Question

You work for an insurance company and are studying the relationship between types of crashes and the vehicle involved. As a part of your study, you randomly selected 3571 vehicle crashes and organize the resulting data as shown in contingency table. At 0.10 can you conclude that the type of crash depends on the type of vehicle?

Type of crash. Car. Pickup. Spot utility

Single vehicle 840. 307. 327

Multiple vehicle 1177 486. 434

Caculate the degrees of freedom

d.f. =

Find the critical value =

Caculate the test statistic, if convient, use technology

Answer 1

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

H₀: Crash and the type of vehicle are independent.
H_a: Crash and the type of vehicle are not independent.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a chi-square test for independence.

Analyze sample data. Applying the chi-square test for independence to sample data, we compute the degrees of freedom, the expected frequency counts, and the chi-square test statistic. Based on the chi-square statistic and the degrees of freedom, we determine the P-value.

DF = (r - 1) * (c - 1) = (2 - 1) * (3 - 1)

D.F = 2
E_r,c = (n_r * n_c) / n


Χ² = 3.162

Χ²_Critical = 4.607.

where DF is the degrees of freedom.

The P-value is the probability that a chi-square statistic having 2 degrees of freedom is more extreme than 4.607.

We use the Chi-Square Distribution Calculator to find P(Χ² > 3.162) = 0.367.

Interpret results. Since the P-value (0.367) is greater than the significance level (0.10), we cannot accept the null hypothesis.

Thus, we conclude that there is no relationship between crash and the type of vehicle.