For Exercises 1 through 10, follow these steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Assume all assumptions have been met. 1. Traffic Accident Fatalities A traffic safety report indicated that for the 21–24 year age group, 31.58% of traffic fatalities were victims who had used a seat belt. Victims who were not wearing a seat belt accounted for 59.83% of the deaths, and the status of the rest was unknown. A study of 120 randomly selected traffic fatalities in a particular region showed that for this age group, 35 of the victims had used a seat belt, 78 had not, and the status of the rest was unknown. At α = 0.05, is there sufficient evidence that the proportions differ from those in the report? Source: New York Times Almanac.

Question

Want to see more full solutions like this?

Answer 1

Textbook Question

Chapter 11, Problem 11.1.1RE

For Exercises 1 through 10, follow these steps.

a. State the hypotheses and identify the claim.

b. Find the critical value(s).

c. Compute the test value.

d. Make the decision.

e. Summarize the results.

Use the traditional method of hypothesis testing unless otherwise specified. Assume all assumptions have been met.

1. Traffic Accident Fatalities A traffic safety report indicated that for the 21–24 year age group, 31.58% of traffic fatalities were victims who had used a seat belt. Victims who were not wearing a seat belt accounted for 59.83% of the deaths, and the status of the rest was unknown. A study of 120 randomly selected traffic fatalities in a particular region showed that for this age group, 35 of the victims had used a seat belt, 78 had not, and the status of the rest was unknown. At α = 0.05, is there sufficient evidence that the proportions differ from those in the report?

Source: New York Times Almanac.

(a)

Expert Solution

To determine

To state: The hypothesis and the claim.

Answer to Problem 11.1.1RE

The null and alternative hypotheses are:

H0 : The proportion is same as the reported proportion.

H1 : The proportion is different from the reported proportion.

And the claim of the test is the observed proportion is different from the reported proportion.

Explanation of Solution

Given info:

The percentage of death and the observed count corresponding to each reason are provided in the question. The level of significance is α=0.05 .

Justification:

The researcher wants to know that whether the proportion of the traffic fatalities corresponding to each reason is different from the reported proportion or not. The null and alternative hypothesis can be defined as:

Null hypothesis:

H0 : The proportion is same as the reported proportion.

Alternative hypothesis:

H1 : The proportion is different from the reported proportion.

In the provided situation, the claim of the study will be the observed proportion corresponding to each reason is different from the reported proportion.

(b)

Expert Solution

To determine

The critical value.

Answer to Problem 11.1.1RE

The required critical value is 5.991.

Explanation of Solution

The required critical value is obtained from the provided chi-square table in the textbook. The number of categories is 3.

The degrees of freedom is calculated as:

Degrees of freedom=Number of categories−1=3−1=2

Therefore, the critical value at α=0.05 level of significance and 2 degrees of freedom is obtained as 5.991.

(c)

Expert Solution

To determine

The value of the test statistic.

Answer to Problem 11.1.1RE

The test statistic value is 1.819.

Explanation of Solution

Calculation:

Software procedure:

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

Enter the data in the Minitab worksheet.
Go to Stat> Tables> Chi-Square Goodness-of-Fit Test (one variable).
Specify the “Observed count”, choose the option “Proportions specified by historic count”, and specify the column where the percentage is written.
Click on OK.

Output using the MINITAB software is given below:

Elementary Statistics: A Step By Step Approach, Chapter 11, Problem 11.1.1RE

Therefore, the obtained value of the test statistic is 1.819.

(d)

Expert Solution

To determine

To make: The decision.

Answer to Problem 11.1.1RE

The null hypothesis will not be rejected.

Explanation of Solution

The obtained value of the test statistic is 1.819 and the critical value is 5.991. As the obtained value of the chi-square statistic is less than the critical value, it can be said that there is not enough evidence to reject the null hypothesis at α=0.05 .

(e)

Expert Solution

To determine

To summarize: The results.

Answer to Problem 11.1.1RE

According to the obtained result, the claim of the study is not true.

