In an effort to predict customer arrivals better, the Avalon grocery store count
ID: 3152862 • Letter: I
Question
In an effort to predict customer arrivals better, the Avalon grocery store counted the number of customers who arrived at the store during randomly selected 10-minute intervals. The following table shows these data. For example, there were 12 10-minute intervals in which no customers arrived, and 27 10-minute intervals in which one customer arrived. Using a = 0.05, perform a chi-square test to determine if the number of customers arriving over a 10-minute interval follows the Poisson probability distribution. Determine the p-value using Excel and interpret its meaning.Explanation / Answer
H0 : Distribution is poisson
H1 : Distribution is not poisson
For each row in the table we next calculate the probability of x (where x = count per minute) for x = 0 through 6 using the Poisson pdf, i.e. f(x) = POISSON(x, 2.2, FALSE), and then multiply this probability by 110 to get the expected number of hits per interval assuming the null hypothesis is true (column D).
we can’t use CHITEST since df the sample size minus 1. In fact, df = k – m – 1 = 7 – 1 – 1 = 5 since there are 7 intervals (k) and the Poisson distribution has 1 unidentified parameter (m), namely the mean. We therefore calculate the chi-square test statistic (in column F) to be 0.638. We next calculate the following:
p-value = CHIDIST(2, df) = CHIDIST(0.638,5) = .986 > .05 =
Based on either of the above inequalities, we retain the null hypothesis, and so with 95% confidence conclude that the observed data follow a Poisson distribution.
(b)
The pvalue = 0.98 < 0.05. Hence the do not reject the null. Thus the distribution can be assumed to be a poisson.
No. of customers frequency Customers x frequency Poisson (0-E)^2/E 0 12 0 12.1883474199 0.0029105464 1 27 27 26.8143643237 0.0012851546 2 29 58 29.4958007561 0.0083340131 3 22 66 21.6302538878 0.0063204153 4 12 48 11.8966396383 0.0008980153 5 5 25 5.2345214408 0.0105072272 6 3 18 1.9193245283 0.6084742095 Total 110 242 Chi-squared 0.6387295814 mean 2.2 P-value 0.9861614017Related Questions
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