The appropriate null and alternate hypotheses.

Question

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Answer 1

Question

Chapter 9.4, Problem 14E

(a)

To determine

The appropriate null and alternate hypotheses.

(a)

Expert Solution

Answer to Problem 14E

Null hypothesis:

H0:p=0.15

That is, 15% of U.S. adults have a great deal of confidence in banks.

Alternative hypothesis:

H1:p<0.15

That is less than 15% of U.S. adults have a great deal of confidence in banks.

Explanation of Solution

Given information:

In a survey of random 1325 adults, 149 adults have a great deal of confidence in banks and other financial institution.

Based on the survey an economist claims that less than 15% of US adults have confidence in banks.

Calculation:

Based on the survey an economist claims that less than 15% of US adults have confidence in banks.

So null and alternate hypotheses would be given as-

Null hypothesis:

H0:p=0.15

That is, 15% of U.S. adults have a great deal of confidence in banks.

Alternative hypothesis:

H1:p<0.15

That is less than 15% of U.S. adults have a great deal of confidence in banks.

(b)

To determine

The test statistic z.

(b)

Expert Solution

Answer to Problem 14E

z=−3.83.

Explanation of Solution

Given information:

In a survey of random 1325 adults, 149 adults have a great deal of confidence in banks and other financial institution.

Based on the survey an economist claims that less than 15% of US adults have confidence in banks.

Formula used:

p^=xnz=p^−p0 p 0(1−p 0)n

Calculation:

In a survey of random 1325 adults, 149 adults have a great deal of confidence in banks and other financial institution. i.e. n=1,325 and x=149

Based on the survey an economist claims that less than 15% of US adults have confidence in banks. i.e. p0=0.15.

So the sample proportion p^.

p^=xn=1491,325=0.1125

Compute the test statistic z by using the formula as given below.

z=p^−p0 p 0(1−p 0)n =0.1125−0.15 0.15(1−0.15)1,325 =−0.03750.0098=−3.83

Thus, the value of test statistic is z=−3.83.

(c)

To determine

Whether the claim made by economist is true for α=0.05.

(c)

Expert Solution

Answer to Problem 14E

The economist's claim is true.

Explanation of Solution

Given information:

In a survey of random 1325 adults, 149 adults have a great deal of confidence in banks and other financial institution.

Based on the survey an economist claims that less than 15% of US adults have confidence in banks.

Calculation:

Obtain the critical value.

Here, the test is left-tailed and the level of significance, α=0.05

From Table 9.1, the critical value for the level of significance α=0.05 is −1.645.

Rejection rule for left-tailed test:

If z≤−zα(=−1.645) then reject the null hypothesis

Conclusion for α=0.05

The value of test statistic is z=−3.83 and the critical value is zα=−1.645.

Here, the value of test statistic is lesser than the critical value.

That is z(=−3.83)≤zα(=−1.645)

Hence, the null hypothesis is rejected.

Thus it can be concluded that less than 15% of U.S. adults have a great deal of confidence in banks. Hence, the economist's claim is true.

(d)

To determine

Whether the claim made by economist is true for α=0.01.

(d)

Expert Solution

Answer to Problem 14E

The executive claim is true.

Explanation of Solution

Given information:

In a survey of random 1325 adults, 149 adults have a great deal of confidence in banks and other financial institution.

Based on the survey an economist claims that less than 15% of US adults have confidence in banks.

Calculation:

Obtain the critical value.

Here, the test is left-tailed and the level of significance, α=0.01

From Table 9.1, the critical value for the level of significance α=0.01 is −2.326.

Rejection rule for left-tailed test:

If z≤−zα(=−2.326) then reject the null hypothesis

Conclusion for α=0.01

The value of test statistic is −3.83 and the critical value is zα=−2.326.

Here, the value of test statistic is lesser than the critical value.

That is z(=−3.83)≤zα(=−2.326)

Hence, the null hypothesis is rejected.

Thus it can be concluded that less than 15% of U.S. adults have a great deal of confidence in banks. Hence, the executive’s claim is true.

