Determine whether there is evidence that the sequence supports the contentions that the sample was selected at random.

Question

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Answer 1

Question

Chapter 16.7, Problem 23E

To determine

Determine whether there is evidence that the sequence supports the contentions that the sample was selected at random.

Expert Solution & Answer

Answer to Problem 23E

There is sufficient evidence to conclude that the sequence supports the contentions that the sample was selected at random at 0.1 level of significance.

Explanation of Solution

The test hypothesis is given as follows:

Null hypothesis:

H0: The sequence is random.

Alternative hypothesis:

H1: The sequence is not random.

It is given that the level of significance is α=0.1.

For the run test, the test statistic will be based on the random variable V (the total number of runs).

In Table A.18, P(V≤v* when H0 is true), where v*=2,3,...20 (number of runs) and the values of n1 and n2 should be less than or equal to 10.

Computation of the test statistic:

N,N,N,N︸,Y,Y︸,N︸,Y,Y︸,N︸,Y︸,N,N,N,N︸

Here, the number of runs is, v=7.

The number of “yes” responses is n1=5.

The number of “no” responses is n2=10.

In this context, the values of n1 and n2 are not greater than 10.

Therefore, the P-value is computed as given below:

P-value=2{P(V≤7 when H0 is true)}

In Appendix A, Table A.18 “P (V ≤ v^∗ when H₀ is true) in the Runs Test”, at v*=7, the probability value corresponding to the column (5,10) is 0.455.

Therefore, the probability of the event is given below:

P-value=2(0.455)=0.91

Decision:

If P-value≤α, reject the null hypothesis H0.
If P-value>α, fail to reject the null hypothesis H0.

Conclusion:

Here, the P-value is greater than the level of significance.

That is, P-value(=0.91)>α(=0.1).

By the rejection rule, fail to reject the null hypothesis.

Therefore, there is no evidence that the sequence was not random at the 0.1 level of significance.

Thus, it can be concluded that the sequence supports the contentions that the sample was selected at random.

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(a) Test the hypothesis. Consider the hypothesis test Ho = : against H₁o < 02. Suppose that the sample sizes aren₁ = 7 and n₂ = 13 and that $² = 22.4 and $22 = 28.2. Use α = 0.05. Ho is not ✓ rejected. 9-9 IV (b) Find a 95% confidence interval on of 102. Round your answer to two decimal places (e.g. 98.76).

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Answer 2

Question

Chapter 16.7, Problem 23E

To determine

Determine whether there is evidence that the sequence supports the contentions that the sample was selected at random.

Expert Solution & Answer

Answer to Problem 23E

There is sufficient evidence to conclude that the sequence supports the contentions that the sample was selected at random at 0.1 level of significance.

Explanation of Solution

The test hypothesis is given as follows:

Null hypothesis:

H0: The sequence is random.

Alternative hypothesis:

H1: The sequence is not random.

It is given that the level of significance is α=0.1.

For the run test, the test statistic will be based on the random variable V (the total number of runs).

In Table A.18, P(V≤v* when H0 is true), where v*=2,3,...20 (number of runs) and the values of n1 and n2 should be less than or equal to 10.

Computation of the test statistic:

N,N,N,N︸,Y,Y︸,N︸,Y,Y︸,N︸,Y︸,N,N,N,N︸

Here, the number of runs is, v=7.

The number of “yes” responses is n1=5.

The number of “no” responses is n2=10.

In this context, the values of n1 and n2 are not greater than 10.

Therefore, the P-value is computed as given below:

P-value=2{P(V≤7 when H0 is true)}

In Appendix A, Table A.18 “P (V ≤ v^∗ when H₀ is true) in the Runs Test”, at v*=7, the probability value corresponding to the column (5,10) is 0.455.

Therefore, the probability of the event is given below:

P-value=2(0.455)=0.91

Decision:

If P-value≤α, reject the null hypothesis H0.
If P-value>α, fail to reject the null hypothesis H0.

Conclusion:

Here, the P-value is greater than the level of significance.

That is, P-value(=0.91)>α(=0.1).

