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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?

5 answers

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.

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

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: \[ \boxed{-3.05} \]

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

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: \[ \boxed{0.0011} \]