p-value is significantly lower than the significance level \(\alpha = 0.05\), we have sufficient evidence to reject the null hypothesis.

Thus, based on the calculated values:

**Final Answers:**

Test statistic is: **-4.245**

p-value is: **0.0000**

Based on this we:
**B. Reject the null hypothesis**