Test whether fatal bicycle accidents are equally likely to occur in each of the 12-months at 0.01 significance level.

Question

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Answer 1

Question

Chapter 12.1, Problem 10E

a.

To determine

Test whether fatal bicycle accidents are equally likely to occur in each of the 12-months at 0.01 significance level.

a.

Expert Solution

Answer to Problem 10E

There is convincing evidence that fatal bicycle accidents are not equally likely to occur in each of the months.

Explanation of Solution

Calculation:

The given data represent the classification of fatal bicycle accidents according to the month in which the accident occurred.

The expected counts are calculated as shown below:

Month	Observed counts	Expected counts
January	38	$71912=59.917$
February	32	$71912=59.917$
March	43	$71912=59.917$
April	59	$71912=59.917$
May	78	$71912=59.917$
June	74	$71912=59.917$
July	98	$71912=59.917$
August	85	$71912=59.917$
September	64	$71912=59.917$
October	66	$71912=59.917$
November	42	$71912=59.917$
December	40	$71912=59.917$
	719	719

The nine-step hypotheses testing procedure to test goodness-of-fit is given below:

1. Consider that the proportion of fatal bicycle accidents occurring in January is $p1$ , the proportion of fatal bicycle accidents occurring in February is $p2$ , the proportion of fatal bicycle accidents occurring in March is $p3$ , the proportion of fatal bicycle accidents occurring in April is $p4$ , the proportion of fatal bicycle accidents occurring in May is $p5$ , the proportion of fatal bicycle accidents occurring in June is $p6$ , the proportion of fatal bicycle accidents occurring in July is $p7$ , the proportion of fatal bicycle accidents occurring in August is $p8$ , the proportion of fatal bicycle accidents occurring in September is $p9$ , the proportion of fatal bicycle accidents occurring in October is $p10$ , the proportion of fatal bicycle accidents occurring in November is $p11$ , and the proportion of fatal bicycle accidents occurring in December is $p12$ .

2. Null hypothesis:

$H0:$ $p1=p2=p3=p4=p5=p6=p7=p8=p9=p10=p11=p12=112$ .

3. Alternative hypothesis:

$Ha:$ At least one of the population proportions is not equal to $112$ .

4. Significance level:

$α=0.01$

5. Test statistic:

$χ2=∑(observed count−expected count)2expected count$

6. Assumptions:

Assume that the 719 accidents included in the study is a random sample from the population of fatal bicycle accidents.
From the table, it is observed that all the expected counts are greater than 5.

7. Calculation:

Software procedure:

Step-by-step procedure to obtain the test statistics and P-value using the MINITAB software:

Choose Stat > Tables > Chi-Square Goodness-of-Fit Test (One Variable).
In Observed counts, enter the column of Number of Accidents.
In Category names, enter the column of Month.
Under Test, select the Equal Proportions.
Click OK.

The output obtained using the MINITAB software is given below:

Bundle: Introduction to Statistics and Data Analysis, 5th + WebAssign Printed Access Card: Peck/Olsen/Devore. 5th Edition, Single-Term, Chapter 12.1, Problem 10E , additional homework tip 1

From the output, $χ2=82.1627$ .

8. P-value:

From the MINITAB output, $df=11$ and P-value is 0.000.

9. Conclusion:

Decision rule:

If P-value is less than or equal to the level of significance, reject the null hypothesis.
Otherwise, do not reject the null hypothesis.

Conclusion:

Here the level of significance is 0.01.

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

That is, $0.000<0.01$ .

Hence, reject the null hypothesis. Therefore, there is convincing evidence that fatal bicycle accidents are not equally likely to occur in each of the months.

b.

To determine

Write the null and alternative hypotheses to determine if some months are riskier than others by taking differing month lengths into account.

b.

Expert Solution

Explanation of Solution

The null hypothesis in Part (a) specifies that fatal accidents are equally likely to occur in any of the 12 months.

The year considered in the study is 2004, which is a leap year. It is known that a leap year contains 366 days, with 29 days in February.

The null and alternative hypotheses to determine if some months are riskier than others by taking differing month lengths and the characteristics of a leap year into account are as follows:

Null hypothesis:

$H0:{p1=31366, p2=29366, p3=31366, p4=30366, p5=31366, p6=30366p7=31366, p8=31366, p9=30366, p10=31366, p11=30366, p12=31366$

Alternative hypothesis:

$Ha:$ At least one of the population proportions is not equal to the specified value.

c.

To determine

Test the hypotheses proposed in Part (b) at 0.05 significance level.

c.

Expert Solution

Answer to Problem 10E

There is convincing evidence that fatal bicycle accidents do not occur in any of the twelve months in proportion to the lengths of the months.

Explanation of Solution

Calculation:

The expected counts are calculated as shown below:

Month	Observed counts	Proportion $pi (in decimal)$	Expected counts
January	38	0.085	$31366×719=60.90$
February	32	0.079	$29366×719=56.97$
March	43	0.085	$31366×719=60.90$
April	59	0.082	$30366×719=58.93$
May	78	0.085	$31366×719=60.90$
June	74	0.082	$30366×719=58.93$
July	98	0.085	$31366×719=60.90$
August	85	0.085	$31366×719=60.90$
September	64	0.082	$30366×719=58.93$
October	66	0.085	$31366×719=60.90$
November	42	0.082	$30366×719=58.93$
December	40	0.085	$31366×719=60.90$
	719	1 (approximately)	719

Significance level:

$α=0.05$

Test statistic:

$χ2=∑(observed count−expected count)2expected count$

Assumptions:

Assume that the 719 accidents included in the study is a random sample from the population of fatal bicycle accidents.
From the table, it is observed that all the expected counts are greater than 5.

Calculation:

Software procedure:

Step-by-step procedure to obtain the test statistic and P-value using the MINITAB software:

Choose Stat > Tables > Chi-Square Goodness-of-Fit Test (One Variable).
In Observed counts, enter the column of Number of Accidents.
In Category names, enter the column of Month.
Under Test, select the column of Proportion in Proportions specified by historical counts.
Click OK.

