To test: The given claim.

Question

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Answer 1

Question

Chapter 15, Problem 1RE

To determine

To test: The given claim.

Expert Solution & Answer

Answer to Problem 1RE

Solution: There is enough sign at the level of significance α=0.05 to show that winning division occurs randomly.

Explanation of Solution

Given: The necessary information is provided.

Explanation: To test the randomness of the data, the requirements for randomness should be satisfied by the data.

The provided sample is an arrangement of observations in the order of existence and the outcomes consist of mutually exclusive classes.

A hypothesis test is conducted through the following steps.

Step 1: The hypotheses are defined as

H0:the sequence of the data is random.H1:the sequence of the data is not random.

Step 2: The significance level alpha is provided as α=0.05.

Step 3: The count of runs r is used to calculate the test statistic. As the sample size of each category is less than 20, the provided problem is a small-sample case.

The total number of winningsbeingsampled be represented by n. Consider that n1 represents the number of winningsby Western Division (W) and n2 represents the number of winnings of Eastern Division (E).

There are 25 winnings out of which 14 winnings are from Western Division (W) and 11 winnings are from Eastern Division (E).

In the provided sequence, the first one winning from Western Division forms a run of length1 as they are of the same type. The next five winnings from Eastern Division form a run of length5 as it is of another type. The next run consists of fourwinnings from Western Division. In this way, the number of runs is 11.

Hence, the values are n=25;n1=14;n2=11;r=11.

The number of runs is obtained as 11. Therefore, the test statistic is r=11.

Step 4: Use the critical values table for the number of runs to determine the intersection of the row that corresponds to n1=14 and the column that corresponds to n2=11. Hence, the lower and the upper critical value are obtained as 8 and 19, respectively.

The test statistic r=11 is not smaller than the inferior critical value 8, and the test statistic r=11 is not larger than the upper critical value 19. Hence, do not discard the null hypothesis.

Conclusion: There isadequateindication to determine that the winning divisionoccurs randomly.

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Answer 2

Question

Chapter 15, Problem 1RE

To determine

To test: The given claim.

Expert Solution & Answer

Answer to Problem 1RE

Solution: There is enough sign at the level of significance α=0.05 to show that winning division occurs randomly.

Explanation of Solution

Given: The necessary information is provided.

Explanation: To test the randomness of the data, the requirements for randomness should be satisfied by the data.

The provided sample is an arrangement of observations in the order of existence and the outcomes consist of mutually exclusive classes.

A hypothesis test is conducted through the following steps.

Step 1: The hypotheses are defined as

H0:the sequence of the data is random.H1:the sequence of the data is not random.

Step 2: The significance level alpha is provided as α=0.05.

Step 3: The count of runs r is used to calculate the test statistic. As the sample size of each category is less than 20, the provided problem is a small-sample case.

The total number of winningsbeingsampled be represented by n. Consider that n1 represents the number of winningsby Western Division (W) and n2 represents the number of winnings of Eastern Division (E).

There are 25 winnings out of which 14 winnings are from Western Division (W) and 11 winnings are from Eastern Division (E).

In the provided sequence, the first one winning from Western Division forms a run of length1 as they are of the same type. The next five winnings from Eastern Division form a run of length5 as it is of another type. The next run consists of fourwinnings from Western Division. In this way, the number of runs is 11.

Hence, the values are n=25;n1=14;n2=11;r=11.

The number of runs is obtained as 11. Therefore, the test statistic is r=11.

Step 4: Use the critical values table for the number of runs to determine the intersection of the row that corresponds to n1=14 and the column that corresponds to n2=11. Hence, the lower and the upper critical value are obtained as 8 and 19, respectively.

The test statistic r=11 is not smaller than the inferior critical value 8, and the test statistic r=11 is not larger than the upper critical value 19. Hence, do not discard the null hypothesis.

Conclusion: There isadequateindication to determine that the winning divisionoccurs randomly.

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Answer 3

Question

Chapter 15, Problem 1RE

To determine

To test: The given claim.

Expert Solution & Answer

Answer to Problem 1RE

Solution: There is enough sign at the level of significance α=0.05 to show that winning division occurs randomly.

Explanation of Solution

Given: The necessary information is provided.

Explanation: To test the randomness of the data, the requirements for randomness should be satisfied by the data.

The provided sample is an arrangement of observations in the order of existence and the outcomes consist of mutually exclusive classes.

A hypothesis test is conducted through the following steps.

Step 1: The hypotheses are defined as

H0:the sequence of the data is random.H1:the sequence of the data is not random.

