A traffic study predicted that accidents would occur at four-way, Y- and three-way intersections in a 3:2:1 ratio. Among 200 accidents at intersections with equal traffic flow, the observed counts were 116, 60 and 24. Carry out a chi-square test of the prediction.

Step-by-step solution

1. 1

   Pick the right test. There is a single categorical variable (intersection type) and a theoretical distribution to compare against, not two variables cross-tabulated. That makes this a chi-square goodness-of-fit test, not a test of independence. The distinction matters because it fixes the degrees of freedom as $k-1$ rather than $(r-1)(c-1)$.

2. 2

   Build the expected counts from the ratio. The total is $116+60+24=200$, and $3:2:1$ splits a whole into sixths:

   $E_1=200\cdot\tfrac{3}{6}=100,\qquad E_2=200\cdot\tfrac{2}{6}=66.\overline{6},\qquad E_3=200\cdot\tfrac{1}{6}=33.\overline{3}.$

   All expected counts exceed $5$, so the chi-square approximation is safe. Keep the exact fractions $\tfrac{200}{3}$ and $\tfrac{100}{3}$ rather than the rounded $66.67$ and $33.33$, or rounding error will creep into the statistic.

3. 3

   Compute each cell's contribution to $\chi^2=\sum\dfrac{(O-E)^2}{E}$.

   $\frac{(116-100)^2}{100}=\frac{256}{100}=2.56,\qquad \frac{\left(60-\tfrac{200}{3}\right)^2}{\tfrac{200}{3}}=\frac{400/9}{200/3}=\frac{2}{3},\qquad \frac{\left(24-\tfrac{100}{3}\right)^2}{\tfrac{100}{3}}=\frac{784/9}{100/3}=\frac{196}{75}.$

4. 4

   Add them up. Over the common denominator $75$:

   $\chi^2=\frac{192}{75}+\frac{50}{75}+\frac{196}{75}=\frac{438}{75}=\frac{146}{25}=5.84.$

   The value is exactly $5.84$, not merely rounded to it. Notice the three-way category, with the smallest expected count, contributes the most ($2.613$) even though its raw shortfall is smaller than the four-way surplus - that is the $1/E$ weighting at work.

5. 5

   Compare with the critical value. With $k=3$ categories, $df=k-1=2$. The critical value at $\alpha=0.05$ is $\chi^2_{0.05,2}=5.991$. For $df=2$ the $p$-value has the closed form $p=e^{-\chi^2/2}=e^{-2.92}\approx 0.054$.

6. 6

   State the conclusion. Since $5.84<5.991$ (equivalently $p\approx 0.054>0.05$), we fail to reject the null hypothesis. The observed counts are consistent with the predicted $3:2:1$ ratio - though only barely, so a larger sample would be worth collecting.