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QuestionHomework /5:9(1,3,4,5)14(1,2)/ 5: 9(1,3,4,5) 14(1,2) Question 11 of 30 (1 point) I Question Attempt 1 of 3 Jonathan 06 6 07 ✓9\checkmark 9 10 11 12 13 14 15 Español 16 17
Confidence in banks: A poll conducted 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%13 \% of U.S\mathrm{U} . \mathrm{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\alpha=0.10 and α=0.01\alpha=0.01 levels of significance and the PP-value method with the TI-84 Plus calculator.
Part: 0/50 / 5
Part 1 of 5 (a) State the appropriate null and alternate hypotheses. H0:□H1:□\begin{array}{l} H_{0}: \square \\ H_{1}: \square \end{array}
This hypothesis test is a (Choose one) ∇\boldsymbol{\nabla} test. Skip Part Check Save For Later O 2024 MaGrow HIIILC. All Rights Resened

Studdy Solution

STEP 1

1. We are conducting a hypothesis test for a proportion.
2. The sample size is 1358 adults, with 149 expressing a great deal of confidence in banks.
3. The economist claims that less than 13% of U.S. adults have a great deal of confidence in banks.
4. We will use the P-value method to test the hypothesis at two significance levels: α=0.10\alpha = 0.10 and α=0.01\alpha = 0.01.

STEP 2

1. State the null and alternative hypotheses.
2. Identify the type of hypothesis test.
3. Calculate the test statistic.
4. Determine the P-value.
5. Compare the P-value to the significance levels and draw a conclusion.

STEP 3

State the null and alternative hypotheses.
The null hypothesis H0H_0 is that the proportion of U.S. adults with a great deal of confidence in banks is equal to 13%:
H0:p=0.13 H_0: p = 0.13
The alternative hypothesis H1H_1 is that the proportion is less than 13%:
H1:p<0.13 H_1: p < 0.13

STEP 4

Identify the type of hypothesis test.
This is a one-tailed test for a proportion.

STEP 5

Calculate the test statistic using the formula for a proportion:
z=p^−p0p0(1−p0)n z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}
Where: - p^=1491358\hat{p} = \frac{149}{1358} - p0=0.13p_0 = 0.13 - n=1358n = 1358
Calculate p^\hat{p}:
p^=1491358≈0.1097 \hat{p} = \frac{149}{1358} \approx 0.1097
Calculate the standard error:
SE=0.13×(1−0.13)1358≈0.0097 \text{SE} = \sqrt{\frac{0.13 \times (1 - 0.13)}{1358}} \approx 0.0097
Calculate the z-score:
z=0.1097−0.130.0097≈−2.09 z = \frac{0.1097 - 0.13}{0.0097} \approx -2.09

STEP 6

Determine the P-value using the z-score.
Using a standard normal distribution table or calculator, find the P-value for z=−2.09z = -2.09.
P-value ≈0.0183\approx 0.0183

STEP 7

Compare the P-value to the significance levels α=0.10\alpha = 0.10 and α=0.01\alpha = 0.01.
- For α=0.10\alpha = 0.10: 0.0183<0.100.0183 < 0.10, reject H0H_0. - For α=0.01\alpha = 0.01: 0.0183>0.010.0183 > 0.01, do not reject H0H_0.
Conclusion: At α=0.10\alpha = 0.10, there is sufficient evidence to support the economist's claim that less than 13% of U.S. adults have a great deal of confidence in banks. However, at α=0.01\alpha = 0.01, there is not sufficient evidence to support the claim.

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