The output obtained using the MINITAB software is given below:

Bundle: Introduction to Statistics and Data Analysis, 5th + WebAssign Printed Access Card: Peck/Olsen/Devore. 5th Edition, Single-Term, Chapter 12.1, Problem 10E , additional homework tip 2

From the output, $χ2=78.1873$ .

8. P-value:

From the MINITAB output, $df=11$ and P-value is 0.000.

9. Conclusion:

Decision rule:

If P-value is less than or equal to the level of significance, reject the null hypothesis.
Otherwise, do not reject the null hypothesis.

Conclusion:

Here the level of significance is 0.05.

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

That is, $0.000<0.05$ .

Hence, reject the null hypothesis.

Therefore, there is convincing evidence that fatal bicycle accidents do not occur in any of the twelve months in proportion to the lengths of the months.

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Answer 2

Question

Chapter 12.1, Problem 10E

a.

To determine

Test whether fatal bicycle accidents are equally likely to occur in each of the 12-months at 0.01 significance level.

a.

Expert Solution

Answer to Problem 10E

There is convincing evidence that fatal bicycle accidents are not equally likely to occur in each of the months.

Explanation of Solution

Calculation:

The given data represent the classification of fatal bicycle accidents according to the month in which the accident occurred.

The expected counts are calculated as shown below:

Month	Observed counts	Expected counts
January	38	$71912=59.917$
February	32	$71912=59.917$
March	43	$71912=59.917$
April	59	$71912=59.917$
May	78	$71912=59.917$
June	74	$71912=59.917$
July	98	$71912=59.917$
August	85	$71912=59.917$
September	64	$71912=59.917$
October	66	$71912=59.917$
November	42	$71912=59.917$
December	40	$71912=59.917$
	719	719

The nine-step hypotheses testing procedure to test goodness-of-fit is given below:

1. Consider that the proportion of fatal bicycle accidents occurring in January is $p1$ , the proportion of fatal bicycle accidents occurring in February is $p2$ , the proportion of fatal bicycle accidents occurring in March is $p3$ , the proportion of fatal bicycle accidents occurring in April is $p4$ , the proportion of fatal bicycle accidents occurring in May is $p5$ , the proportion of fatal bicycle accidents occurring in June is $p6$ , the proportion of fatal bicycle accidents occurring in July is $p7$ , the proportion of fatal bicycle accidents occurring in August is $p8$ , the proportion of fatal bicycle accidents occurring in September is $p9$ , the proportion of fatal bicycle accidents occurring in October is $p10$ , the proportion of fatal bicycle accidents occurring in November is $p11$ , and the proportion of fatal bicycle accidents occurring in December is $p12$ .

2. Null hypothesis:

$H0:$ $p1=p2=p3=p4=p5=p6=p7=p8=p9=p10=p11=p12=112$ .

3. Alternative hypothesis:

$Ha:$ At least one of the population proportions is not equal to $112$ .

4. Significance level:

$α=0.01$

5. Test statistic:

$χ2=∑(observed count−expected count)2expected count$

6. Assumptions:

Assume that the 719 accidents included in the study is a random sample from the population of fatal bicycle accidents.
From the table, it is observed that all the expected counts are greater than 5.

7. Calculation:

Software procedure:

Step-by-step procedure to obtain the test statistics and P-value using the MINITAB software:

Choose Stat > Tables > Chi-Square Goodness-of-Fit Test (One Variable).
In Observed counts, enter the column of Number of Accidents.
In Category names, enter the column of Month.
Under Test, select the Equal Proportions.
Click OK.

The output obtained using the MINITAB software is given below:

Bundle: Introduction to Statistics and Data Analysis, 5th + WebAssign Printed Access Card: Peck/Olsen/Devore. 5th Edition, Single-Term, Chapter 12.1, Problem 10E , additional homework tip 1

From the output, $χ2=82.1627$ .

8. P-value:

From the MINITAB output, $df=11$ and P-value is 0.000.

9. Conclusion:

Decision rule:

If P-value is less than or equal to the level of significance, reject the null hypothesis.
Otherwise, do not reject the null hypothesis.

Conclusion:

Here the level of significance is 0.01.

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

That is, $0.000<0.01$ .

Hence, reject the null hypothesis. Therefore, there is convincing evidence that fatal bicycle accidents are not equally likely to occur in each of the months.

b.

To determine

Write the null and alternative hypotheses to determine if some months are riskier than others by taking differing month lengths into account.

b.

Expert Solution

Explanation of Solution

The null hypothesis in Part (a) specifies that fatal accidents are equally likely to occur in any of the 12 months.

The year considered in the study is 2004, which is a leap year. It is known that a leap year contains 366 days, with 29 days in February.

The null and alternative hypotheses to determine if some months are riskier than others by taking differing month lengths and the characteristics of a leap year into account are as follows:

Null hypothesis:

$H0:{p1=31366, p2=29366, p3=31366, p4=30366, p5=31366, p6=30366p7=31366, p8=31366, p9=30366, p10=31366, p11=30366, p12=31366$

Alternative hypothesis:

$Ha:$ At least one of the population proportions is not equal to the specified value.

c.

To determine

Test the hypotheses proposed in Part (b) at 0.05 significance level.

c.

Expert Solution

Answer to Problem 10E

There is convincing evidence that fatal bicycle accidents do not occur in any of the twelve months in proportion to the lengths of the months.