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Answer 2

Question

Chapter 9.4, Problem 14E

(a)

To determine

The appropriate null and alternate hypotheses.

(a)

Expert Solution

Answer to Problem 14E

Null hypothesis:

H0:p=0.15

That is, 15% of U.S. adults have a great deal of confidence in banks.

Alternative hypothesis:

H1:p<0.15

That is less than 15% of U.S. adults have a great deal of confidence in banks.

Explanation of Solution

Given information:

In a survey of random 1325 adults, 149 adults have a great deal of confidence in banks and other financial institution.

Based on the survey an economist claims that less than 15% of US adults have confidence in banks.

Calculation:

Based on the survey an economist claims that less than 15% of US adults have confidence in banks.

So null and alternate hypotheses would be given as-

Null hypothesis:

H0:p=0.15

That is, 15% of U.S. adults have a great deal of confidence in banks.

Alternative hypothesis:

H1:p<0.15

That is less than 15% of U.S. adults have a great deal of confidence in banks.

(b)

To determine

The test statistic z.

(b)

Expert Solution

Answer to Problem 14E

z=−3.83.

Explanation of Solution

Given information:

In a survey of random 1325 adults, 149 adults have a great deal of confidence in banks and other financial institution.

Based on the survey an economist claims that less than 15% of US adults have confidence in banks.

Formula used:

p^=xnz=p^−p0 p 0(1−p 0)n

Calculation:

In a survey of random 1325 adults, 149 adults have a great deal of confidence in banks and other financial institution. i.e. n=1,325 and x=149

Based on the survey an economist claims that less than 15% of US adults have confidence in banks. i.e. p0=0.15.

So the sample proportion p^.

p^=xn=1491,325=0.1125

Compute the test statistic z by using the formula as given below.

z=p^−p0 p 0(1−p 0)n =0.1125−0.15 0.15(1−0.15)1,325 =−0.03750.0098=−3.83

Thus, the value of test statistic is z=−3.83.

(c)

To determine

Whether the claim made by economist is true for α=0.05.

(c)

Expert Solution

Answer to Problem 14E

The economist's claim is true.

Explanation of Solution

Given information:

In a survey of random 1325 adults, 149 adults have a great deal of confidence in banks and other financial institution.

Based on the survey an economist claims that less than 15% of US adults have confidence in banks.

Calculation:

Obtain the critical value.

Here, the test is left-tailed and the level of significance, α=0.05

From Table 9.1, the critical value for the level of significance α=0.05 is −1.645.

Rejection rule for left-tailed test:

If z≤−zα(=−1.645) then reject the null hypothesis

Conclusion for α=0.05

The value of test statistic is z=−3.83 and the critical value is zα=−1.645.

Here, the value of test statistic is lesser than the critical value.

That is z(=−3.83)≤zα(=−1.645)

Hence, the null hypothesis is rejected.

Thus it can be concluded that less than 15% of U.S. adults have a great deal of confidence in banks. Hence, the economist's claim is true.

(d)

To determine

Whether the claim made by economist is true for α=0.01.

(d)

Expert Solution

Answer to Problem 14E

The executive claim is true.

Explanation of Solution

Given information:

In a survey of random 1325 adults, 149 adults have a great deal of confidence in banks and other financial institution.

Based on the survey an economist claims that less than 15% of US adults have confidence in banks.

Calculation:

Obtain the critical value.

Here, the test is left-tailed and the level of significance, α=0.01

From Table 9.1, the critical value for the level of significance α=0.01 is −2.326.

Rejection rule for left-tailed test:

If z≤−zα(=−2.326) then reject the null hypothesis

Conclusion for α=0.01

The value of test statistic is −3.83 and the critical value is zα=−2.326.

Here, the value of test statistic is lesser than the critical value.

That is z(=−3.83)≤zα(=−2.326)

Hence, the null hypothesis is rejected.

Thus it can be concluded that less than 15% of U.S. adults have a great deal of confidence in banks. Hence, the executive’s claim is true.

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Answer 3

Question

Chapter 9.4, Problem 14E

(a)

To determine

The appropriate null and alternate hypotheses.

