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

Question

Want to see more full solutions like this?

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.

Want to see more full solutions like this?

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

Students have asked these similar questions

a) When two variables are correlated, can the researcher be sure that one variable causes the other? If YES , why? If NO , why? b) What is meant by the statement that two variables are related? Discuss.

SCIE 211 Lab 3: Graphing and DataWorksheetPre-lab Questions:1. When should you use each of the following types of graphs? Fill answers in the table below.Type of Graph Used to showLine graphScatter plotBar graphHistogramPie Chart2. Several ways in which we can be fooled or misled by a graph were identified in the Lab 3Introduction. Find two examples of misleading graphs on the Internet and paste them below. Besure to identify why each graph is misleading. Data Charts:Circumference vs. Diameter for circular objectsDiameter Can 1 (cm) Can 2 (cm) Can 3 (cm)Trial 1Trial 2Trial 3MeanCircumference Can 1 (cm) Can 2 (cm) Can 3 (cm)Trial 1Trial 2Trial 3MeanScatter Plot Graph – Circumference Vs. DiameterIdentify 2 points of the Trendline.Y1 = ________ Y2 = _________X1 = ________ X2 = _________Calculate the Slope of the Trendline = Post-lab Questions:1. Answer the questions below. You will need to use the following equation to answer…

The U.S. Bureau of Labor Statistics reports that 11.3% of U.S. workers belong to unions (BLS website, January 2014). Suppose a sample of 400 U.S. workers is collected in 2014 to determine whether union efforts to organize have increased union membership. a. Formulate the hypotheses that can be used to determine whether union membership increased in 2014.H 0: p H a: p b. If the sample results show that 52 of the workers belonged to unions, what is the p-value for your hypothesis test (to 4 decimals)?

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.

Want to see more full solutions like this?

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

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.

Want to see more full solutions like this?

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

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.

Want to see more full solutions like this?

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

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.