91Ó°ÊÓ

The report "Fatality Facts 2004: Bicydes" (Insurance Institute, 2004) included the following table classifying 715 fatal bicycle accidents according to time of day the accident occurred. $$ \begin{array}{lc} \text { Time of Day } & \text { Number of Accidents } \\ \hline \text { Midnight to } 3 \text { A.M. } & 38 \\ \text { 3 A.M. to } 6 \text { A.M. } & 29 \\ \text { 6 A.M. to } 9 \text { A.M. } & 66 \\ \text { 9A.M. to Noon } & 77 \\ \text { Noon to } 3 \text { P.M. } & 99 \\ \text { 3 P.M. to } 6 \text { r.M. } & 127 \\ \text { 6 P.M. to 9 p.M. } & 166 \\ \text { 9 P.M. to Midnight } & 113 \\ \hline \end{array} $$ a. Assume it is reasonable to regard the 715 bicycle accidents summarized in the table as a random sample of fatal bicycle accidents in 2004 . 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 \(.05\). 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: \(H_{0}: p_{1}=1 / 3, p_{2}=2 / 3\), where \(p_{1}\) is the proportion of accidents occurring between midnight and noon and \(p_{2}\) 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.)

Step by step solution

01

Understanding the Data

First, study the given data that categorizes 715 fatal bike accidents according to the time of day. Divide total accidents of each time period by the total number of accidents to get the observed proportions.

02

Performing the Hypothesis Test (Part a)

Now, we'll test the assumption of equal likelihood of accidents to happen during every time period. The null hypothesis \( H_{0} \) is: all time intervals have equal probabilities. The alternative hypothesis \( H_{A} \) is: not all time intervals have equal probabilities. Use the Chi-square test to test these hypotheses.

03

Calculating the Chi-square Statistic

Compute the Chi-square test statistic using the formula: \[ \chi^{2} = \sum \frac {(O_i - E_i)^{2}} {E_i} \]Where \( O_i \) refers to observed frequency and \( E_i \) refers to expected frequency. Find these values and compute the test statistic.

04

Checking the Test Statistic Against the Chi-square Distribution

Next, determine the critical value of \( \chi^{2} \) for 7 degrees of freedom (‘number of time periods’ - 1 = 8 - 1 = 7) at a significance level of .05. Compare the test statistic to this value.

05

Performing the Hypothesis Test (Part b)

The safety officer's hypothesis is that the proportion of accidents from midnight to noon, \(p_{1}\), equals 1/3, and the proportion from noon to midnight, \(p_{2}\), equals 2/3. Consider the null hypothesis \( H_{0} \) as \(p_{1} = 1 / 3, p_{2} = 2 / 3\), and the alternative hypothesis \( H_{A} \) is: \( p_{1} \neq 1 / 3, p_{2} \neq 2 / 3 \). We'll use a one-sample z-test to test these hypothesis.

06

Calculating the Test Statistic for the Second Test

Calculate the test statistic for the one-sample z-test using the formula:\[ Z = \frac {(\hat{p} - p_0)}{\sqrt{\frac {p_0 * (1 - p_0)}{n}}} \]Where \( p_0 \) is the proportion under the null hypothesis, \( \hat{p} \) is the sample proportion, and \( n \) is the sample size. Compute the test statistic for each part of the day.

07

Checking the Test Statistic Against the Normal Distribution

Determine z(critical) for the significance level of .05 (which equals 1.96 for a one-tail test) and compare it to the calculated test statistic.

08

Drawing the Conclusion

Depending on the comparisons between the test statistics and the critical values in two tests, draw a conclusion about each hypothesis test.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chi-square test

The Chi-square test is a statistical method used to determine if there is a significant difference between observed and expected frequencies in one or more categories of a contingency table. In the context of the exercise, this test is applied to understand whether fatal bicycle accidents occur with equal frequency during different times of the day.

To perform the test, we calculate the chi-square statistic, which quantifies the discrepancies between observed frequencies (actual number of accidents in each time period) and expected frequencies (what we would expect if accidents were evenly distributed across all time periods). The formula to compute the chi-square statistic is: \[ \chi^{2} = \sum \frac {(O_i - E_i)^{2}} {E_i} \]
where \( O_i \) is the observed frequency and \( E_i \) is the expected frequency for each category. A high chi-square statistic indicates a higher likelihood that the observed distribution does not match the expected one under the null hypothesis of equal likelihood across time periods.

Test statistic

A test statistic is a value computed from sample data, which is used in hypothesis testing. It helps to determine whether to reject the null hypothesis. In both Chi-square and z-test, the test statistic serves a similar purpose but is calculated differently based on the type of data and distribution involved.

For the Chi-square test, as mentioned previously, the test statistic is calculated using the sum of squared differences between observed and expected frequencies. In case of a z-test, which is used when the sample size is large enough to approximate a normal distribution, the test statistic is given by:\[ Z = \frac {(\hat{p} - p_0)}{\sqrt{\frac {p_0 * (1 - p_0)}{n}}} \]
In this exercise, the test statistic for the z-test helps to evaluate the safety officer’s claim about the proportion of accidents in the two halves of the day.

Significance level

The significance level, typically represented by \( \alpha \), is a threshold used to decide whether the observed data is statistically significant. It represents the probability of rejecting the null hypothesis when it is actually true – known as a Type I error.

In most social and behavioral science research, the significance level is set at 0.05, implying that there is a 5% probability of committing a Type I error. During hypothesis testing, if the p-value (the probability of observing the test statistic as extreme as, or more extreme than, the one calculated from the sample data) is less than the significance level, we reject the null hypothesis, asserting that the sample provides enough evidence against it.

Z-test

A z-test is used to determine whether there is a significant difference between sample observations and a known population parameter. It’s appropriate when we can assume that the sample is normally distributed and either the population variance is known, or the sample size is large (typically n>30).

In the exercise, the z-test is chosen to assess the validity of the safety officer's hypothesis concerning the proportions of bicycle accidents occurring at different halves of the day. We compute the z-test statistic using the formula provided and check it against the standard normal distribution. Comparing the computed z value to the critical value corresponding to our chosen significance level allows us to conclude whether the safety officer's hypothesis should be accepted or rejected.