Explanation of Solution

Calculation:

The expected counts are calculated as shown below:

Month	Observed counts	Proportion $pi (in decimal)$	Expected counts
January	38	0.085	$31366×719=60.90$
February	32	0.079	$29366×719=56.97$
March	43	0.085	$31366×719=60.90$
April	59	0.082	$30366×719=58.93$
May	78	0.085	$31366×719=60.90$
June	74	0.082	$30366×719=58.93$
July	98	0.085	$31366×719=60.90$
August	85	0.085	$31366×719=60.90$
September	64	0.082	$30366×719=58.93$
October	66	0.085	$31366×719=60.90$
November	42	0.082	$30366×719=58.93$
December	40	0.085	$31366×719=60.90$
	719	1 (approximately)	719

Significance level:

$α=0.05$

Test statistic:

$χ2=∑(observed count−expected count)2expected count$

Assumptions:

Assume that the 719 accidents included in the study is a random sample from the population of fatal bicycle accidents.
From the table, it is observed that all the expected counts are greater than 5.

Calculation:

Software procedure:

Step-by-step procedure to obtain the test statistic and P-value using the MINITAB software:

Choose Stat > Tables > Chi-Square Goodness-of-Fit Test (One Variable).
In Observed counts, enter the column of Number of Accidents.
In Category names, enter the column of Month.
Under Test, select the column of Proportion in Proportions specified by historical counts.
Click OK.

The output obtained using the MINITAB software is given below:

Bundle: Introduction to Statistics and Data Analysis, 5th + WebAssign Printed Access Card: Peck/Olsen/Devore. 5th Edition, Single-Term, Chapter 12.1, Problem 10E , additional homework tip 2

From the output, $χ2=78.1873$ .

8. P-value:

From the MINITAB output, $df=11$ and P-value is 0.000.

9. Conclusion:

Decision rule:

If P-value is less than or equal to the level of significance, reject the null hypothesis.
Otherwise, do not reject the null hypothesis.

Conclusion:

Here the level of significance is 0.05.

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

That is, $0.000<0.05$ .

Hence, reject the null hypothesis.

Therefore, there is convincing evidence that fatal bicycle accidents do not occur in any of the twelve months in proportion to the lengths of the months.

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Answer 3

Question

Chapter 12.1, Problem 10E

a.

To determine

Test whether fatal bicycle accidents are equally likely to occur in each of the 12-months at 0.01 significance level.

a.

Expert Solution

Answer to Problem 10E

There is convincing evidence that fatal bicycle accidents are not equally likely to occur in each of the months.

Explanation of Solution

Calculation:

The given data represent the classification of fatal bicycle accidents according to the month in which the accident occurred.

The expected counts are calculated as shown below:

Month	Observed counts	Expected counts
January	38	$71912=59.917$
February	32	$71912=59.917$
March	43	$71912=59.917$
April	59	$71912=59.917$
May	78	$71912=59.917$
June	74	$71912=59.917$
July	98	$71912=59.917$
August	85	$71912=59.917$
September	64	$71912=59.917$
October	66	$71912=59.917$
November	42	$71912=59.917$
December	40	$71912=59.917$
	719	719

The nine-step hypotheses testing procedure to test goodness-of-fit is given below:

1. Consider that the proportion of fatal bicycle accidents occurring in January is $p1$ , the proportion of fatal bicycle accidents occurring in February is $p2$ , the proportion of fatal bicycle accidents occurring in March is $p3$ , the proportion of fatal bicycle accidents occurring in April is $p4$ , the proportion of fatal bicycle accidents occurring in May is $p5$ , the proportion of fatal bicycle accidents occurring in June is $p6$ , the proportion of fatal bicycle accidents occurring in July is $p7$ , the proportion of fatal bicycle accidents occurring in August is $p8$ , the proportion of fatal bicycle accidents occurring in September is $p9$ , the proportion of fatal bicycle accidents occurring in October is $p10$ , the proportion of fatal bicycle accidents occurring in November is $p11$ , and the proportion of fatal bicycle accidents occurring in December is $p12$ .

2. Null hypothesis:

$H0:$ $p1=p2=p3=p4=p5=p6=p7=p8=p9=p10=p11=p12=112$ .

3. Alternative hypothesis:

$Ha:$ At least one of the population proportions is not equal to $112$ .

4. Significance level:

$α=0.01$

5. Test statistic:

$χ2=∑(observed count−expected count)2expected count$

6. Assumptions:

Assume that the 719 accidents included in the study is a random sample from the population of fatal bicycle accidents.
From the table, it is observed that all the expected counts are greater than 5.

7. Calculation:

Software procedure:

Step-by-step procedure to obtain the test statistics and P-value using the MINITAB software:

Choose Stat > Tables > Chi-Square Goodness-of-Fit Test (One Variable).
In Observed counts, enter the column of Number of Accidents.
In Category names, enter the column of Month.
Under Test, select the Equal Proportions.
Click OK.

The output obtained using the MINITAB software is given below:

Bundle: Introduction to Statistics and Data Analysis, 5th + WebAssign Printed Access Card: Peck/Olsen/Devore. 5th Edition, Single-Term, Chapter 12.1, Problem 10E , additional homework tip 1

From the output, $χ2=82.1627$ .

8. P-value:

From the MINITAB output, $df=11$ and P-value is 0.000.

9. Conclusion:

Decision rule:

If P-value is less than or equal to the level of significance, reject the null hypothesis.
Otherwise, do not reject the null hypothesis.

Conclusion:

Here the level of significance is 0.01.

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

That is, $0.000<0.01$ .

Hence, reject the null hypothesis. Therefore, there is convincing evidence that fatal bicycle accidents are not equally likely to occur in each of the months.

b.

To determine

Write the null and alternative hypotheses to determine if some months are riskier than others by taking differing month lengths into account.

b.

Expert Solution

Explanation of Solution

The null hypothesis in Part (a) specifies that fatal accidents are equally likely to occur in any of the 12 months.

The year considered in the study is 2004, which is a leap year. It is known that a leap year contains 366 days, with 29 days in February.

The null and alternative hypotheses to determine if some months are riskier than others by taking differing month lengths and the characteristics of a leap year into account are as follows:

Null hypothesis:

$H0:{p1=31366, p2=29366, p3=31366, p4=30366, p5=31366, p6=30366p7=31366, p8=31366, p9=30366, p10=31366, p11=30366, p12=31366$

Alternative hypothesis:

$Ha:$ At least one of the population proportions is not equal to the specified value.

c.