(a)

Expert Solution

Answer to Problem 14E

Null hypothesis:

H0:p=0.15

That is, 15% of U.S. adults have a great deal of confidence in banks.

Alternative hypothesis:

H1:p<0.15

That is less than 15% of U.S. adults have a great deal of confidence in banks.

Explanation of Solution

Given information:

In a survey of random 1325 adults, 149 adults have a great deal of confidence in banks and other financial institution.

Based on the survey an economist claims that less than 15% of US adults have confidence in banks.

Calculation:

Based on the survey an economist claims that less than 15% of US adults have confidence in banks.

So null and alternate hypotheses would be given as-

Null hypothesis:

H0:p=0.15

That is, 15% of U.S. adults have a great deal of confidence in banks.

Alternative hypothesis:

H1:p<0.15

That is less than 15% of U.S. adults have a great deal of confidence in banks.

(b)

To determine

The test statistic z.

(b)

Expert Solution

Answer to Problem 14E

z=−3.83.

Explanation of Solution

Given information:

In a survey of random 1325 adults, 149 adults have a great deal of confidence in banks and other financial institution.

Based on the survey an economist claims that less than 15% of US adults have confidence in banks.

Formula used:

p^=xnz=p^−p0 p 0(1−p 0)n

Calculation:

In a survey of random 1325 adults, 149 adults have a great deal of confidence in banks and other financial institution. i.e. n=1,325 and x=149

Based on the survey an economist claims that less than 15% of US adults have confidence in banks. i.e. p0=0.15.

So the sample proportion p^.

p^=xn=1491,325=0.1125

Compute the test statistic z by using the formula as given below.

z=p^−p0 p 0(1−p 0)n =0.1125−0.15 0.15(1−0.15)1,325 =−0.03750.0098=−3.83

Thus, the value of test statistic is z=−3.83.

(c)

To determine

Whether the claim made by economist is true for α=0.05.

(c)

Expert Solution

Answer to Problem 14E

The economist's claim is true.

Explanation of Solution

Given information:

In a survey of random 1325 adults, 149 adults have a great deal of confidence in banks and other financial institution.

Based on the survey an economist claims that less than 15% of US adults have confidence in banks.

Calculation:

Obtain the critical value.

Here, the test is left-tailed and the level of significance, α=0.05

From Table 9.1, the critical value for the level of significance α=0.05 is −1.645.

Rejection rule for left-tailed test:

If z≤−zα(=−1.645) then reject the null hypothesis

Conclusion for α=0.05

The value of test statistic is z=−3.83 and the critical value is zα=−1.645.

Here, the value of test statistic is lesser than the critical value.

That is z(=−3.83)≤zα(=−1.645)

Hence, the null hypothesis is rejected.

Thus it can be concluded that less than 15% of U.S. adults have a great deal of confidence in banks. Hence, the economist's claim is true.

(d)

To determine

Whether the claim made by economist is true for α=0.01.

(d)

Expert Solution

Answer to Problem 14E

The executive claim is true.

Explanation of Solution

Given information:

In a survey of random 1325 adults, 149 adults have a great deal of confidence in banks and other financial institution.

Based on the survey an economist claims that less than 15% of US adults have confidence in banks.

Calculation:

Obtain the critical value.

Here, the test is left-tailed and the level of significance, α=0.01

From Table 9.1, the critical value for the level of significance α=0.01 is −2.326.

Rejection rule for left-tailed test:

If z≤−zα(=−2.326) then reject the null hypothesis

Conclusion for α=0.01

The value of test statistic is −3.83 and the critical value is zα=−2.326.

Here, the value of test statistic is lesser than the critical value.

That is z(=−3.83)≤zα(=−2.326)

Hence, the null hypothesis is rejected.

Thus it can be concluded that less than 15% of U.S. adults have a great deal of confidence in banks. Hence, the executive’s claim is true.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 9.4, Problem 14E

(a)

To determine

The appropriate null and alternate hypotheses.

