A researcher wanted to determine whether certain accidents were uniformly distributed over the days of the week. The data show the day of the week for n=309 randomly selected accidents. Is there reason to believe that the accident occurs with equal frequency with respect to the day of the week at the a= 0.05 level of significance? Click the icon to view the table. H₁: P₁ P₂ == P7=7 1 Ho: P1 P2 == P7 = 7 H₁: At least one proportion is different from the others. Compute the expected counts for day of the week. Day of the Week Observed Count Sunday 44 Expected Count 44.14 Monday 43 44.14 28 44.14 Tuesday Wednesday 41 - X 44.14 Distribution of accidents 43 44.14 Thursday Friday Saturday 50 44.14 60 44.14 (Round to two decimal places as needed.) Day of the Sunday Monday Tuesday Wednesday Thursday Friday Saturday Week Frequency 44 43 28 41 43 50 60 What is the test statistic? x = 12.660 (Round to three decimal places as needed.) What is the P-value of the test? Print Done P-value = 0.049 (Round to three decimal places as needed.) Based on the results, do the accidents follow a uniform distribution? A. Do not reject Ho, because the calculated P-value is greater than the given a level of significance. O B. Do not reject Ho, because the calculated P-value is less than the given a level of significance. c. Reject Ho. because the calculated P-value is less than the given a level of significance. OD. Reject H. because the calculated P-value is greater than the given a level of significance. D.
A researcher wanted to determine whether certain accidents were uniformly distributed over the days of the week. The data show the day of the week for n=309 randomly selected accidents. Is there reason to believe that the accident occurs with equal frequency with respect to the day of the week at the a= 0.05 level of significance? Click the icon to view the table. H₁: P₁ P₂ == P7=7 1 Ho: P1 P2 == P7 = 7 H₁: At least one proportion is different from the others. Compute the expected counts for day of the week. Day of the Week Observed Count Sunday 44 Expected Count 44.14 Monday 43 44.14 28 44.14 Tuesday Wednesday 41 - X 44.14 Distribution of accidents 43 44.14 Thursday Friday Saturday 50 44.14 60 44.14 (Round to two decimal places as needed.) Day of the Sunday Monday Tuesday Wednesday Thursday Friday Saturday Week Frequency 44 43 28 41 43 50 60 What is the test statistic? x = 12.660 (Round to three decimal places as needed.) What is the P-value of the test? Print Done P-value = 0.049 (Round to three decimal places as needed.) Based on the results, do the accidents follow a uniform distribution? A. Do not reject Ho, because the calculated P-value is greater than the given a level of significance. O B. Do not reject Ho, because the calculated P-value is less than the given a level of significance. c. Reject Ho. because the calculated P-value is less than the given a level of significance. OD. Reject H. because the calculated P-value is greater than the given a level of significance. D.
MATLAB: An Introduction with Applications
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Author:Amos Gilat
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How would you solve for the test statistic and p-value in this case? Thank you.
![# Distribution of Accidents Over the Days of the Week
A researcher wanted to determine whether certain accidents were uniformly distributed over the days of the week. The data show the day of the week for \(n = 309\) randomly selected accidents. Is there reason to believe that the accident occurs with equal frequency with respect to the day of the week at the \(\alpha = 0.05\) level of significance?
## Hypothesis
- **Null Hypothesis (\(H_0\))**: \(p_1 = p_2 = ... = p_7 = \frac{1}{7}\)
- **Alternative Hypothesis (\(H_1\))**: At least one proportion is different from the others.
## Observed and Expected Counts
To compute the expected counts for each day of the week:
| Day of the Week | Observed Count | Expected Count |
|-----------------|-----------------|----------------|
| Sunday | 44 | 44.14 |
| Monday | 43 | 44.14 |
| Tuesday | 28 | 44.14 |
| Wednesday | 41 | 44.14 |
| Thursday | 43 | 44.14 |
| Friday | 50 | 44.14 |
| Saturday | 60 | 44.14 |
*(All expected counts are rounded to two decimal places as needed.)*
## Statistical Test
To determine whether to reject the null hypothesis, we use the Chi-square test statistic:
\[ \chi^2_0 = 12.660 \]
*(Round to three decimal places as needed.)*
## P-value
The P-value of the test is:
\[ \text{P-value} = 0.049 \]
*(Round to three decimal places as needed.)*
## Conclusion
Based on the results, do the accidents follow a uniform distribution?
### Decision Rule
- If the P-value is less than the given \(\alpha\) level of significance, reject \(H_0\).
- If the P-value is greater than the given \(\alpha\) level of significance, do not reject \(H_0\).
Given that the P-value of 0.049 is less than the \(\alpha\) level of significance of 0.05:
- **Reject \(H_0\)** because the calculated P-value is](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcbaaf3a5-97a4-44d4-82b4-c685cbb2fa1a%2F36e6a182-6704-4ffc-add4-df92f3a1bb17%2F0unovkp_processed.png&w=3840&q=75)
Transcribed Image Text:# Distribution of Accidents Over the Days of the Week
A researcher wanted to determine whether certain accidents were uniformly distributed over the days of the week. The data show the day of the week for \(n = 309\) randomly selected accidents. Is there reason to believe that the accident occurs with equal frequency with respect to the day of the week at the \(\alpha = 0.05\) level of significance?
## Hypothesis
- **Null Hypothesis (\(H_0\))**: \(p_1 = p_2 = ... = p_7 = \frac{1}{7}\)
- **Alternative Hypothesis (\(H_1\))**: At least one proportion is different from the others.
## Observed and Expected Counts
To compute the expected counts for each day of the week:
| Day of the Week | Observed Count | Expected Count |
|-----------------|-----------------|----------------|
| Sunday | 44 | 44.14 |
| Monday | 43 | 44.14 |
| Tuesday | 28 | 44.14 |
| Wednesday | 41 | 44.14 |
| Thursday | 43 | 44.14 |
| Friday | 50 | 44.14 |
| Saturday | 60 | 44.14 |
*(All expected counts are rounded to two decimal places as needed.)*
## Statistical Test
To determine whether to reject the null hypothesis, we use the Chi-square test statistic:
\[ \chi^2_0 = 12.660 \]
*(Round to three decimal places as needed.)*
## P-value
The P-value of the test is:
\[ \text{P-value} = 0.049 \]
*(Round to three decimal places as needed.)*
## Conclusion
Based on the results, do the accidents follow a uniform distribution?
### Decision Rule
- If the P-value is less than the given \(\alpha\) level of significance, reject \(H_0\).
- If the P-value is greater than the given \(\alpha\) level of significance, do not reject \(H_0\).
Given that the P-value of 0.049 is less than the \(\alpha\) level of significance of 0.05:
- **Reject \(H_0\)** because the calculated P-value is
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