A rare type of heredity change causes the bacterium in E. coli to become resistant to the drug strepto- mycin. This type of change, called mutation, can be detected by plating many bacteria on petri dishes containing an antibiotic medium. Any colonies that grow on this medium result from a single mutant cell. A sample of n 200 petri dishes of streptomycin agar were each plated with 106 bacteria, and the numbers of colonies were counted on each dish. The observed results were that 110 dishes had 0 colonies, 61 had 1, 17 had 2, 9 had 3, 3 dishes had 4 colonies, and no dishes had more than 4 colonies. Let X equal the number of colonies per dish. Test the hypothesis that X has a Poisson distribution. (a) Compute i as an estimate of λ. (b) Set up classes (categories) for the a-values so that the expected number of observations in each class is at least 5 (using z as an estimate of λ) (c) Using the classes from (b), compute the test statistic and give the critical region at a = 0.01

(a) The sample mean can be used as an estimate of λ: 0.95.

(b) The expected number of observations in each class are

Class 0: 18.2

Class 1: 86.5

Class 2: 163.8

Class 3 or more:  31.5

(c) The distribution of X is not Poisson because we reject the null hypothesis that X has a Poisson distribution with parameter λ = 0.95.

(a) The sample mean can be used as an estimate of λ:

i = (110×0 + 61×1 + 17×2 + 9×3 + 3×4) / 200 = 0.95

(b) We can use the Poisson distribution to estimate the expected number of observations in each class. Let z = i = 0.95 be the estimated value of λ. Then the classes can be set up as follows:

Class 0: X = 0

Class 1: X = 1

Class 2: X = 2

Class 3 or more: X ≥ 3

Using the Poisson distribution, we can calculate the expected number of observations in each class:

Class 0: P(X=0; λ=z) × n = e^(-z) × z^0 / 0! × 200 = 18.2

Class 1: P(X=1; λ=z) × n = e^(-z) × z^1 / 1! × 200 = 86.5

Class 2: P(X=2; λ=z) × n = e^(-z) × z^2 / 2! × 200 = 163.8

Class 3 or more: P(X≥3; λ=z) × n = 1 - P(X=0; λ=z) - P(X=1; λ=z) - P(X=2; λ=z) = 31.5

(c) To test the hypothesis that X has a Poisson distribution with parameter λ = 0.95, we can use the chi-squared goodness-of-fit test. The test statistic is given by:

χ^2 = Σ (Oi - Ei)^2 / Ei

where Oi is the observed frequency in the i-th class and Ei is the expected frequency in the i-th class. Using the classes from (b), we can calculate the test statistic:

χ^2 = [(110-18.2)^2 / 18.2] + [(61-86.5)^2 / 86.5] + [(17-163.8)^2 / 163.8] + [(9-31.5)^2 / 31.5] = 137.52

The critical value of chi-squared for 3 degrees of freedom and a significance level of 0.01 is 11.345. Since the calculated test statistic (137.52) is greater than the critical value (11.345), we reject the null hypothesis that X has a Poisson distribution with parameter λ = 0.95. Therefore, there is evidence that the distribution of X is not Poisson.

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