(a)

Expert Solution

Answer to Problem 14E

Null hypothesis:

H0:p=0.15

That is, 15% of U.S. adults have a great deal of confidence in banks.

Alternative hypothesis:

H1:p<0.15

That is less than 15% of U.S. adults have a great deal of confidence in banks.

Explanation of Solution

Given information:

In a survey of random 1325 adults, 149 adults have a great deal of confidence in banks and other financial institution.

Based on the survey an economist claims that less than 15% of US adults have confidence in banks.

Calculation:

Based on the survey an economist claims that less than 15% of US adults have confidence in banks.

So null and alternate hypotheses would be given as-

Null hypothesis:

H0:p=0.15

That is, 15% of U.S. adults have a great deal of confidence in banks.

Alternative hypothesis:

H1:p<0.15

That is less than 15% of U.S. adults have a great deal of confidence in banks.

(b)

To determine

The test statistic z.

(b)

Expert Solution

Answer to Problem 14E

z=−3.83.

Explanation of Solution

Given information:

In a survey of random 1325 adults, 149 adults have a great deal of confidence in banks and other financial institution.

Based on the survey an economist claims that less than 15% of US adults have confidence in banks.

Formula used:

p^=xnz=p^−p0 p 0(1−p 0)n

Calculation:

In a survey of random 1325 adults, 149 adults have a great deal of confidence in banks and other financial institution. i.e. n=1,325 and x=149

Based on the survey an economist claims that less than 15% of US adults have confidence in banks. i.e. p0=0.15.

So the sample proportion p^.

p^=xn=1491,325=0.1125

Compute the test statistic z by using the formula as given below.

z=p^−p0 p 0(1−p 0)n =0.1125−0.15 0.15(1−0.15)1,325 =−0.03750.0098=−3.83

Thus, the value of test statistic is z=−3.83.

(c)

To determine

Whether the claim made by economist is true for α=0.05.

(c)

Expert Solution

Answer to Problem 14E

The economist's claim is true.

Explanation of Solution

Given information:

In a survey of random 1325 adults, 149 adults have a great deal of confidence in banks and other financial institution.

Based on the survey an economist claims that less than 15% of US adults have confidence in banks.

Calculation:

Obtain the critical value.

Here, the test is left-tailed and the level of significance, α=0.05

From Table 9.1, the critical value for the level of significance α=0.05 is −1.645.

Rejection rule for left-tailed test:

If z≤−zα(=−1.645) then reject the null hypothesis

Conclusion for α=0.05

The value of test statistic is z=−3.83 and the critical value is zα=−1.645.

Here, the value of test statistic is lesser than the critical value.

That is z(=−3.83)≤zα(=−1.645)

Hence, the null hypothesis is rejected.

Thus it can be concluded that less than 15% of U.S. adults have a great deal of confidence in banks. Hence, the economist's claim is true.

(d)

To determine

Whether the claim made by economist is true for α=0.01.

(d)

Expert Solution

Answer to Problem 14E

The executive claim is true.

Explanation of Solution

Given information:

In a survey of random 1325 adults, 149 adults have a great deal of confidence in banks and other financial institution.

Based on the survey an economist claims that less than 15% of US adults have confidence in banks.

Calculation:

Obtain the critical value.

Here, the test is left-tailed and the level of significance, α=0.01

From Table 9.1, the critical value for the level of significance α=0.01 is −2.326.

Rejection rule for left-tailed test:

If z≤−zα(=−2.326) then reject the null hypothesis

Conclusion for α=0.01

The value of test statistic is −3.83 and the critical value is zα=−2.326.

Here, the value of test statistic is lesser than the critical value.

That is z(=−3.83)≤zα(=−2.326)

Hence, the null hypothesis is rejected.

Thus it can be concluded that less than 15% of U.S. adults have a great deal of confidence in banks. Hence, the executive’s claim is true.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 9.4, Problem 14E

(a)

To determine

The appropriate null and alternate hypotheses.

(a)

Expert Solution

Answer to Problem 14E

Null hypothesis:

H0:p=0.15

That is, 15% of U.S. adults have a great deal of confidence in banks.