Explanation of Solution

The null hypothesis is not rejected. On the basis of the obtained result, it can be concluded that the result of the proportion of the traffic fatalities is not different from the reported proportion at α=0.05 . That is, the claim is not true.

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

ged the variance for group 1) Different groups of male stalk-eyed flies were raised on different diets: a high nutrient corn diet vs. a low nutrient cotton wool diet. Investigators wanted to see if diet quality influenced eye-stalk length. They obtained the following data: d Diet Sample Mean Eye-stalk Length Variance in Eye-stalk d size, n (mm) Length (mm²) Corn (group 1) 21 2.05 0.0558 Cotton (group 2) 24 1.54 0.0812 =205-1.54-05T a) Construct a 95% confidence interval for the difference in mean eye-stalk length between the two diets (e.g., use group 1 - group 2).

An article in Business Week discussed the large spread between the federal funds rate and the average credit card rate. The table below is a frequency distribution of the credit card rate charged by the top 100 issuers. Credit Card Rates Credit Card Rate Frequency 18% -23% 19 17% -17.9% 16 16% -16.9% 31 15% -15.9% 26 14% -14.9% Copy Data 8 Step 1 of 2: Calculate the average credit card rate charged by the top 100 issuers based on the frequency distribution. Round your answer to two decimal places.

Please could you check my answers

Answer 2

Textbook Question

Chapter 11, Problem 11.1.1RE

For Exercises 1 through 10, follow these steps.

a. State the hypotheses and identify the claim.

b. Find the critical value(s).

c. Compute the test value.

d. Make the decision.

e. Summarize the results.

Use the traditional method of hypothesis testing unless otherwise specified. Assume all assumptions have been met.

1. Traffic Accident Fatalities A traffic safety report indicated that for the 21–24 year age group, 31.58% of traffic fatalities were victims who had used a seat belt. Victims who were not wearing a seat belt accounted for 59.83% of the deaths, and the status of the rest was unknown. A study of 120 randomly selected traffic fatalities in a particular region showed that for this age group, 35 of the victims had used a seat belt, 78 had not, and the status of the rest was unknown. At α = 0.05, is there sufficient evidence that the proportions differ from those in the report?

Source: New York Times Almanac.

(a)

Expert Solution

To determine

To state: The hypothesis and the claim.

Answer to Problem 11.1.1RE

The null and alternative hypotheses are:

H0 : The proportion is same as the reported proportion.

H1 : The proportion is different from the reported proportion.

And the claim of the test is the observed proportion is different from the reported proportion.

Explanation of Solution

Given info:

The percentage of death and the observed count corresponding to each reason are provided in the question. The level of significance is α=0.05 .

Justification:

The researcher wants to know that whether the proportion of the traffic fatalities corresponding to each reason is different from the reported proportion or not. The null and alternative hypothesis can be defined as:

Null hypothesis:

H0 : The proportion is same as the reported proportion.

Alternative hypothesis:

H1 : The proportion is different from the reported proportion.

In the provided situation, the claim of the study will be the observed proportion corresponding to each reason is different from the reported proportion.

(b)

Expert Solution

To determine

The critical value.

Answer to Problem 11.1.1RE

The required critical value is 5.991.

Explanation of Solution

The required critical value is obtained from the provided chi-square table in the textbook. The number of categories is 3.

The degrees of freedom is calculated as:

Degrees of freedom=Number of categories−1=3−1=2

Therefore, the critical value at α=0.05 level of significance and 2 degrees of freedom is obtained as 5.991.

(c)

Expert Solution

To determine

The value of the test statistic.

Answer to Problem 11.1.1RE

The test statistic value is 1.819.

Explanation of Solution

Calculation:

Software procedure:

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

Enter the data in the Minitab worksheet.
Go to Stat> Tables> Chi-Square Goodness-of-Fit Test (one variable).
Specify the “Observed count”, choose the option “Proportions specified by historic count”, and specify the column where the percentage is written.
Click on OK.

Output using the MINITAB software is given below:

Elementary Statistics: A Step By Step Approach, Chapter 11, Problem 11.1.1RE

Therefore, the obtained value of the test statistic is 1.819.