To determine

Test the hypotheses proposed in Part (b) at 0.05 significance level.

c.

Expert Solution

Answer to Problem 10E

There is convincing evidence that fatal bicycle accidents do not occur in any of the twelve months in proportion to the lengths of the months.

Explanation of Solution

Calculation:

The expected counts are calculated as shown below:

Month	Observed counts	Proportion $pi (in decimal)$	Expected counts
January	38	0.085	$31366×719=60.90$
February	32	0.079	$29366×719=56.97$
March	43	0.085	$31366×719=60.90$
April	59	0.082	$30366×719=58.93$
May	78	0.085	$31366×719=60.90$
June	74	0.082	$30366×719=58.93$
July	98	0.085	$31366×719=60.90$
August	85	0.085	$31366×719=60.90$
September	64	0.082	$30366×719=58.93$
October	66	0.085	$31366×719=60.90$
November	42	0.082	$30366×719=58.93$
December	40	0.085	$31366×719=60.90$
	719	1 (approximately)	719

Significance level:

$α=0.05$

Test statistic:

$χ2=∑(observed count−expected count)2expected count$

Assumptions:

Assume that the 719 accidents included in the study is a random sample from the population of fatal bicycle accidents.
From the table, it is observed that all the expected counts are greater than 5.

Calculation:

Software procedure:

Step-by-step procedure to obtain the test statistic and P-value using the MINITAB software:

Choose Stat > Tables > Chi-Square Goodness-of-Fit Test (One Variable).
In Observed counts, enter the column of Number of Accidents.
In Category names, enter the column of Month.
Under Test, select the column of Proportion in Proportions specified by historical counts.
Click OK.

The output obtained using the MINITAB software is given below:

Bundle: Introduction to Statistics and Data Analysis, 5th + WebAssign Printed Access Card: Peck/Olsen/Devore. 5th Edition, Single-Term, Chapter 12.1, Problem 10E , additional homework tip 2

From the output, $χ2=78.1873$ .

8. P-value:

From the MINITAB output, $df=11$ and P-value is 0.000.

9. Conclusion:

Decision rule:

If P-value is less than or equal to the level of significance, reject the null hypothesis.
Otherwise, do not reject the null hypothesis.

Conclusion:

Here the level of significance is 0.05.

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

That is, $0.000<0.05$ .

Hence, reject the null hypothesis.

Therefore, there is convincing evidence that fatal bicycle accidents do not occur in any of the twelve months in proportion to the lengths of the months.

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Answer 4

Question

Chapter 12.1, Problem 10E

a.

To determine

Test whether fatal bicycle accidents are equally likely to occur in each of the 12-months at 0.01 significance level.

a.

Expert Solution

Answer to Problem 10E

There is convincing evidence that fatal bicycle accidents are not equally likely to occur in each of the months.

Explanation of Solution

Calculation:

The given data represent the classification of fatal bicycle accidents according to the month in which the accident occurred.

The expected counts are calculated as shown below:

Month	Observed counts	Expected counts
January	38	$71912=59.917$
February	32	$71912=59.917$
March	43	$71912=59.917$
April	59	$71912=59.917$
May	78	$71912=59.917$
June	74	$71912=59.917$
July	98	$71912=59.917$
August	85	$71912=59.917$
September	64	$71912=59.917$
October	66	$71912=59.917$
November	42	$71912=59.917$
December	40	$71912=59.917$
	719	719

The nine-step hypotheses testing procedure to test goodness-of-fit is given below:

1. Consider that the proportion of fatal bicycle accidents occurring in January is $p1$ , the proportion of fatal bicycle accidents occurring in February is $p2$ , the proportion of fatal bicycle accidents occurring in March is $p3$ , the proportion of fatal bicycle accidents occurring in April is $p4$ , the proportion of fatal bicycle accidents occurring in May is $p5$ , the proportion of fatal bicycle accidents occurring in June is $p6$ , the proportion of fatal bicycle accidents occurring in July is $p7$ , the proportion of fatal bicycle accidents occurring in August is $p8$ , the proportion of fatal bicycle accidents occurring in September is $p9$ , the proportion of fatal bicycle accidents occurring in October is $p10$ , the proportion of fatal bicycle accidents occurring in November is $p11$ , and the proportion of fatal bicycle accidents occurring in December is $p12$ .

2. Null hypothesis:

$H0:$ $p1=p2=p3=p4=p5=p6=p7=p8=p9=p10=p11=p12=112$ .

3. Alternative hypothesis:

$Ha:$ At least one of the population proportions is not equal to $112$ .

4. Significance level:

$α=0.01$

5. Test statistic:

$χ2=∑(observed count−expected count)2expected count$

6. Assumptions:

Assume that the 719 accidents included in the study is a random sample from the population of fatal bicycle accidents.
From the table, it is observed that all the expected counts are greater than 5.

7. Calculation:

Software procedure:

Step-by-step procedure to obtain the test statistics and P-value using the MINITAB software:

Choose Stat > Tables > Chi-Square Goodness-of-Fit Test (One Variable).
In Observed counts, enter the column of Number of Accidents.
In Category names, enter the column of Month.
Under Test, select the Equal Proportions.
Click OK.

The output obtained using the MINITAB software is given below:

Bundle: Introduction to Statistics and Data Analysis, 5th + WebAssign Printed Access Card: Peck/Olsen/Devore. 5th Edition, Single-Term, Chapter 12.1, Problem 10E , additional homework tip 1

From the output, $χ2=82.1627$ .

8. P-value:

From the MINITAB output, $df=11$ and P-value is 0.000.

9. Conclusion:

Decision rule:

If P-value is less than or equal to the level of significance, reject the null hypothesis.
Otherwise, do not reject the null hypothesis.

Conclusion:

Here the level of significance is 0.01.

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

That is, $0.000<0.01$ .

Hence, reject the null hypothesis. Therefore, there is convincing evidence that fatal bicycle accidents are not equally likely to occur in each of the months.

b.