By the rejection rule, fail to reject the null hypothesis.

Therefore, there is no evidence that the sequence was not random at the 0.1 level of significance.

Thus, it can be concluded that the sequence supports the contentions that the sample was selected at random.

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Answer 3

Question

Chapter 16.7, Problem 23E

To determine

Determine whether there is evidence that the sequence supports the contentions that the sample was selected at random.

Expert Solution & Answer

Answer to Problem 23E

There is sufficient evidence to conclude that the sequence supports the contentions that the sample was selected at random at 0.1 level of significance.

Explanation of Solution

The test hypothesis is given as follows:

Null hypothesis:

H0: The sequence is random.

Alternative hypothesis:

H1: The sequence is not random.

It is given that the level of significance is α=0.1.

For the run test, the test statistic will be based on the random variable V (the total number of runs).

In Table A.18, P(V≤v* when H0 is true), where v*=2,3,...20 (number of runs) and the values of n1 and n2 should be less than or equal to 10.

Computation of the test statistic:

N,N,N,N︸,Y,Y︸,N︸,Y,Y︸,N︸,Y︸,N,N,N,N︸

Here, the number of runs is, v=7.

The number of “yes” responses is n1=5.

The number of “no” responses is n2=10.

In this context, the values of n1 and n2 are not greater than 10.

Therefore, the P-value is computed as given below:

P-value=2{P(V≤7 when H0 is true)}

In Appendix A, Table A.18 “P (V ≤ v^∗ when H₀ is true) in the Runs Test”, at v*=7, the probability value corresponding to the column (5,10) is 0.455.

Therefore, the probability of the event is given below:

P-value=2(0.455)=0.91

Decision:

If P-value≤α, reject the null hypothesis H0.
If P-value>α, fail to reject the null hypothesis H0.

Conclusion:

Here, the P-value is greater than the level of significance.

That is, P-value(=0.91)>α(=0.1).

By the rejection rule, fail to reject the null hypothesis.

Therefore, there is no evidence that the sequence was not random at the 0.1 level of significance.

Thus, it can be concluded that the sequence supports the contentions that the sample was selected at random.

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Answer 4

Question

Chapter 16.7, Problem 23E

To determine

Determine whether there is evidence that the sequence supports the contentions that the sample was selected at random.

Expert Solution & Answer

Answer to Problem 23E

There is sufficient evidence to conclude that the sequence supports the contentions that the sample was selected at random at 0.1 level of significance.

Explanation of Solution

The test hypothesis is given as follows:

Null hypothesis:

H0: The sequence is random.

Alternative hypothesis:

H1: The sequence is not random.

It is given that the level of significance is α=0.1.

For the run test, the test statistic will be based on the random variable V (the total number of runs).

In Table A.18, P(V≤v* when H0 is true), where v*=2,3,...20 (number of runs) and the values of n1 and n2 should be less than or equal to 10.

Computation of the test statistic:

N,N,N,N︸,Y,Y︸,N︸,Y,Y︸,N︸,Y︸,N,N,N,N︸

Here, the number of runs is, v=7.

The number of “yes” responses is n1=5.

The number of “no” responses is n2=10.

In this context, the values of n1 and n2 are not greater than 10.

Therefore, the P-value is computed as given below:

P-value=2{P(V≤7 when H0 is true)}

In Appendix A, Table A.18 “P (V ≤ v^∗ when H₀ is true) in the Runs Test”, at v*=7, the probability value corresponding to the column (5,10) is 0.455.

Therefore, the probability of the event is given below:

P-value=2(0.455)=0.91

Decision:

If P-value≤α, reject the null hypothesis H0.
If P-value>α, fail to reject the null hypothesis H0.

Conclusion:

Here, the P-value is greater than the level of significance.

That is, P-value(=0.91)>α(=0.1).

By the rejection rule, fail to reject the null hypothesis.

Therefore, there is no evidence that the sequence was not random at the 0.1 level of significance.

Thus, it can be concluded that the sequence supports the contentions that the sample was selected at random.