(d)

Expert Solution

To determine

To make: The decision.

Answer to Problem 11.1.1RE

The null hypothesis will not be rejected.

Explanation of Solution

The obtained value of the test statistic is 1.819 and the critical value is 5.991. As the obtained value of the chi-square statistic is less than the critical value, it can be said that there is not enough evidence to reject the null hypothesis at α=0.05 .

(e)

Expert Solution

To determine

To summarize: The results.

Answer to Problem 11.1.1RE

According to the obtained result, the claim of the study is not true.

Explanation of Solution

The null hypothesis is not rejected. On the basis of the obtained result, it can be concluded that the result of the proportion of the traffic fatalities is not different from the reported proportion at α=0.05 . That is, the claim is not true.

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

Textbook Question

Chapter 11, Problem 11.1.1RE

For Exercises 1 through 10, follow these steps.

a. State the hypotheses and identify the claim.

b. Find the critical value(s).

c. Compute the test value.

d. Make the decision.

e. Summarize the results.

Use the traditional method of hypothesis testing unless otherwise specified. Assume all assumptions have been met.

1. Traffic Accident Fatalities A traffic safety report indicated that for the 21–24 year age group, 31.58% of traffic fatalities were victims who had used a seat belt. Victims who were not wearing a seat belt accounted for 59.83% of the deaths, and the status of the rest was unknown. A study of 120 randomly selected traffic fatalities in a particular region showed that for this age group, 35 of the victims had used a seat belt, 78 had not, and the status of the rest was unknown. At α = 0.05, is there sufficient evidence that the proportions differ from those in the report?

Source: New York Times Almanac.

(a)

Expert Solution

To determine

To state: The hypothesis and the claim.

Answer to Problem 11.1.1RE

The null and alternative hypotheses are:

H0 : The proportion is same as the reported proportion.

H1 : The proportion is different from the reported proportion.

And the claim of the test is the observed proportion is different from the reported proportion.

Explanation of Solution

Given info:

The percentage of death and the observed count corresponding to each reason are provided in the question. The level of significance is α=0.05 .

Justification:

The researcher wants to know that whether the proportion of the traffic fatalities corresponding to each reason is different from the reported proportion or not. The null and alternative hypothesis can be defined as:

Null hypothesis:

H0 : The proportion is same as the reported proportion.

Alternative hypothesis:

H1 : The proportion is different from the reported proportion.

In the provided situation, the claim of the study will be the observed proportion corresponding to each reason is different from the reported proportion.

(b)

Expert Solution

To determine

The critical value.

Answer to Problem 11.1.1RE

The required critical value is 5.991.

Explanation of Solution

The required critical value is obtained from the provided chi-square table in the textbook. The number of categories is 3.

The degrees of freedom is calculated as:

Degrees of freedom=Number of categories−1=3−1=2

Therefore, the critical value at α=0.05 level of significance and 2 degrees of freedom is obtained as 5.991.

(c)

Expert Solution

To determine

The value of the test statistic.

Answer to Problem 11.1.1RE

The test statistic value is 1.819.

Explanation of Solution

Calculation:

Software procedure:

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

Enter the data in the Minitab worksheet.
Go to Stat> Tables> Chi-Square Goodness-of-Fit Test (one variable).
Specify the “Observed count”, choose the option “Proportions specified by historic count”, and specify the column where the percentage is written.
Click on OK.

Output using the MINITAB software is given below:

Elementary Statistics: A Step By Step Approach, Chapter 11, Problem 11.1.1RE

Therefore, the obtained value of the test statistic is 1.819.

(d)

Expert Solution

To determine

To make: The decision.

Answer to Problem 11.1.1RE

The null hypothesis will not be rejected.

Explanation of Solution

The obtained value of the test statistic is 1.819 and the critical value is 5.991. As the obtained value of the chi-square statistic is less than the critical value, it can be said that there is not enough evidence to reject the null hypothesis at α=0.05 .