To determine

Write the null and alternative hypotheses to determine if some months are riskier than others by taking differing month lengths into account.

b.

Expert Solution

Explanation of Solution

The null hypothesis in Part (a) specifies that fatal accidents are equally likely to occur in any of the 12 months.

The year considered in the study is 2004, which is a leap year. It is known that a leap year contains 366 days, with 29 days in February.

The null and alternative hypotheses to determine if some months are riskier than others by taking differing month lengths and the characteristics of a leap year into account are as follows:

Null hypothesis:

$H0:{p1=31366, p2=29366, p3=31366, p4=30366, p5=31366, p6=30366p7=31366, p8=31366, p9=30366, p10=31366, p11=30366, p12=31366$

Alternative hypothesis:

$Ha:$ At least one of the population proportions is not equal to the specified value.

c.

To determine

Test the hypotheses proposed in Part (b) at 0.05 significance level.

c.

Expert Solution

Answer to Problem 10E

There is convincing evidence that fatal bicycle accidents do not occur in any of the twelve months in proportion to the lengths of the months.

Explanation of Solution

Calculation:

The expected counts are calculated as shown below:

Month	Observed counts	Proportion $pi (in decimal)$	Expected counts
January	38	0.085	$31366×719=60.90$
February	32	0.079	$29366×719=56.97$
March	43	0.085	$31366×719=60.90$
April	59	0.082	$30366×719=58.93$
May	78	0.085	$31366×719=60.90$
June	74	0.082	$30366×719=58.93$
July	98	0.085	$31366×719=60.90$
August	85	0.085	$31366×719=60.90$
September	64	0.082	$30366×719=58.93$
October	66	0.085	$31366×719=60.90$
November	42	0.082	$30366×719=58.93$
December	40	0.085	$31366×719=60.90$
	719	1 (approximately)	719

Significance level:

$α=0.05$

Test statistic:

$χ2=∑(observed count−expected count)2expected count$

Assumptions:

Assume that the 719 accidents included in the study is a random sample from the population of fatal bicycle accidents.
From the table, it is observed that all the expected counts are greater than 5.

Calculation:

Software procedure:

Step-by-step procedure to obtain the test statistic and P-value using the MINITAB software:

Choose Stat > Tables > Chi-Square Goodness-of-Fit Test (One Variable).
In Observed counts, enter the column of Number of Accidents.
In Category names, enter the column of Month.
Under Test, select the column of Proportion in Proportions specified by historical counts.
Click OK.

The output obtained using the MINITAB software is given below:

Bundle: Introduction to Statistics and Data Analysis, 5th + WebAssign Printed Access Card: Peck/Olsen/Devore. 5th Edition, Single-Term, Chapter 12.1, Problem 10E , additional homework tip 2

From the output, $χ2=78.1873$ .

8. P-value:

From the MINITAB output, $df=11$ and P-value is 0.000.

9. Conclusion:

Decision rule:

If P-value is less than or equal to the level of significance, reject the null hypothesis.
Otherwise, do not reject the null hypothesis.

Conclusion:

Here the level of significance is 0.05.

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

That is, $0.000<0.05$ .

Hence, reject the null hypothesis.

Therefore, there is convincing evidence that fatal bicycle accidents do not occur in any of the twelve months in proportion to the lengths of the months.

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Answer 5

Chapter 12.1, Problem 10E

a.

To determine

Test whether fatal bicycle accidents are equally likely to occur in each of the 12-months at 0.01 significance level.

a.

Expert Solution

Answer to Problem 10E

There is convincing evidence that fatal bicycle accidents are not equally likely to occur in each of the months.

Explanation of Solution

Calculation:

The given data represent the classification of fatal bicycle accidents according to the month in which the accident occurred.

The expected counts are calculated as shown below:

Month	Observed counts	Expected counts
January	38	$71912=59.917$
February	32	$71912=59.917$
March	43	$71912=59.917$
April	59	$71912=59.917$
May	78	$71912=59.917$
June	74	$71912=59.917$
July	98	$71912=59.917$
August	85	$71912=59.917$
September	64	$71912=59.917$
October	66	$71912=59.917$
November	42	$71912=59.917$
December	40	$71912=59.917$
	719	719

The nine-step hypotheses testing procedure to test goodness-of-fit is given below:

1. Consider that the proportion of fatal bicycle accidents occurring in January is $p1$ , the proportion of fatal bicycle accidents occurring in February is $p2$ , the proportion of fatal bicycle accidents occurring in March is $p3$ , the proportion of fatal bicycle accidents occurring in April is $p4$ , the proportion of fatal bicycle accidents occurring in May is $p5$ , the proportion of fatal bicycle accidents occurring in June is $p6$ , the proportion of fatal bicycle accidents occurring in July is $p7$ , the proportion of fatal bicycle accidents occurring in August is $p8$ , the proportion of fatal bicycle accidents occurring in September is $p9$ , the proportion of fatal bicycle accidents occurring in October is $p10$ , the proportion of fatal bicycle accidents occurring in November is $p11$ , and the proportion of fatal bicycle accidents occurring in December is $p12$ .

2. Null hypothesis:

$H0:$ $p1=p2=p3=p4=p5=p6=p7=p8=p9=p10=p11=p12=112$ .

3. Alternative hypothesis:

$Ha:$ At least one of the population proportions is not equal to $112$ .

4. Significance level:

$α=0.01$

5. Test statistic:

$χ2=∑(observed count−expected count)2expected count$

6. Assumptions:

Assume that the 719 accidents included in the study is a random sample from the population of fatal bicycle accidents.
From the table, it is observed that all the expected counts are greater than 5.

7. Calculation:

Software procedure:

Step-by-step procedure to obtain the test statistics and P-value using the MINITAB software:

Choose Stat > Tables > Chi-Square Goodness-of-Fit Test (One Variable).
In Observed counts, enter the column of Number of Accidents.
In Category names, enter the column of Month.
Under Test, select the Equal Proportions.
Click OK.