Step 2: The significance level alpha is provided as α=0.05.

Step 3: The count of runs r is used to calculate the test statistic. As the sample size of each category is less than 20, the provided problem is a small-sample case.

The total number of winningsbeingsampled be represented by n. Consider that n1 represents the number of winningsby Western Division (W) and n2 represents the number of winnings of Eastern Division (E).

There are 25 winnings out of which 14 winnings are from Western Division (W) and 11 winnings are from Eastern Division (E).

In the provided sequence, the first one winning from Western Division forms a run of length1 as they are of the same type. The next five winnings from Eastern Division form a run of length5 as it is of another type. The next run consists of fourwinnings from Western Division. In this way, the number of runs is 11.

Hence, the values are n=25;n1=14;n2=11;r=11.

The number of runs is obtained as 11. Therefore, the test statistic is r=11.

Step 4: Use the critical values table for the number of runs to determine the intersection of the row that corresponds to n1=14 and the column that corresponds to n2=11. Hence, the lower and the upper critical value are obtained as 8 and 19, respectively.

The test statistic r=11 is not smaller than the inferior critical value 8, and the test statistic r=11 is not larger than the upper critical value 19. Hence, do not discard the null hypothesis.

Conclusion: There isadequateindication to determine that the winning divisionoccurs randomly.

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Answer 4

Question

Chapter 15, Problem 1RE

To determine

To test: The given claim.

Expert Solution & Answer

Answer to Problem 1RE

Solution: There is enough sign at the level of significance α=0.05 to show that winning division occurs randomly.

Explanation of Solution

Given: The necessary information is provided.

Explanation: To test the randomness of the data, the requirements for randomness should be satisfied by the data.

The provided sample is an arrangement of observations in the order of existence and the outcomes consist of mutually exclusive classes.

A hypothesis test is conducted through the following steps.

Step 1: The hypotheses are defined as

H0:the sequence of the data is random.H1:the sequence of the data is not random.

Step 2: The significance level alpha is provided as α=0.05.

Step 3: The count of runs r is used to calculate the test statistic. As the sample size of each category is less than 20, the provided problem is a small-sample case.

The total number of winningsbeingsampled be represented by n. Consider that n1 represents the number of winningsby Western Division (W) and n2 represents the number of winnings of Eastern Division (E).

There are 25 winnings out of which 14 winnings are from Western Division (W) and 11 winnings are from Eastern Division (E).

In the provided sequence, the first one winning from Western Division forms a run of length1 as they are of the same type. The next five winnings from Eastern Division form a run of length5 as it is of another type. The next run consists of fourwinnings from Western Division. In this way, the number of runs is 11.

Hence, the values are n=25;n1=14;n2=11;r=11.

The number of runs is obtained as 11. Therefore, the test statistic is r=11.

Step 4: Use the critical values table for the number of runs to determine the intersection of the row that corresponds to n1=14 and the column that corresponds to n2=11. Hence, the lower and the upper critical value are obtained as 8 and 19, respectively.

The test statistic r=11 is not smaller than the inferior critical value 8, and the test statistic r=11 is not larger than the upper critical value 19. Hence, do not discard the null hypothesis.

Conclusion: There isadequateindication to determine that the winning divisionoccurs randomly.

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Answer 5

Chapter 15, Problem 1RE

To determine

To test: The given claim.

Expert Solution & Answer

Answer to Problem 1RE

Solution: There is enough sign at the level of significance α=0.05 to show that winning division occurs randomly.

Explanation of Solution

Given: The necessary information is provided.

Explanation: To test the randomness of the data, the requirements for randomness should be satisfied by the data.

The provided sample is an arrangement of observations in the order of existence and the outcomes consist of mutually exclusive classes.

A hypothesis test is conducted through the following steps.

Step 1: The hypotheses are defined as

H0:the sequence of the data is random.H1:the sequence of the data is not random.

Step 2: The significance level alpha is provided as α=0.05.

Step 3: The count of runs r is used to calculate the test statistic. As the sample size of each category is less than 20, the provided problem is a small-sample case.

The total number of winningsbeingsampled be represented by n. Consider that n1 represents the number of winningsby Western Division (W) and n2 represents the number of winnings of Eastern Division (E).

There are 25 winnings out of which 14 winnings are from Western Division (W) and 11 winnings are from Eastern Division (E).

In the provided sequence, the first one winning from Western Division forms a run of length1 as they are of the same type. The next five winnings from Eastern Division form a run of length5 as it is of another type. The next run consists of fourwinnings from Western Division. In this way, the number of runs is 11.