(e)

Expert Solution

To determine

To summarize: The results.

Answer to Problem 11.1.1RE

According to the obtained result, the claim of the study is not true.

Explanation of Solution

The null hypothesis is not rejected. On the basis of the obtained result, it can be concluded that the result of the proportion of the traffic fatalities is not different from the reported proportion at α=0.05 . That is, the claim is not true.

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

Textbook Question

Chapter 11, Problem 11.1.1RE

For Exercises 1 through 10, follow these steps.

a. State the hypotheses and identify the claim.

b. Find the critical value(s).

c. Compute the test value.

d. Make the decision.

e. Summarize the results.

Use the traditional method of hypothesis testing unless otherwise specified. Assume all assumptions have been met.

1. Traffic Accident Fatalities A traffic safety report indicated that for the 21–24 year age group, 31.58% of traffic fatalities were victims who had used a seat belt. Victims who were not wearing a seat belt accounted for 59.83% of the deaths, and the status of the rest was unknown. A study of 120 randomly selected traffic fatalities in a particular region showed that for this age group, 35 of the victims had used a seat belt, 78 had not, and the status of the rest was unknown. At α = 0.05, is there sufficient evidence that the proportions differ from those in the report?

Source: New York Times Almanac.

(a)

Expert Solution

To determine

To state: The hypothesis and the claim.

Answer to Problem 11.1.1RE

The null and alternative hypotheses are:

H0 : The proportion is same as the reported proportion.

H1 : The proportion is different from the reported proportion.

And the claim of the test is the observed proportion is different from the reported proportion.

Explanation of Solution

Given info:

The percentage of death and the observed count corresponding to each reason are provided in the question. The level of significance is α=0.05 .

Justification:

The researcher wants to know that whether the proportion of the traffic fatalities corresponding to each reason is different from the reported proportion or not. The null and alternative hypothesis can be defined as:

Null hypothesis:

H0 : The proportion is same as the reported proportion.

Alternative hypothesis:

H1 : The proportion is different from the reported proportion.

In the provided situation, the claim of the study will be the observed proportion corresponding to each reason is different from the reported proportion.

(b)

Expert Solution

To determine

The critical value.

Answer to Problem 11.1.1RE

The required critical value is 5.991.

Explanation of Solution

The required critical value is obtained from the provided chi-square table in the textbook. The number of categories is 3.

The degrees of freedom is calculated as:

Degrees of freedom=Number of categories−1=3−1=2

Therefore, the critical value at α=0.05 level of significance and 2 degrees of freedom is obtained as 5.991.

(c)

Expert Solution

To determine

The value of the test statistic.

Answer to Problem 11.1.1RE

The test statistic value is 1.819.

Explanation of Solution

Calculation:

Software procedure:

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

Enter the data in the Minitab worksheet.
Go to Stat> Tables> Chi-Square Goodness-of-Fit Test (one variable).
Specify the “Observed count”, choose the option “Proportions specified by historic count”, and specify the column where the percentage is written.
Click on OK.

Output using the MINITAB software is given below:

Elementary Statistics: A Step By Step Approach, Chapter 11, Problem 11.1.1RE

Therefore, the obtained value of the test statistic is 1.819.

(d)

Expert Solution

To determine

To make: The decision.

Answer to Problem 11.1.1RE

The null hypothesis will not be rejected.

Explanation of Solution

The obtained value of the test statistic is 1.819 and the critical value is 5.991. As the obtained value of the chi-square statistic is less than the critical value, it can be said that there is not enough evidence to reject the null hypothesis at α=0.05 .

(e)

Expert Solution

To determine

To summarize: The results.

Answer to Problem 11.1.1RE

According to the obtained result, the claim of the study is not true.

Explanation of Solution

The null hypothesis is not rejected. On the basis of the obtained result, it can be concluded that the result of the proportion of the traffic fatalities is not different from the reported proportion at α=0.05 . That is, the claim is not true.

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

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

To determine

To state: The hypothesis and the claim.

Answer to Problem 11.1.1RE

The null and alternative hypotheses are:

H0 : The proportion is same as the reported proportion.

