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Math  /  Data & Statistics

QuestionConfidence in banks: A poll conducted asked a random sample of 1268 adults in the United States how much confidence they had in banks and other financial institutions. A total of 144 adults said that they had a great deal of confidence. An economist claims that less than 11% 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=0.05\alpha = 0.05 and α=0.10\alpha = 0.10 levels of significance and the PP-value method with the TI-84 Plus calculator.
Part 1 of 5
(a) State the appropriate null and alternate hypotheses.
H0:p=0.11H_0: p = 0.11 H1:p<0.11H_1: p < 0.11
This hypothesis test is a left-tailed test.
Part: 1/5
Part 2 of 5
(b) Compute the value of the test statistic. Round the answer to at least two decimal places. z=z =

Studdy Solution

STEP 1

1. The sample size is n=1268 n = 1268 .
2. The number of adults with a great deal of confidence is x=144 x = 144 .
3. The claimed population proportion is p0=0.11 p_0 = 0.11 .
4. We are conducting a left-tailed test.

STEP 2

1. Calculate the sample proportion.
2. Compute the standard error.
3. Calculate the test statistic z z .

STEP 3

Calculate the sample proportion p^ \hat{p} :
p^=xn=1441268\hat{p} = \frac{x}{n} = \frac{144}{1268}
Calculate p^ \hat{p} :
p^≈0.1136\hat{p} \approx 0.1136

STEP 4

Compute the standard error SE SE using the formula:
SE=p0(1−p0)nSE = \sqrt{\frac{p_0(1 - p_0)}{n}}
Substitute the known values:
SE=0.11×(1−0.11)1268SE = \sqrt{\frac{0.11 \times (1 - 0.11)}{1268}}
Calculate SE SE :
SE≈0.0089SE \approx 0.0089

STEP 5

Calculate the test statistic z z using the formula:
z=p^−p0SEz = \frac{\hat{p} - p_0}{SE}
Substitute the known values:
z=0.1136−0.110.0089z = \frac{0.1136 - 0.11}{0.0089}
Calculate z z :
z≈0.4045z \approx 0.4045
The value of the test statistic is:
z≈0.40 z \approx 0.40

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