Confidence in banks: A news poll conducted in 2012 asked a random sample of 1220 adults in the United States how much confidence they had in banks and other financial institutions. A total of 156 adults said that they had a great deal of confidence. An economist claims that greater than \( \mathbf{1 2 \%} \) of U.S. adults have a great deal of confidence in banks. Can you conclude that the economist's claim is true? Use both \( \alpha=0.10 \) and \( \alpha=0.01 \) levels of significance and the criticat vatue method with the table. Part: \( 0 / 5 \) Part 1 of 5 State the appropriate null and alternate hypotheses. \[ \begin{array}{l} H_{0}: p=0.12 \\ H_{1}: p>0.12 \end{array} \] This hypothesis test is a \( \square \) right-talled test. ■<0 \( \square>\square \) \( \square=\square \) \( \square \geqslant \square \) p This hypothesis test is a ght-tailed \( \square \) Part: \( 1 / 5 \) Part 2 of 5 Find the critical values. Round the answer to three decimal places. For \( \alpha=0.10 \), the critical value is \( \square \) 1.282 For \( \alpha=0.01 \), the critical value is 2.326 . Part: \( 2 / 5 \) Part 3 of 5 Compute the test statistic. Do not round intermediate calculations. Round the answer to two decimal places. \[ z=0.86 \] Part: \( 3 / 5 \) Part 4 of 5 Determine whether to reject \( H_{0} \). At the \( \alpha=0.10 \) level, (Choose one) \( \mathbf{~} \) the null hypothesis \( H_{0} \). reiect At the \( \alpha=0.01 \) level, do not reject Lunoose biet the null hypothesis \( H_{0} \).

Let's go through the hypothesis testing step by step. ### Part 1: State the Hypotheses The null and alternative hypotheses are correctly stated as: \[ \begin{array}{l} H_{0}: p = 0.12 \\ H_{1}: p > 0.12 \end{array} \] This hypothesis test is a **right-tailed test**. ### Part 2: Find the Critical Values To find the critical values for the significance levels \( \alpha = 0.10 \) and \( \alpha = 0.01 \): 1. For \( \alpha = 0.10 \): - The critical value from the z-table for a right-tailed test at \( \alpha = 0.10 \) is approximately \( z = 1.282 \). 2. For \( \alpha = 0.01 \): - The critical value from the z-table for a right-tailed test at \( \alpha = 0.01 \) is approximately \( z = 2.326 \). ### Part 3: Compute the Test Statistic To compute the test statistic \( z \), we use the formula: \[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} \] Where: - \( \hat{p} = \frac{x}{n} \) is the sample proportion, - \( p_0 = 0.12 \) is the hypothesized population proportion, - \( n = 1220 \) is the sample size, - \( x = 156 \) is the number of adults with a great deal of confidence. First, calculate \( \hat{p} \): \[ \hat{p} = \frac{156}{1220} \approx 0.1287 \] Now, substitute the values into the formula: \[ z = \frac{0.1287 - 0.12}{\sqrt{\frac{0.12(1 - 0.12)}{1220}}} \] Calculating the denominator: \[ \sqrt{\frac{0.12 \times 0.88}{1220}} \approx \sqrt{\frac{0.1056}{1220}} \approx \sqrt{0.0000865} \approx 0.0093 \] Now, calculate \( z \): \[ z = \frac{0.1287 - 0.12}{0.0093} \approx \frac{0.0087}{0.0093} \approx 0.9355 \] Rounding to two decimal places, we get: \[ z \approx 0.94 \] ### Part 4: Determine Whether to Reject \( H_{0} \) Now we compare the test statistic \( z \) with the critical values: 1. For \( \alpha = 0.10 \): - Critical value: \( 1.282 \) - Test statistic: \( 0.94 \) - Since \( 0.94 < 1.282 \), we **do not reject** \( H_{0} \). 2. For \( \alpha = 0.01 \): - Critical value: \( 2.326 \) - Test statistic: \( 0.94 \) - Since \( 0.94 < 2.326 \), we **do not reject** \( H_{0} \). ### Conclusion At both significance levels \( \alpha = 0.10 \) and \( \alpha = 0.01 \), we do not have enough evidence to support the economist's claim that greater than \( 12\% \) of U.S. adults have a great deal of confidence in banks.