Confidence in banks: A poll conducted asked a random sample of 1276 adults in the United States how much confidence they had in banks and other financial institutions. A total of 150 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=.01  and a=.05 levels of significance and the P-value method with the TI-84 Plus calculator.

To determine whether the economist's claim that less than 14% of U.S. adults have a great deal of confidence in banks is true, we will conduct a hypothesis test for a population proportion using the given sample data. We will use both \(\alpha = 0.01\) and \(\alpha = 0.05\) significance levels and the P-value method.

Here's the step-by-step process to solve this problem:

### Step 1: State the Hypotheses
The null hypothesis (\(H_0\)) and the alternative hypothesis (\(H_a\)) are:
- \(H_0: p \geq 0.14\)
- \(H_a: p < 0.14\)

Where \(p\) is the proportion of U.S. adults who have a great deal of confidence in banks.

### Step 2: Calculate the Test Statistic
We will use the sample proportion (\(\hat{p}\)) and compare it to the hypothesized proportion (0.14).

The sample proportion is:
\[
\hat{p} = \frac{150}{1276} \approx 0.1175
\]

Next, we calculate the standard error (SE) using the hypothesized proportion \(p_0 = 0.14\):
\[
SE = \sqrt{\frac{p_0 (1 - p_0)}{n}} = \sqrt{\frac{0.14 \cdot 0.86}{1276}} \approx 0.0097
\]

The test statistic (z) is then calculated as:
\[
z = \frac{\hat{p} - p_0}{SE} = \frac{0.1175 - 0.14}{0.0097} \approx -2.31
\]

### Step 3: Determine the P-value
Using the TI-84 Plus calculator, we can find the P-value for the test statistic \(z = -2.31\).

1. Press `2nd` and then `VARS` to access the DISTR menu.
2. Choose `normalcdf(`.
3. Enter the lower bound as `-1E99`, the upper bound as `-2.31`, and the mean and standard deviation as `0` and `1`, respectively: `normalcdf(-1E99, -2.31, 0, 1)`.
4. Press `ENTER` to get the P-value.

The P-value is approximately 0.0104.

### Step 4: Make a Decision
Compare the P-value to the significance levels \(\alpha = 0.01\) and \(\alpha = 0.05\):

- For \(\alpha = 0.05\): The P-value (0.0104) is less than 0.05, so we reject the null hypothesis.
- For \(\alpha = 0.01\): The P-value (0.0104) is greater than 0.01, so we fail to reject the null hypothesis.

### Conclusion
- At \(\alpha = 0.05\): We reject the null hypothesis and conclude that there is sufficient evidence to support the economist's claim that less than 14% of U.S. adults have a great deal of confidence in banks.
- At \(\alpha = 0.01\): We fail to reject the null hypothesis, meaning we do not have sufficient evidence to support the economist's claim at this stricter significance level.

Would you like further details on this process or have any questions?

### Additional Related Questions:
1. How would the results change if the sample size were larger?
2. What if the proportion who had a great deal of confidence was higher, say 200 out of 1276?
3. Can we use this result to generalize about the confidence in banks globally?
4. What are the assumptions underlying this hypothesis test?
5. How does the choice of significance level affect the conclusions we draw?
6. What is the power of this test, and how could it be calculated?
7. How could we use a confidence interval to check the economist's claim?
8. What are some potential sources of bias in this poll?

### Tip:
When conducting hypothesis tests, always ensure that the assumptions (such as random sampling and appropriate sample size) are met to validate the test results.

Explore a hypothesis test for population proportion using sample data on confidence in banks among U.S. adults. Includes step-by-step calculation of test statistic, P-value determination, and conclusion at significance levels of 0.01 and 0.05.

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