Math Problem Statement

Confidence in banks: A news poll conducted in 2012

 asked a random sample of 1358

 adults in the United States how much confidence they had in banks and other financial institutions. A total of 149

 adults said that they had a great deal of confidence. An economist claims that less than 13

%

 of U.S. adults have a great deal of confidence in banks. Can you conclude that the economist's claim is true? Use both =

α

0.10

 and =

α

0.01

 levels of significance and the critical value method with the table.

Part: 0 / 5

0 of 5 Parts Complete

Part 1 of 5

State the appropriate null and alternate hypotheses.

H

0

H

1

This hypothesis test is a ▼

(Choose one)

 test.

Part: 1 / 5

1 of 5 Parts Complete

Part 2 of 5

Find the critical values. Round the answer to three decimal places.

For =

α

0.10

, the critical value is 

.

For =

α

0.01

, the critical value is 

.

Part: 2 / 5

2 of 5 Parts Complete

Part 3 of 5

Compute the test statistic. Do not round intermediate calculations. Round the answer to two decimal places.

=

z

Solution

Let's go through the solution step-by-step.

Part 1: State the Hypotheses

The problem states that an economist claims that less than 13% of U.S. adults have a great deal of confidence in banks. This means we are testing whether the population proportion pp is less than 0.13.

  • Null Hypothesis (H0H_0): p=0.13p = 0.13
  • Alternative Hypothesis (H1H_1): p<0.13p < 0.13

This is a left-tailed test since we are testing if the proportion is less than the hypothesized value.

Part 2: Find the Critical Values

For a left-tailed test, we need to find the critical zz-values for both significance levels, α=0.10\alpha = 0.10 and α=0.01\alpha = 0.01.

  1. For α=0.10\alpha = 0.10, the critical value zαz_{\alpha} corresponds to the 10th percentile of the standard normal distribution.
  2. For α=0.01\alpha = 0.01, the critical value zαz_{\alpha} corresponds to the 1st percentile of the standard normal distribution.

Using the standard normal table:

  • For α=0.10\alpha = 0.10: z0.101.282z_{0.10} \approx -1.282
  • For α=0.01\alpha = 0.01: z0.012.326z_{0.01} \approx -2.326

Part 3: Compute the Test Statistic

  1. Sample Proportion p^\hat{p}: p^=14913580.1097\hat{p} = \frac{149}{1358} \approx 0.1097

  2. Standard Error SESE: SE=p(1p)n=0.13×(10.13)13580.0094SE = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.13 \times (1 - 0.13)}{1358}} \approx 0.0094

  3. Test Statistic zz: z=p^pSE=0.10970.130.00942.16z = \frac{\hat{p} - p}{SE} = \frac{0.1097 - 0.13}{0.0094} \approx -2.16

Now that we have the test statistic z2.16z \approx -2.16, we can compare it to our critical values.

Would you like me to proceed with the conclusions for both significance levels?


Follow-up Questions

  1. How do you interpret the results of a hypothesis test?
  2. What does a left-tailed test signify in this context?
  3. Why is it essential to specify the significance level in hypothesis testing?
  4. How do we interpret the critical value in a hypothesis test?
  5. What steps would we take if the test were two-tailed?

Tip

Always check if the assumptions of the hypothesis test are met, such as sample size and randomness.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Hypothesis Testing
Proportion Testing
Confidence Levels
Significance Levels

Formulas

Sample proportion \( \hat{p} = \frac{x}{n} \)
Standard Error \( SE = \sqrt{\frac{p(1 - p)}{n}} \)
Test Statistic \( z = \frac{\hat{p} - p}{SE} \)

Theorems

Normal Approximation for Proportion Testing

Suitable Grade Level

Undergraduate (Statistics)