H1 : The proportion is different from the reported proportion.

And the claim of the test is the observed proportion is different from the reported proportion.

Explanation of Solution

Given info:

The percentage of death and the observed count corresponding to each reason are provided in the question. The level of significance is α=0.05 .

Justification:

The researcher wants to know that whether the proportion of the traffic fatalities corresponding to each reason is different from the reported proportion or not. The null and alternative hypothesis can be defined as:

Null hypothesis:

H0 : The proportion is same as the reported proportion.

Alternative hypothesis:

H1 : The proportion is different from the reported proportion.

In the provided situation, the claim of the study will be the observed proportion corresponding to each reason is different from the reported proportion.

(b)

Expert Solution

To determine

The critical value.

Answer to Problem 11.1.1RE

The required critical value is 5.991.

Explanation of Solution

The required critical value is obtained from the provided chi-square table in the textbook. The number of categories is 3.

The degrees of freedom is calculated as:

Degrees of freedom=Number of categories−1=3−1=2

Therefore, the critical value at α=0.05 level of significance and 2 degrees of freedom is obtained as 5.991.

(c)

Expert Solution

To determine

The value of the test statistic.

Answer to Problem 11.1.1RE

The test statistic value is 1.819.

Explanation of Solution

Calculation:

Software procedure:

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

Enter the data in the Minitab worksheet.
Go to Stat> Tables> Chi-Square Goodness-of-Fit Test (one variable).
Specify the “Observed count”, choose the option “Proportions specified by historic count”, and specify the column where the percentage is written.
Click on OK.

Output using the MINITAB software is given below:

Elementary Statistics: A Step By Step Approach, Chapter 11, Problem 11.1.1RE

Therefore, the obtained value of the test statistic is 1.819.

(d)

Expert Solution

To determine

To make: The decision.

Answer to Problem 11.1.1RE

The null hypothesis will not be rejected.

Explanation of Solution

The obtained value of the test statistic is 1.819 and the critical value is 5.991. As the obtained value of the chi-square statistic is less than the critical value, it can be said that there is not enough evidence to reject the null hypothesis at α=0.05 .

(e)

Expert Solution

To determine

To summarize: The results.

Answer to Problem 11.1.1RE

According to the obtained result, the claim of the study is not true.

Explanation of Solution

The null hypothesis is not rejected. On the basis of the obtained result, it can be concluded that the result of the proportion of the traffic fatalities is not different from the reported proportion at α=0.05 . That is, the claim is not true.

Answer 8

(a)

Expert Solution

To determine

To state: The hypothesis and the claim.

Answer to Problem 11.1.1RE

The null and alternative hypotheses are:

H0 : The proportion is same as the reported proportion.

H1 : The proportion is different from the reported proportion.

And the claim of the test is the observed proportion is different from the reported proportion.

Explanation of Solution

Given info:

The percentage of death and the observed count corresponding to each reason are provided in the question. The level of significance is α=0.05 .

Justification:

The researcher wants to know that whether the proportion of the traffic fatalities corresponding to each reason is different from the reported proportion or not. The null and alternative hypothesis can be defined as:

Null hypothesis:

H0 : The proportion is same as the reported proportion.

Alternative hypothesis:

H1 : The proportion is different from the reported proportion.

In the provided situation, the claim of the study will be the observed proportion corresponding to each reason is different from the reported proportion.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

To state: The hypothesis and the claim.

Answer 13

Answer to Problem 11.1.1RE

The null and alternative hypotheses are:

H0 : The proportion is same as the reported proportion.

H1 : The proportion is different from the reported proportion.

And the claim of the test is the observed proportion is different from the reported proportion.

Answer 14

Explanation of Solution

Given info:

The percentage of death and the observed count corresponding to each reason are provided in the question. The level of significance is α=0.05 .

Justification:

The researcher wants to know that whether the proportion of the traffic fatalities corresponding to each reason is different from the reported proportion or not. The null and alternative hypothesis can be defined as:

Null hypothesis:

H0 : The proportion is same as the reported proportion.

