A particular report included the following table classifying 713 fatal bicycle accidents according to time of day the accident occurred.
Time of Day Number of Accidents
Midnight to 3 a.m. 36
3 a.m. to 6 a.m. 29
6 a.m. to 9 a.m. 65
9 a.m. to Noon 78
Noon to 3 p.m. 98
3 p.m. to 6 p.m. 127
6 p.m. to 9 p.m. 165
9 p.m. to Midnight 115
(a) Assume it is reasonable to regard the 713 bicycle accidents summarized in the table as a random sample of fatal bicycle accidents in that year. Do these data support the hypothesis that fatal bicycle accidents are not equally likely to occur in each of the 3-hour time periods used to construct the table? Test the relevant hypotheses using a significance level of .05. (Round your χ2 value to two decimal places, and round your P-value to three decimal places.)
χ2 =
P-value =
What can you conclude?
There is sufficient evidence to reject H0. There is insufficient evidence to reject H0.
(b) Suppose a safety office proposes that bicycle fatalities are twice as likely to occur between noon and midnight as during midnight to noon and suggests the following hypothesis: H0: p1 = 1/3, p2 = 2/3, where p1 is the proportion of accidents occurring between midnight and noon and p2 is the proportion occurring between noon and midnight. Do the given data provide evidence against this hypothesis, or are the data consistent with it? Justify your answer with an appropriate test. (Hint: Use the data to construct a one-way table with just two time categories. Use α = 0.05. Round your χ2 value to two decimal places, and round your P-value to three decimal places.)
χ2 =
P-value =
What can you conclude?
There is sufficient evidence to reject H0. There is insufficient evidence to reject H0.
For part (a), to test the hypothesis that fatal bicycle accidents are equally likely to occur in each 3-hour time period, we will use the chi-square test of goodness-of-fit.
Step 1: State the null and alternative hypotheses:
H0: Fatal bicycle accidents are equally likely to occur in each 3-hour time period.
Ha: Fatal bicycle accidents are not equally likely to occur in each 3-hour time period.
Step 2: Calculate the expected frequencies:
To calculate the expected frequencies, we need to assume that the null hypothesis is true, meaning that each time period has an equal probability of an accident. Since there are eight time periods, the expected frequency for each time period is 713/8 = 89.125.
Step 3: Calculate the chi-square test statistic:
The chi-square test statistic is calculated using the formula:
χ2 = ∑ (Observed frequency - Expected frequency)^2 / Expected frequency
Calculating the chi-square test statistic for the given data:
χ2 = [(36-89.125)^2/89.125] + [(29-89.125)^2/89.125] + ... + [(115-89.125)^2/89.125]
Step 4: Calculate the degrees of freedom:
The degrees of freedom is calculated as (number of categories - 1). In this case, there are 8 time periods, so df = 8 - 1 = 7.
Step 5: Determine the critical value:
Using a significance level of 0.05 and the degrees of freedom calculated in Step 4, consult a chi-square distribution table or use a calculator to find the critical value. In this case, the critical value is 14.067.
Step 6: Calculate the p-value:
Using the chi-square distribution table or a calculator, find the p-value associated with the calculated chi-square test statistic.
Step 7: Compare the p-value to the significance level:
If the p-value is less than the significance level (0.05), we reject the null hypothesis. If the p-value is greater than or equal to the significance level, we fail to reject the null hypothesis.
For part (b), to test the new hypothesis proposed by the safety office, we will use the chi-square test of independence.
Step 1: State the null and alternative hypotheses:
H0: p1 = 1/3, p2 = 2/3 (The proportion of accidents occurring between midnight and noon is 1/3, and the proportion occurring between noon and midnight is 2/3).
Ha: The proportions are not equal.
Step 2: Construct a one-way table:
Combining the time periods from midnight to noon and from noon to midnight, we have two categories: midnight to noon (65 + 78 + 98 + 36 = 277) and noon to midnight (127 + 165 + 115 + 29 = 436).
Step 3: Calculate the expected frequencies:
Under the null hypothesis, we assume the proportions are 1/3 and 2/3. Therefore, the expected frequencies are 713 * (1/3) = 237.67 for midnight to noon and 713 * (2/3) = 475.33 for noon to midnight.
Step 4: Calculate the chi-square test statistic:
The chi-square test statistic is calculated using the formula:
χ2 = ∑ (Observed frequency - Expected frequency)^2 / Expected frequency
Calculating the chi-square test statistic for the given data:
χ2 = [(277-237.67)^2/237.67] + [(436-475.33)^2/475.33]
Step 5: Calculate the degrees of freedom:
The degrees of freedom is calculated as (number of categories - 1). In this case, there are 2 categories, so df = 2 - 1 = 1.
Step 6: Determine the critical value:
Using a significance level of 0.05 and the degrees of freedom calculated in Step 5, consult a chi-square distribution table or use a calculator to find the critical value. In this case, the critical value is 3.841.
Step 7: Calculate the p-value:
Using the chi-square distribution table or a calculator, find the p-value associated with the calculated chi-square test statistic.
Step 8: Compare the p-value to the significance level:
If the p-value is less than the significance level (0.05), we reject the null hypothesis. If the p-value is greater than or equal to the significance level, we fail to reject the null hypothesis.
After performing the calculations and comparing the p-values to the significance level, based on the results, we can conclude whether there is sufficient evidence to reject the null hypothesis or not.