Alternative hypothesis:

H1:p<0.15

That is less than 15% of U.S. adults have a great deal of confidence in banks.

Explanation of Solution

Given information:

In a survey of random 1325 adults, 149 adults have a great deal of confidence in banks and other financial institution.

Based on the survey an economist claims that less than 15% of US adults have confidence in banks.

Calculation:

Based on the survey an economist claims that less than 15% of US adults have confidence in banks.

So null and alternate hypotheses would be given as-

Null hypothesis:

H0:p=0.15

That is, 15% of U.S. adults have a great deal of confidence in banks.

Alternative hypothesis:

H1:p<0.15

That is less than 15% of U.S. adults have a great deal of confidence in banks.

(b)

To determine

The test statistic z.

(b)

Expert Solution

Answer to Problem 14E

z=−3.83.

Explanation of Solution

Given information:

In a survey of random 1325 adults, 149 adults have a great deal of confidence in banks and other financial institution.

Based on the survey an economist claims that less than 15% of US adults have confidence in banks.

Formula used:

p^=xnz=p^−p0 p 0(1−p 0)n

Calculation:

In a survey of random 1325 adults, 149 adults have a great deal of confidence in banks and other financial institution. i.e. n=1,325 and x=149

Based on the survey an economist claims that less than 15% of US adults have confidence in banks. i.e. p0=0.15.

So the sample proportion p^.

p^=xn=1491,325=0.1125

Compute the test statistic z by using the formula as given below.

z=p^−p0 p 0(1−p 0)n =0.1125−0.15 0.15(1−0.15)1,325 =−0.03750.0098=−3.83

Thus, the value of test statistic is z=−3.83.

(c)

To determine

Whether the claim made by economist is true for α=0.05.

(c)

Expert Solution

Answer to Problem 14E

The economist's claim is true.

Explanation of Solution

Given information:

In a survey of random 1325 adults, 149 adults have a great deal of confidence in banks and other financial institution.

Based on the survey an economist claims that less than 15% of US adults have confidence in banks.

Calculation:

Obtain the critical value.

Here, the test is left-tailed and the level of significance, α=0.05

From Table 9.1, the critical value for the level of significance α=0.05 is −1.645.

Rejection rule for left-tailed test:

If z≤−zα(=−1.645) then reject the null hypothesis

Conclusion for α=0.05

The value of test statistic is z=−3.83 and the critical value is zα=−1.645.

Here, the value of test statistic is lesser than the critical value.

That is z(=−3.83)≤zα(=−1.645)

Hence, the null hypothesis is rejected.

Thus it can be concluded that less than 15% of U.S. adults have a great deal of confidence in banks. Hence, the economist's claim is true.

(d)

To determine

Whether the claim made by economist is true for α=0.01.

(d)

Expert Solution

Answer to Problem 14E

The executive claim is true.

Explanation of Solution

Given information:

In a survey of random 1325 adults, 149 adults have a great deal of confidence in banks and other financial institution.

Based on the survey an economist claims that less than 15% of US adults have confidence in banks.

Calculation:

Obtain the critical value.

Here, the test is left-tailed and the level of significance, α=0.01

From Table 9.1, the critical value for the level of significance α=0.01 is −2.326.

Rejection rule for left-tailed test:

If z≤−zα(=−2.326) then reject the null hypothesis

Conclusion for α=0.01

The value of test statistic is −3.83 and the critical value is zα=−2.326.

Here, the value of test statistic is lesser than the critical value.

That is z(=−3.83)≤zα(=−2.326)

Hence, the null hypothesis is rejected.

Thus it can be concluded that less than 15% of U.S. adults have a great deal of confidence in banks. Hence, the executive’s claim is true.

Answer 6

Chapter 9.4, Problem 14E

(a)

To determine

The appropriate null and alternate hypotheses.

(a)

Expert Solution

Answer to Problem 14E

Null hypothesis:

H0:p=0.15

That is, 15% of U.S. adults have a great deal of confidence in banks.