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Answer 5

Chapter 16.7, Problem 23E

To determine

Determine whether there is evidence that the sequence supports the contentions that the sample was selected at random.

Expert Solution & Answer

Answer to Problem 23E

There is sufficient evidence to conclude that the sequence supports the contentions that the sample was selected at random at 0.1 level of significance.

Explanation of Solution

The test hypothesis is given as follows:

Null hypothesis:

H0: The sequence is random.

Alternative hypothesis:

H1: The sequence is not random.

It is given that the level of significance is α=0.1.

For the run test, the test statistic will be based on the random variable V (the total number of runs).

In Table A.18, P(V≤v* when H0 is true), where v*=2,3,...20 (number of runs) and the values of n1 and n2 should be less than or equal to 10.

Computation of the test statistic:

N,N,N,N︸,Y,Y︸,N︸,Y,Y︸,N︸,Y︸,N,N,N,N︸

Here, the number of runs is, v=7.

The number of “yes” responses is n1=5.

The number of “no” responses is n2=10.

In this context, the values of n1 and n2 are not greater than 10.

Therefore, the P-value is computed as given below:

P-value=2{P(V≤7 when H0 is true)}

In Appendix A, Table A.18 “P (V ≤ v^∗ when H₀ is true) in the Runs Test”, at v*=7, the probability value corresponding to the column (5,10) is 0.455.

Therefore, the probability of the event is given below:

P-value=2(0.455)=0.91

Decision:

If P-value≤α, reject the null hypothesis H0.
If P-value>α, fail to reject the null hypothesis H0.

Conclusion:

Here, the P-value is greater than the level of significance.

That is, P-value(=0.91)>α(=0.1).

By the rejection rule, fail to reject the null hypothesis.

Therefore, there is no evidence that the sequence was not random at the 0.1 level of significance.

Thus, it can be concluded that the sequence supports the contentions that the sample was selected at random.

Answer 6

Chapter 16.7, Problem 23E

To determine

Determine whether there is evidence that the sequence supports the contentions that the sample was selected at random.

Expert Solution & Answer

Answer to Problem 23E

There is sufficient evidence to conclude that the sequence supports the contentions that the sample was selected at random at 0.1 level of significance.

Explanation of Solution

The test hypothesis is given as follows:

Null hypothesis:

H0: The sequence is random.

Alternative hypothesis:

H1: The sequence is not random.

It is given that the level of significance is α=0.1.

For the run test, the test statistic will be based on the random variable V (the total number of runs).

In Table A.18, P(V≤v* when H0 is true), where v*=2,3,...20 (number of runs) and the values of n1 and n2 should be less than or equal to 10.

Computation of the test statistic:

N,N,N,N︸,Y,Y︸,N︸,Y,Y︸,N︸,Y︸,N,N,N,N︸

Here, the number of runs is, v=7.

The number of “yes” responses is n1=5.

The number of “no” responses is n2=10.

In this context, the values of n1 and n2 are not greater than 10.

Therefore, the P-value is computed as given below:

P-value=2{P(V≤7 when H0 is true)}

In Appendix A, Table A.18 “P (V ≤ v^∗ when H₀ is true) in the Runs Test”, at v*=7, the probability value corresponding to the column (5,10) is 0.455.

Therefore, the probability of the event is given below:

P-value=2(0.455)=0.91

Decision:

If P-value≤α, reject the null hypothesis H0.
If P-value>α, fail to reject the null hypothesis H0.

Conclusion:

Here, the P-value is greater than the level of significance.

That is, P-value(=0.91)>α(=0.1).

By the rejection rule, fail to reject the null hypothesis.

Therefore, there is no evidence that the sequence was not random at the 0.1 level of significance.

Thus, it can be concluded that the sequence supports the contentions that the sample was selected at random.

Answer 7

Chapter 16.7, Problem 23E

To determine

Determine whether there is evidence that the sequence supports the contentions that the sample was selected at random.

Answer 8

Expert Solution & Answer

Answer to Problem 23E

There is sufficient evidence to conclude that the sequence supports the contentions that the sample was selected at random at 0.1 level of significance.