The output obtained using the MINITAB software is given below:

Bundle: Introduction to Statistics and Data Analysis, 5th + WebAssign Printed Access Card: Peck/Olsen/Devore. 5th Edition, Single-Term, Chapter 12.1, Problem 10E , additional homework tip 1

From the output, $χ2=82.1627$ .

8. P-value:

From the MINITAB output, $df=11$ and P-value is 0.000.

9. Conclusion:

Decision rule:

If P-value is less than or equal to the level of significance, reject the null hypothesis.
Otherwise, do not reject the null hypothesis.

Conclusion:

Here the level of significance is 0.01.

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

That is, $0.000<0.01$ .

Hence, reject the null hypothesis. Therefore, there is convincing evidence that fatal bicycle accidents are not equally likely to occur in each of the months.

b.

To determine

Write the null and alternative hypotheses to determine if some months are riskier than others by taking differing month lengths into account.

b.

Expert Solution

Explanation of Solution

The null hypothesis in Part (a) specifies that fatal accidents are equally likely to occur in any of the 12 months.

The year considered in the study is 2004, which is a leap year. It is known that a leap year contains 366 days, with 29 days in February.

The null and alternative hypotheses to determine if some months are riskier than others by taking differing month lengths and the characteristics of a leap year into account are as follows:

Null hypothesis:

$H0:{p1=31366, p2=29366, p3=31366, p4=30366, p5=31366, p6=30366p7=31366, p8=31366, p9=30366, p10=31366, p11=30366, p12=31366$

Alternative hypothesis:

$Ha:$ At least one of the population proportions is not equal to the specified value.

c.

To determine

Test the hypotheses proposed in Part (b) at 0.05 significance level.

c.

Expert Solution

Answer to Problem 10E

There is convincing evidence that fatal bicycle accidents do not occur in any of the twelve months in proportion to the lengths of the months.

Explanation of Solution

Calculation:

The expected counts are calculated as shown below:

Month	Observed counts	Proportion $pi (in decimal)$	Expected counts
January	38	0.085	$31366×719=60.90$
February	32	0.079	$29366×719=56.97$
March	43	0.085	$31366×719=60.90$
April	59	0.082	$30366×719=58.93$
May	78	0.085	$31366×719=60.90$
June	74	0.082	$30366×719=58.93$
July	98	0.085	$31366×719=60.90$
August	85	0.085	$31366×719=60.90$
September	64	0.082	$30366×719=58.93$
October	66	0.085	$31366×719=60.90$
November	42	0.082	$30366×719=58.93$
December	40	0.085	$31366×719=60.90$
	719	1 (approximately)	719

Significance level:

$α=0.05$

Test statistic:

$χ2=∑(observed count−expected count)2expected count$

Assumptions:

Assume that the 719 accidents included in the study is a random sample from the population of fatal bicycle accidents.
From the table, it is observed that all the expected counts are greater than 5.

Calculation:

Software procedure:

Step-by-step procedure to obtain the test statistic and P-value using the MINITAB software:

Choose Stat > Tables > Chi-Square Goodness-of-Fit Test (One Variable).
In Observed counts, enter the column of Number of Accidents.
In Category names, enter the column of Month.
Under Test, select the column of Proportion in Proportions specified by historical counts.
Click OK.

The output obtained using the MINITAB software is given below:

Bundle: Introduction to Statistics and Data Analysis, 5th + WebAssign Printed Access Card: Peck/Olsen/Devore. 5th Edition, Single-Term, Chapter 12.1, Problem 10E , additional homework tip 2

From the output, $χ2=78.1873$ .

8. P-value:

From the MINITAB output, $df=11$ and P-value is 0.000.

9. Conclusion:

Decision rule:

If P-value is less than or equal to the level of significance, reject the null hypothesis.
Otherwise, do not reject the null hypothesis.

Conclusion:

Here the level of significance is 0.05.

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

That is, $0.000<0.05$ .

Hence, reject the null hypothesis.

Therefore, there is convincing evidence that fatal bicycle accidents do not occur in any of the twelve months in proportion to the lengths of the months.

Answer 6

Chapter 12.1, Problem 10E

a.

To determine

Test whether fatal bicycle accidents are equally likely to occur in each of the 12-months at 0.01 significance level.

a.

Expert Solution

Answer to Problem 10E

There is convincing evidence that fatal bicycle accidents are not equally likely to occur in each of the months.

Explanation of Solution

Calculation:

The given data represent the classification of fatal bicycle accidents according to the month in which the accident occurred.

The expected counts are calculated as shown below:

Month	Observed counts	Expected counts
January	38	$71912=59.917$
February	32	$71912=59.917$
March	43	$71912=59.917$
April	59	$71912=59.917$
May	78	$71912=59.917$
June	74	$71912=59.917$
July	98	$71912=59.917$
August	85	$71912=59.917$
September	64	$71912=59.917$
October	66	$71912=59.917$
November	42	$71912=59.917$
December	40	$71912=59.917$
	719	719

The nine-step hypotheses testing procedure to test goodness-of-fit is given below:

1. Consider that the proportion of fatal bicycle accidents occurring in January is $p1$ , the proportion of fatal bicycle accidents occurring in February is $p2$ , the proportion of fatal bicycle accidents occurring in March is $p3$ , the proportion of fatal bicycle accidents occurring in April is $p4$ , the proportion of fatal bicycle accidents occurring in May is $p5$ , the proportion of fatal bicycle accidents occurring in June is $p6$ , the proportion of fatal bicycle accidents occurring in July is $p7$ , the proportion of fatal bicycle accidents occurring in August is $p8$ , the proportion of fatal bicycle accidents occurring in September is $p9$ , the proportion of fatal bicycle accidents occurring in October is $p10$ , the proportion of fatal bicycle accidents occurring in November is $p11$ , and the proportion of fatal bicycle accidents occurring in December is $p12$ .

2. Null hypothesis:

$H0:$ $p1=p2=p3=p4=p5=p6=p7=p8=p9=p10=p11=p12=112$ .

