For the week of February 6, 2010, the CDC reported a total of 68 cases of mumps,
ID: 3175734 • Letter: F
Question
For the week of February 6, 2010, the CDC reported a total of 68 cases of mumps, whereas the historical weekly average was only 11 cases, based on data from the past five years.
(a) Assuming the cases of mumps are random and independent, the number X of weekly mumps cases in the United States has approximately what distribution?
(b) Based on this model, find the probabilities: P( X 15 ), P( X 30 ), and P( X 67 ).
What is the probability that there would be 68 or more cases of mumps in one week in the United States?
(c) Does the data from February 2010 support the idea that the mumps cases observed at the time are random and independent with a mean of 11 cases a week? Explain your answer.
Explanation / Answer
Solution
Back-up Theory
Let X = number of mumps cases in a week. Then,
X ~ Poisson(), where = mean (average) number of mumps cases in a week
If a random variable X ~ Poisson(), i.e., X has Poisson Distribution with mean then
probability mass function (pmf) of X is given by P(X = x) = e – .x/(x!) …………..(1)
where x = 0, 1, 2, ……. ,
Values of p(x) for various values of and x can be obtained by using Excel Function.
Now, to work out the solution,
‘historical weekly average was only 11 cases, based on data from the past five years’ should be taken to infer = 11 per week
Part (a)
The number, X, of weekly mumps cases in the United States has Poisson distribution with parameter = 11 per week.
ANSWER
Part (b) [using Excel Function]
P( X 15 ) = 0.907396
P( X 30 ) = 0.999999
P( X 67 ) = 1.000000
Probability that there would be 68 or more cases of mumps in one week in the United States
= P(X 68) = 1 - P( X 67 ) = 1 - 1.000000 [from earlier Answer]
= 0
ANSWERS in bold.
Part (c)
To answer this question, we need to test the null hypothesis, H0: = 11 Vs HA: > 11.
The test statistic is: Z = (observed number of cases – 11)/11 = (68 - 11)/11 = 17.18675
Under H0, Z has Standard Normal Distribution, by virtue of Central Limit Theorem.
We will take 5% level of significance.
Upper 5% point of Standard Normal Distribution is 1.645.
Since the calculated value of the statistic > 1.645, H0 is rejected.
=> there is no evidence to suggest that the data from February 2010 support the idea that the mumps cases observed at the time are random and independent with a mean of 11 cases a week. ANSWER