Alternative hypothesis:

H1:p<0.15

That is less than 15% of U.S. adults have a great deal of confidence in banks.

Explanation of Solution

Given information:

In a survey of random 1325 adults, 149 adults have a great deal of confidence in banks and other financial institution.

Based on the survey an economist claims that less than 15% of US adults have confidence in banks.

Calculation:

Based on the survey an economist claims that less than 15% of US adults have confidence in banks.

So null and alternate hypotheses would be given as-

Null hypothesis:

H0:p=0.15

That is, 15% of U.S. adults have a great deal of confidence in banks.

Alternative hypothesis:

H1:p<0.15

That is less than 15% of U.S. adults have a great deal of confidence in banks.

Answer 7

Chapter 9.4, Problem 14E

(a)

To determine

The appropriate null and alternate hypotheses.

Answer 8

(a)

Expert Solution

Answer to Problem 14E

Null hypothesis:

H0:p=0.15

That is, 15% of U.S. adults have a great deal of confidence in banks.

Alternative hypothesis:

H1:p<0.15

That is less than 15% of U.S. adults have a great deal of confidence in banks.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given information:

In a survey of random 1325 adults, 149 adults have a great deal of confidence in banks and other financial institution.

Based on the survey an economist claims that less than 15% of US adults have confidence in banks.

Calculation:

Based on the survey an economist claims that less than 15% of US adults have confidence in banks.

So null and alternate hypotheses would be given as-

Null hypothesis:

H0:p=0.15

That is, 15% of U.S. adults have a great deal of confidence in banks.

Alternative hypothesis:

H1:p<0.15

That is less than 15% of U.S. adults have a great deal of confidence in banks.

Answer 13

(b)

To determine

The test statistic z.

(b)

Expert Solution

Answer to Problem 14E

z=−3.83.

Explanation of Solution

Given information:

In a survey of random 1325 adults, 149 adults have a great deal of confidence in banks and other financial institution.

Based on the survey an economist claims that less than 15% of US adults have confidence in banks.

Formula used:

p^=xnz=p^−p0 p 0(1−p 0)n

Calculation:

In a survey of random 1325 adults, 149 adults have a great deal of confidence in banks and other financial institution. i.e. n=1,325 and x=149

Based on the survey an economist claims that less than 15% of US adults have confidence in banks. i.e. p0=0.15.

So the sample proportion p^.

p^=xn=1491,325=0.1125

Compute the test statistic z by using the formula as given below.

z=p^−p0 p 0(1−p 0)n =0.1125−0.15 0.15(1−0.15)1,325 =−0.03750.0098=−3.83

Thus, the value of test statistic is z=−3.83.

Answer 14

(b)

To determine

The test statistic z.

Answer 15

(b)

Expert Solution

Answer to Problem 14E

z=−3.83.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given information:

In a survey of random 1325 adults, 149 adults have a great deal of confidence in banks and other financial institution.

Based on the survey an economist claims that less than 15% of US adults have confidence in banks.

Formula used:

p^=xnz=p^−p0 p 0(1−p 0)n

Calculation:

In a survey of random 1325 adults, 149 adults have a great deal of confidence in banks and other financial institution. i.e. n=1,325 and x=149

Based on the survey an economist claims that less than 15% of US adults have confidence in banks. i.e. p0=0.15.

So the sample proportion p^.

p^=xn=1491,325=0.1125

Compute the test statistic z by using the formula as given below.

z=p^−p0 p 0(1−p 0)n =0.1125−0.15 0.15(1−0.15)1,325 =−0.03750.0098=−3.83

Thus, the value of test statistic is z=−3.83.

Answer 20

(c)

To determine

Whether the claim made by economist is true for α=0.05.

(c)

Expert Solution

Answer to Problem 14E

The economist's claim is true.

Explanation of Solution

Given information:

In a survey of random 1325 adults, 149 adults have a great deal of confidence in banks and other financial institution.

Based on the survey an economist claims that less than 15% of US adults have confidence in banks.

Calculation:

Obtain the critical value.

Here, the test is left-tailed and the level of significance, α=0.05

From Table 9.1, the critical value for the level of significance α=0.05 is −1.645.