3. Alternative hypothesis:

$Ha:$ At least one of the population proportions is not equal to $112$ .

4. Significance level:

$α=0.01$

5. Test statistic:

$χ2=∑(observed count−expected count)2expected count$

6. Assumptions:

Assume that the 719 accidents included in the study is a random sample from the population of fatal bicycle accidents.
From the table, it is observed that all the expected counts are greater than 5.

7. Calculation:

Software procedure:

Step-by-step procedure to obtain the test statistics and P-value using the MINITAB software:

Choose Stat > Tables > Chi-Square Goodness-of-Fit Test (One Variable).
In Observed counts, enter the column of Number of Accidents.
In Category names, enter the column of Month.
Under Test, select the Equal Proportions.
Click OK.

The output obtained using the MINITAB software is given below:

Bundle: Introduction to Statistics and Data Analysis, 5th + WebAssign Printed Access Card: Peck/Olsen/Devore. 5th Edition, Single-Term, Chapter 12.1, Problem 10E , additional homework tip 1

From the output, $χ2=82.1627$ .

8. P-value:

From the MINITAB output, $df=11$ and P-value is 0.000.

9. Conclusion:

Decision rule:

If P-value is less than or equal to the level of significance, reject the null hypothesis.
Otherwise, do not reject the null hypothesis.

Conclusion:

Here the level of significance is 0.01.

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

That is, $0.000<0.01$ .

Hence, reject the null hypothesis. Therefore, there is convincing evidence that fatal bicycle accidents are not equally likely to occur in each of the months.

Answer 7

Chapter 12.1, Problem 10E

a.

To determine

Test whether fatal bicycle accidents are equally likely to occur in each of the 12-months at 0.01 significance level.

Answer 8

a.

Expert Solution

Answer to Problem 10E

There is convincing evidence that fatal bicycle accidents are not equally likely to occur in each of the months.

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Calculation:

The given data represent the classification of fatal bicycle accidents according to the month in which the accident occurred.

The expected counts are calculated as shown below:

Month	Observed counts	Expected counts
January	38	$71912=59.917$
February	32	$71912=59.917$
March	43	$71912=59.917$
April	59	$71912=59.917$
May	78	$71912=59.917$
June	74	$71912=59.917$
July	98	$71912=59.917$
August	85	$71912=59.917$
September	64	$71912=59.917$
October	66	$71912=59.917$
November	42	$71912=59.917$
December	40	$71912=59.917$
	719	719

The nine-step hypotheses testing procedure to test goodness-of-fit is given below:

1. Consider that the proportion of fatal bicycle accidents occurring in January is $p1$ , the proportion of fatal bicycle accidents occurring in February is $p2$ , the proportion of fatal bicycle accidents occurring in March is $p3$ , the proportion of fatal bicycle accidents occurring in April is $p4$ , the proportion of fatal bicycle accidents occurring in May is $p5$ , the proportion of fatal bicycle accidents occurring in June is $p6$ , the proportion of fatal bicycle accidents occurring in July is $p7$ , the proportion of fatal bicycle accidents occurring in August is $p8$ , the proportion of fatal bicycle accidents occurring in September is $p9$ , the proportion of fatal bicycle accidents occurring in October is $p10$ , the proportion of fatal bicycle accidents occurring in November is $p11$ , and the proportion of fatal bicycle accidents occurring in December is $p12$ .

2. Null hypothesis:

$H0:$ $p1=p2=p3=p4=p5=p6=p7=p8=p9=p10=p11=p12=112$ .

3. Alternative hypothesis:

$Ha:$ At least one of the population proportions is not equal to $112$ .

4. Significance level:

$α=0.01$

5. Test statistic:

$χ2=∑(observed count−expected count)2expected count$

6. Assumptions:

Assume that the 719 accidents included in the study is a random sample from the population of fatal bicycle accidents.
From the table, it is observed that all the expected counts are greater than 5.

7. Calculation:

Software procedure:

Step-by-step procedure to obtain the test statistics and P-value using the MINITAB software:

Choose Stat > Tables > Chi-Square Goodness-of-Fit Test (One Variable).
In Observed counts, enter the column of Number of Accidents.
In Category names, enter the column of Month.
Under Test, select the Equal Proportions.
Click OK.

The output obtained using the MINITAB software is given below:

Bundle: Introduction to Statistics and Data Analysis, 5th + WebAssign Printed Access Card: Peck/Olsen/Devore. 5th Edition, Single-Term, Chapter 12.1, Problem 10E , additional homework tip 1

From the output, $χ2=82.1627$ .

8. P-value:

From the MINITAB output, $df=11$ and P-value is 0.000.

9. Conclusion:

Decision rule:

If P-value is less than or equal to the level of significance, reject the null hypothesis.
Otherwise, do not reject the null hypothesis.

Conclusion:

Here the level of significance is 0.01.

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

That is, $0.000<0.01$ .

Hence, reject the null hypothesis. Therefore, there is convincing evidence that fatal bicycle accidents are not equally likely to occur in each of the months.

Answer 13

b.

To determine

Write the null and alternative hypotheses to determine if some months are riskier than others by taking differing month lengths into account.

b.

Expert Solution

Explanation of Solution

The null hypothesis in Part (a) specifies that fatal accidents are equally likely to occur in any of the 12 months.

The year considered in the study is 2004, which is a leap year. It is known that a leap year contains 366 days, with 29 days in February.

The null and alternative hypotheses to determine if some months are riskier than others by taking differing month lengths and the characteristics of a leap year into account are as follows:

Null hypothesis:

$H0:{p1=31366, p2=29366, p3=31366, p4=30366, p5=31366, p6=30366p7=31366, p8=31366, p9=30366, p10=31366, p11=30366, p12=31366$

Alternative hypothesis:

$Ha:$ At least one of the population proportions is not equal to the specified value.

Answer 14

b.

