Question

A poll conducted asked a random sample of 1342 adults in the United States how much confidence they had in banks and other financial institutions. A total of 143 adults said that they had a great deal of confidence. An economist claims that less than 14% of U.S. adults have a great deal of confidence in banks. Can you conclude that the economist's claim is true? Use both a=0.10 and a=0.01 levels of significance and the P-value method.
Hv0: [?]
Hv1: [?]
is this two tailed, left tailed or right tailed?

Answer 1

(b) Compute the value of the test statistic. Round the answer to at least two decimal places.

Answer 2

(c) Compute the P-value. Round the answer to at least four decimal places.

Answer 3

To test the economist's claim that less than 14% of U.S. adults have a great deal of confidence in banks, we set up the following hypotheses:

**Null Hypothesis (H0):** The proportion of U.S. adults who have a great deal of confidence in banks is 14% or more.
\[ H_0: p \geq 0.14 \]

**Alternative Hypothesis (H1):** The proportion of U.S. adults who have a great deal of confidence in banks is less than 14%.
\[ H_1: p < 0.14 \]

Since the economist is claiming that the proportion is *less than* a specified value (14%), this is a **left-tailed test**.

### Testing at the significance levels \( \alpha = 0.10 \) and \( \alpha = 0.01 \)

To conduct the test, we need to calculate the sample proportion, the standard error, and then the test statistic.

1. **Calculate the Sample Proportion (\( \hat{p} \))**:
\[ \hat{p} = \frac{x}{n} = \frac{143}{1342} \approx 0.1065 \]

2. **Standard Error (SE)**:
\[ SE = \sqrt{\frac{p_0(1 - p_0)}{n}} \]
Where \( p_0 = 0.14 \):
\[ SE = \sqrt{\frac{0.14 \times (1 - 0.14)}{1342}} \approx \sqrt{\frac{0.14 \times 0.86}{1342}} \approx \sqrt{\frac{0.1204}{1342}} \approx 0.00943 \]

3. **Calculate the Z-Score**:
\[ Z = \frac{\hat{p} - p_0}{SE} = \frac{0.1065 - 0.14}{0.00943} \approx \frac{-0.0335}{0.00943} \approx -3.55 \]

4. **Find the P-value**:
Using standard normal distribution tables or calculators, find the P-value corresponding to \( Z = -3.55 \). The P-value is approximately 0.0002.

### Conclusion of the test:
- For \( \alpha = 0.10 \): Since \( P\text{-value} \approx 0.0002 < 0.10 \), we reject the null hypothesis.
- For \( \alpha = 0.01 \): Since \( P\text{-value} \approx 0.0002 < 0.01 \), we reject the null hypothesis.

### Final Conclusion:
At both significance levels (0.10 and 0.01), we can conclude that there is enough evidence to support the economist's claim that less than 14% of U.S. adults have a great deal of confidence in banks and other financial institutions.

Answer 4

To compute the value of the test statistic, let's summarize the relevant calculations:

1. **Sample Size \( n \)**:
\[
n = 1342
\]

2. **Number of adults with a great deal of confidence \( x \)**:
\[
x = 143
\]

3. **Sample Proportion \( \hat{p} \)**:
\[
\hat{p} = \frac{x}{n} = \frac{143}{1342} \approx 0.1065
\]

4. **Null Hypothesis Proportion \( p_0 \)**:
\[
p_0 = 0.14
\]

5. **Standard Error (SE)**:
\[
SE = \sqrt{\frac{p_0(1 - p_0)}{n}} = \sqrt{\frac{0.14 \times (1 - 0.14)}{1342}} = \sqrt{\frac{0.14 \times 0.86}{1342}} \approx \sqrt{0.0001204} \approx 0.01098
\]

6. **Calculate the Z-Score (Test Statistic)**:
\[
Z = \frac{\hat{p} - p_0}{SE} = \frac{0.1065 - 0.14}{0.01098} \approx \frac{-0.0335}{0.01098} \approx -3.05
\]

### Final Result:
The value of the test statistic \( Z \) (rounded to at least two decimal places) is:
\[
\boxed{-3.05}
\]

Answer 5

To compute the P-value associated with the test statistic \( Z \), we typically refer to the standard normal (Z) distribution.

Given that we calculated the test statistic as \( Z = -3.05 \), we now need to find the P-value for this left-tailed test.

1. **Using Standard Normal Distribution Tables or a Calculator**:
- The P-value is the probability that a standard normal random variable is less than \( Z = -3.05 \).

Using standard Z-tables or a normal distribution calculator, we can find:

\[
\text{P-value} = P(Z < -3.05)
\]

Using a standard normal distribution calculator, the P-value can be found to be approximately:

\[
\text{P-value} \approx 0.0011
\]

### Final Result:
The P-value, rounded to at least four decimal places, is:
\[
\boxed{0.0011}
\]