Hence, the values are n=25;n1=14;n2=11;r=11.

The number of runs is obtained as 11. Therefore, the test statistic is r=11.

Step 4: Use the critical values table for the number of runs to determine the intersection of the row that corresponds to n1=14 and the column that corresponds to n2=11. Hence, the lower and the upper critical value are obtained as 8 and 19, respectively.

The test statistic r=11 is not smaller than the inferior critical value 8, and the test statistic r=11 is not larger than the upper critical value 19. Hence, do not discard the null hypothesis.

Conclusion: There isadequateindication to determine that the winning divisionoccurs randomly.

Answer 6

Chapter 15, Problem 1RE

To determine

To test: The given claim.

Expert Solution & Answer

Answer to Problem 1RE

Solution: There is enough sign at the level of significance α=0.05 to show that winning division occurs randomly.

Explanation of Solution

Given: The necessary information is provided.

Explanation: To test the randomness of the data, the requirements for randomness should be satisfied by the data.

The provided sample is an arrangement of observations in the order of existence and the outcomes consist of mutually exclusive classes.

A hypothesis test is conducted through the following steps.

Step 1: The hypotheses are defined as

H0:the sequence of the data is random.H1:the sequence of the data is not random.

Step 2: The significance level alpha is provided as α=0.05.

Step 3: The count of runs r is used to calculate the test statistic. As the sample size of each category is less than 20, the provided problem is a small-sample case.

The total number of winningsbeingsampled be represented by n. Consider that n1 represents the number of winningsby Western Division (W) and n2 represents the number of winnings of Eastern Division (E).

There are 25 winnings out of which 14 winnings are from Western Division (W) and 11 winnings are from Eastern Division (E).

In the provided sequence, the first one winning from Western Division forms a run of length1 as they are of the same type. The next five winnings from Eastern Division form a run of length5 as it is of another type. The next run consists of fourwinnings from Western Division. In this way, the number of runs is 11.

Hence, the values are n=25;n1=14;n2=11;r=11.

The number of runs is obtained as 11. Therefore, the test statistic is r=11.

Step 4: Use the critical values table for the number of runs to determine the intersection of the row that corresponds to n1=14 and the column that corresponds to n2=11. Hence, the lower and the upper critical value are obtained as 8 and 19, respectively.

The test statistic r=11 is not smaller than the inferior critical value 8, and the test statistic r=11 is not larger than the upper critical value 19. Hence, do not discard the null hypothesis.

Conclusion: There isadequateindication to determine that the winning divisionoccurs randomly.

Answer 7

Chapter 15, Problem 1RE

To determine

To test: The given claim.

Answer 8

Expert Solution & Answer

Answer to Problem 1RE

Solution: There is enough sign at the level of significance α=0.05 to show that winning division occurs randomly.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given: The necessary information is provided.

Explanation: To test the randomness of the data, the requirements for randomness should be satisfied by the data.

The provided sample is an arrangement of observations in the order of existence and the outcomes consist of mutually exclusive classes.

A hypothesis test is conducted through the following steps.

Step 1: The hypotheses are defined as

H0:the sequence of the data is random.H1:the sequence of the data is not random.

Step 2: The significance level alpha is provided as α=0.05.

Step 3: The count of runs r is used to calculate the test statistic. As the sample size of each category is less than 20, the provided problem is a small-sample case.

The total number of winningsbeingsampled be represented by n. Consider that n1 represents the number of winningsby Western Division (W) and n2 represents the number of winnings of Eastern Division (E).

There are 25 winnings out of which 14 winnings are from Western Division (W) and 11 winnings are from Eastern Division (E).

In the provided sequence, the first one winning from Western Division forms a run of length1 as they are of the same type. The next five winnings from Eastern Division form a run of length5 as it is of another type. The next run consists of fourwinnings from Western Division. In this way, the number of runs is 11.

Hence, the values are n=25;n1=14;n2=11;r=11.

The number of runs is obtained as 11. Therefore, the test statistic is r=11.

Step 4: Use the critical values table for the number of runs to determine the intersection of the row that corresponds to n1=14 and the column that corresponds to n2=11. Hence, the lower and the upper critical value are obtained as 8 and 19, respectively.

The test statistic r=11 is not smaller than the inferior critical value 8, and the test statistic r=11 is not larger than the upper critical value 19. Hence, do not discard the null hypothesis.

Conclusion: There isadequateindication to determine that the winning divisionoccurs randomly.