To determine

Write the null and alternative hypotheses to determine if some months are riskier than others by taking differing month lengths into account.

Answer 15

b.

Expert Solution

Explanation of Solution

The null hypothesis in Part (a) specifies that fatal accidents are equally likely to occur in any of the 12 months.

The year considered in the study is 2004, which is a leap year. It is known that a leap year contains 366 days, with 29 days in February.

The null and alternative hypotheses to determine if some months are riskier than others by taking differing month lengths and the characteristics of a leap year into account are as follows:

Null hypothesis:

$H0:{p1=31366, p2=29366, p3=31366, p4=30366, p5=31366, p6=30366p7=31366, p8=31366, p9=30366, p10=31366, p11=30366, p12=31366$

Alternative hypothesis:

$Ha:$ At least one of the population proportions is not equal to the specified value.

Answer 16

b.

Expert Solution

Answer 17

b.

Expert Solution

Answer 18

Expert Solution

Answer 19

c.

To determine

Test the hypotheses proposed in Part (b) at 0.05 significance level.

c.

Expert Solution

Answer to Problem 10E

There is convincing evidence that fatal bicycle accidents do not occur in any of the twelve months in proportion to the lengths of the months.

Explanation of Solution

Calculation:

The expected counts are calculated as shown below:

Month	Observed counts	Proportion $pi (in decimal)$	Expected counts
January	38	0.085	$31366×719=60.90$
February	32	0.079	$29366×719=56.97$
March	43	0.085	$31366×719=60.90$
April	59	0.082	$30366×719=58.93$
May	78	0.085	$31366×719=60.90$
June	74	0.082	$30366×719=58.93$
July	98	0.085	$31366×719=60.90$
August	85	0.085	$31366×719=60.90$
September	64	0.082	$30366×719=58.93$
October	66	0.085	$31366×719=60.90$
November	42	0.082	$30366×719=58.93$
December	40	0.085	$31366×719=60.90$
	719	1 (approximately)	719

Significance level:

$α=0.05$

Test statistic:

$χ2=∑(observed count−expected count)2expected count$

Assumptions:

Assume that the 719 accidents included in the study is a random sample from the population of fatal bicycle accidents.
From the table, it is observed that all the expected counts are greater than 5.

Calculation:

Software procedure:

Step-by-step procedure to obtain the test statistic and P-value using the MINITAB software:

Choose Stat > Tables > Chi-Square Goodness-of-Fit Test (One Variable).
In Observed counts, enter the column of Number of Accidents.
In Category names, enter the column of Month.
Under Test, select the column of Proportion in Proportions specified by historical counts.
Click OK.

The output obtained using the MINITAB software is given below:

Bundle: Introduction to Statistics and Data Analysis, 5th + WebAssign Printed Access Card: Peck/Olsen/Devore. 5th Edition, Single-Term, Chapter 12.1, Problem 10E , additional homework tip 2

From the output, $χ2=78.1873$ .

8. P-value:

From the MINITAB output, $df=11$ and P-value is 0.000.

9. Conclusion:

Decision rule:

If P-value is less than or equal to the level of significance, reject the null hypothesis.
Otherwise, do not reject the null hypothesis.

Conclusion:

Here the level of significance is 0.05.

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

That is, $0.000<0.05$ .

Hence, reject the null hypothesis.

Therefore, there is convincing evidence that fatal bicycle accidents do not occur in any of the twelve months in proportion to the lengths of the months.

Answer 20

c.

To determine

Test the hypotheses proposed in Part (b) at 0.05 significance level.

Answer 21

c.

Expert Solution

Answer to Problem 10E

There is convincing evidence that fatal bicycle accidents do not occur in any of the twelve months in proportion to the lengths of the months.

Answer 22

c.

Expert Solution

Answer 23

c.

Expert Solution

Answer 24

Expert Solution

Answer 25

Explanation of Solution

Calculation:

The expected counts are calculated as shown below:

Month	Observed counts	Proportion $pi (in decimal)$	Expected counts
January	38	0.085	$31366×719=60.90$
February	32	0.079	$29366×719=56.97$
March	43	0.085	$31366×719=60.90$
April	59	0.082	$30366×719=58.93$
May	78	0.085	$31366×719=60.90$
June	74	0.082	$30366×719=58.93$
July	98	0.085	$31366×719=60.90$
August	85	0.085	$31366×719=60.90$
September	64	0.082	$30366×719=58.93$
October	66	0.085	$31366×719=60.90$
November	42	0.082	$30366×719=58.93$
December	40	0.085	$31366×719=60.90$
	719	1 (approximately)	719

Significance level:

$α=0.05$

Test statistic:

$χ2=∑(observed count−expected count)2expected count$

Assumptions:

Assume that the 719 accidents included in the study is a random sample from the population of fatal bicycle accidents.
From the table, it is observed that all the expected counts are greater than 5.

Calculation:

Software procedure:

Step-by-step procedure to obtain the test statistic and P-value using the MINITAB software:

Choose Stat > Tables > Chi-Square Goodness-of-Fit Test (One Variable).
In Observed counts, enter the column of Number of Accidents.
In Category names, enter the column of Month.
Under Test, select the column of Proportion in Proportions specified by historical counts.
Click OK.

The output obtained using the MINITAB software is given below:

Bundle: Introduction to Statistics and Data Analysis, 5th + WebAssign Printed Access Card: Peck/Olsen/Devore. 5th Edition, Single-Term, Chapter 12.1, Problem 10E , additional homework tip 2

From the output, $χ2=78.1873$ .

8. P-value:

From the MINITAB output, $df=11$ and P-value is 0.000.

9. Conclusion:

Decision rule:

If P-value is less than or equal to the level of significance, reject the null hypothesis.
Otherwise, do not reject the null hypothesis.

Conclusion:

Here the level of significance is 0.05.

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

That is, $0.000<0.05$ .

Hence, reject the null hypothesis.

Therefore, there is convincing evidence that fatal bicycle accidents do not occur in any of the twelve months in proportion to the lengths of the months.