Question

A traffic safety company publishes reports about motorcycle fatalities and helmet use. In the first accompanying data​ table, the distribution shows the proportion of fatalities by location of injury for motorcycle accidents. The second data table shows the location of injury and fatalities for 2061 riders not wearing a helmet.

Location of Injury   Probability    Injuries/fatalities not wearing helmet
Multiple Locations   0.570   1030
Head   0.310   863
Neck   0.030   40
Thorax   0.060   82
Abdomen/Lumbar/Spine   0.030   46

(a) Does the distribution of fatal injuries for riders not wearing a helmet follow the distribution for all​ riders? Use α=0.10 level of significance. What are the null and alternative​ hypotheses?

A.

H0​: The distribution of fatal injuries for riders not wearing a helmet follows the same distribution for all other riders.

H1​: The distribution of fatal injuries for riders not wearing a helmet does not follow the same distribution for all other riders.

B.

H0​: The distribution of fatal injuries for riders not wearing a helmet does not follow the same distribution for all other riders.

H1​: The distribution of fatal injuries for riders not wearing a helmet does follow the same distribution for all other riders.

C.

None of these.

Answer 1

Solution:-

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

A)

H₀​: The distribution of fatal injuries for riders not wearing a helmet follows the same distribution for all other riders.

H₁​: The distribution of fatal injuries for riders not wearing a helmet does not follow the same distribution for all other riders.

Formulate an analysis plan. For this analysis, the significance level is 0.10. Using sample data, we will conduct a chi-square goodness of fit test of the null hypothesis.

Analyze sample data. Applying the chi-square goodness of fit test 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 = k - 1 = 5 - 1
D.F = 4
(E_i) = n * p_i

X² = 122.23

where DF is the degrees of freedom, k is the number of levels of the categorical variable, n is the number of observations in the sample, E_i is the expected frequency count for level i, O_i is the observed frequency count for level i, and X² is the chi-square test statistic.

The P-value is the probability that a chi-square statistic having 4 degrees of freedom is more extreme than 122.23.

We use the Chi-Square Distribution Calculator to find P(X² > 122.23) = less than 0.0001

Interpret results. Since the P-value (almost 0) is less than the significance level (0.10), we have to reject the null hypothesis.