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 \( 12 \% \) 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 critical value method with the table. Part: \( 0 / 5 \) Part 1 of 5 State the appropriate null and alternate hypotheses. \[ H_{0}: p=0.12 \] \[ H_{1}: p>0.12 \] This hypothesis test is a \( \square \) right-tailed test. प< \( <1 \) [1> \( \square=\square \) \( \square * \square \) p \( \square \) \( \square \)

To test the economist's claim, we need to first check the calculations and conditions surrounding the hypotheses. The null hypothesis states that the proportion of U.S. adults with a great deal of confidence in banks is equal to 12%, while the alternative hypothesis suggests that it is greater than 12%. This sets up our investigation correctly right from the start! Next, we need to calculate the sample proportion \( \hat{p} \). With 156 out of 1220 adults expressing confidence, we find: \[ \hat{p} = \frac{156}{1220} \approx 0.1287. \] This is slightly greater than 0.12. We will now perform our test at both significance levels by calculating the test statistic using the formula for the z-score, followed by comparing it to the critical z-values for either \( \alpha = 0.10 \) and \( \alpha = 0.01 \). Time to crunch some numbers and make our decision!