Alternative hypothesis:

H1 : The proportion is different from the reported proportion.

In the provided situation, the claim of the study will be the observed proportion corresponding to each reason is different from the reported proportion.

Answer 15

(b)

Expert Solution

To determine

The critical value.

Answer to Problem 11.1.1RE

The required critical value is 5.991.

Explanation of Solution

The required critical value is obtained from the provided chi-square table in the textbook. The number of categories is 3.

The degrees of freedom is calculated as:

Degrees of freedom=Number of categories−1=3−1=2

Therefore, the critical value at α=0.05 level of significance and 2 degrees of freedom is obtained as 5.991.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

The critical value.

Answer 20

Answer to Problem 11.1.1RE

The required critical value is 5.991.

Answer 21

Explanation of Solution

The required critical value is obtained from the provided chi-square table in the textbook. The number of categories is 3.

The degrees of freedom is calculated as:

Degrees of freedom=Number of categories−1=3−1=2

Therefore, the critical value at α=0.05 level of significance and 2 degrees of freedom is obtained as 5.991.

Answer 22

(c)

Expert Solution

To determine

The value of the test statistic.

Answer to Problem 11.1.1RE

The test statistic value is 1.819.

Explanation of Solution

Calculation:

Software procedure:

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

Enter the data in the Minitab worksheet.
Go to Stat> Tables> Chi-Square Goodness-of-Fit Test (one variable).
Specify the “Observed count”, choose the option “Proportions specified by historic count”, and specify the column where the percentage is written.
Click on OK.

Output using the MINITAB software is given below:

Elementary Statistics: A Step By Step Approach, Chapter 11, Problem 11.1.1RE

Therefore, the obtained value of the test statistic is 1.819.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

To determine

The value of the test statistic.

Answer 27

Answer to Problem 11.1.1RE

The test statistic value is 1.819.

Answer 28

Explanation of Solution

Calculation:

Software procedure:

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

Enter the data in the Minitab worksheet.
Go to Stat> Tables> Chi-Square Goodness-of-Fit Test (one variable).
Specify the “Observed count”, choose the option “Proportions specified by historic count”, and specify the column where the percentage is written.
Click on OK.

Output using the MINITAB software is given below:

Elementary Statistics: A Step By Step Approach, Chapter 11, Problem 11.1.1RE

Therefore, the obtained value of the test statistic is 1.819.

Answer 29

(d)

Expert Solution

To determine

To make: The decision.

Answer to Problem 11.1.1RE

The null hypothesis will not be rejected.

Explanation of Solution

The obtained value of the test statistic is 1.819 and the critical value is 5.991. As the obtained value of the chi-square statistic is less than the critical value, it can be said that there is not enough evidence to reject the null hypothesis at α=0.05 .

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

To determine

To make: The decision.

Answer 34

Answer to Problem 11.1.1RE

The null hypothesis will not be rejected.

Answer 35

Explanation of Solution

The obtained value of the test statistic is 1.819 and the critical value is 5.991. As the obtained value of the chi-square statistic is less than the critical value, it can be said that there is not enough evidence to reject the null hypothesis at α=0.05 .

Answer 36

(e)

Expert Solution

To determine

To summarize: The results.

Answer to Problem 11.1.1RE

According to the obtained result, the claim of the study is not true.

Explanation of Solution

The null hypothesis is not rejected. On the basis of the obtained result, it can be concluded that the result of the proportion of the traffic fatalities is not different from the reported proportion at α=0.05 . That is, the claim is not true.

Answer 37

(e)

Expert Solution

Answer 38

(e)

Expert Solution

Answer 39

Expert Solution

Answer 40

To determine

To summarize: The results.

Answer 41

Answer to Problem 11.1.1RE

According to the obtained result, the claim of the study is not true.

Answer 42

Explanation of Solution

The null hypothesis is not rejected. On the basis of the obtained result, it can be concluded that the result of the proportion of the traffic fatalities is not different from the reported proportion at α=0.05 . That is, the claim is not true.