Sami Schmitt believes that the number of cars arriving at his Scrub and Shine Car Wash follows a Poisson distribution.He collected a random sample and constructed the following frequency distribution to test his hypothesis. $$\begin{array} { | l | l | l | l | l | l | l | }
\hline \text { Cars per 15-rrirute iriterval } & 0 & 1 & 2 & 3 & 4 & \geq 5 \
\hline \text { Observed frequency } & 5 & 15&17 & 12& 10 & 8 \
\hline
\end{array}$$

Using$\alpha$ = 0.05, the critical value of chi-square for this goodness-of-fit test is ____.

Question

Sami Schmitt believes that the number of cars arriving at his Scrub and Shine Car Wash follows a Poisson distribution.He collected a random sample and constructed the following frequency distribution to test his hypothesis. $$\begin{array} { | l | l | l | l | l | l | l | } 
\hline \text { Cars per 15-rrirute iriterval } & 0 & 1 & 2 & 3 & 4 & \geq 5 \
\hline \text { Observed frequency } &  5 & 15&17 & 12& 10 & 8 \
\hline
\end{array}$$

Using$\alpha$ = 0.05, the critical value of chi-square for this goodness-of-fit test is ____.

Quizplus · Accepted Answer

The degrees of freedom (df) for a chi-square goodness-of-fit test are calculated as the number of categories minus 1. However, when the expected frequency in any category is less than 5, categories are combined to meet this requirement. In this case, there are 6 categories (0, 1, 2, 3, 4, ≥5 cars per 15-minute interval). Since the last category is already a combination of all occurrences of 5 or more cars, no further combination is needed based on the information provided. Thus, df = 6 - 1 = 5. For a significance level (alpha) of 0.05 and 5 degrees of freedom, the critical value of chi-square is 11.07, not 9.49. My initial response was incorrect; the correct critical value for df = 5 at alpha = 0.05 is indeed 11.07, not 9.49.

Sami Schmitt Believes That the Number of Cars Arriving at His