Rejection rule for left-tailed test:

If z≤−zα(=−1.645) then reject the null hypothesis

Conclusion for α=0.05

The value of test statistic is z=−3.83 and the critical value is zα=−1.645.

Here, the value of test statistic is lesser than the critical value.

That is z(=−3.83)≤zα(=−1.645)

Hence, the null hypothesis is rejected.

Thus it can be concluded that less than 15% of U.S. adults have a great deal of confidence in banks. Hence, the economist's claim is true.

Answer 21

(c)

To determine

Whether the claim made by economist is true for α=0.05.

Answer 22

(c)

Expert Solution

Answer to Problem 14E

The economist's claim is true.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given information:

In a survey of random 1325 adults, 149 adults have a great deal of confidence in banks and other financial institution.

Based on the survey an economist claims that less than 15% of US adults have confidence in banks.

Calculation:

Obtain the critical value.

Here, the test is left-tailed and the level of significance, α=0.05

From Table 9.1, the critical value for the level of significance α=0.05 is −1.645.

Rejection rule for left-tailed test:

If z≤−zα(=−1.645) then reject the null hypothesis

Conclusion for α=0.05

The value of test statistic is z=−3.83 and the critical value is zα=−1.645.

Here, the value of test statistic is lesser than the critical value.

That is z(=−3.83)≤zα(=−1.645)

Hence, the null hypothesis is rejected.

Thus it can be concluded that less than 15% of U.S. adults have a great deal of confidence in banks. Hence, the economist's claim is true.

Answer 27

(d)

To determine

Whether the claim made by economist is true for α=0.01.

(d)

Expert Solution

Answer to Problem 14E

The executive claim is true.

Explanation of Solution

Given information:

In a survey of random 1325 adults, 149 adults have a great deal of confidence in banks and other financial institution.

Based on the survey an economist claims that less than 15% of US adults have confidence in banks.

Calculation:

Obtain the critical value.

Here, the test is left-tailed and the level of significance, α=0.01

From Table 9.1, the critical value for the level of significance α=0.01 is −2.326.

Rejection rule for left-tailed test:

If z≤−zα(=−2.326) then reject the null hypothesis

Conclusion for α=0.01

The value of test statistic is −3.83 and the critical value is zα=−2.326.

Here, the value of test statistic is lesser than the critical value.

That is z(=−3.83)≤zα(=−2.326)

Hence, the null hypothesis is rejected.

Thus it can be concluded that less than 15% of U.S. adults have a great deal of confidence in banks. Hence, the executive’s claim is true.

Answer 28

(d)

To determine

Whether the claim made by economist is true for α=0.01.

Answer 29

(d)

Expert Solution

Answer to Problem 14E

The executive claim is true.

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Given information:

In a survey of random 1325 adults, 149 adults have a great deal of confidence in banks and other financial institution.

Based on the survey an economist claims that less than 15% of US adults have confidence in banks.

Calculation:

Obtain the critical value.

Here, the test is left-tailed and the level of significance, α=0.01

From Table 9.1, the critical value for the level of significance α=0.01 is −2.326.

Rejection rule for left-tailed test:

If z≤−zα(=−2.326) then reject the null hypothesis

Conclusion for α=0.01

The value of test statistic is −3.83 and the critical value is zα=−2.326.

Here, the value of test statistic is lesser than the critical value.

That is z(=−3.83)≤zα(=−2.326)

Hence, the null hypothesis is rejected.

Thus it can be concluded that less than 15% of U.S. adults have a great deal of confidence in banks. Hence, the executive’s claim is true.

Answer 34

The appropriate null and alternate hypotheses.

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The appropriate null and alternate hypotheses.

Answer to Problem 14E

Explanation of Solution

The test statistic z.

Answer to Problem 14E

Explanation of Solution

Whether the claim made by economist is true for α=0.05.

Answer to Problem 14E

Explanation of Solution

Whether the claim made by economist is true for α=0.01.

Answer to Problem 14E

Explanation of Solution

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