INTRODUCTION TO STATISTICAL THINKING Directions: Complete the following question

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INTRODUCTION TO STATISTICAL THINKING

Directions: Complete the following questions. The most important part of statistics is the thought process, so make sure that you explain your answers, but be careful with statistics. The following statistics/probability problems may intrigue you and you may be surprised. The answers are not always as you might think. Please answer them as well as you can by using common logic.

1. There are 23 people at a party. Explain what the probability is that any two of them share the same birthday.

2. A cold and flu study is looking at how two different medications work on sore throats and fever. Results are as follows:

Sore throat - Medication A: Success rate - 90% (101 out of 112 trials were successful)

Sore throat - Medication B: Success rate - 83% (252 out of 305 trials were successful)

Fever - Medication A: Success rate - 71% (205 out of 288 trials were successful)

Fever - Medication B: Success rate - 68% (65 out of 95 trials were successful)

Analyze the data and explain which one would be the better medication for both a sore throat and a fever.

3. The United States employed a statistician to examine damaged planes returning from bombing missions over Germany in World War II. He found that the number of returned planes that had damage to the fuselage was far greater than those that had damage to the engines. His recommendation was to enhance the reinforcement of the engines rather than the fuselages. If damage to the fuselage was far more common, explain why he made this recommendation.

Please answer all three questions please.

Explanation / Answer

1.The birthday problem or birthday paradox[1] concerns the probability that, in a set of randomly chosen people, some pair of them will have the samebirthday. By the pigeonhole principle, the probability reaches 100% when the number of people reaches 367 (since there are only 366 possible birthdays, including February 29). However, 99.9% probability is reached with just 70 people, and 50% probability with 23 people. These conclusions are based on the assumption that each day of the year (except February 29) is equally probable for a birthday.

The birthday problem is to find the probability that, in a group of N people, there is at least one pair of people who have the same birthday. See "Same birthday as you" further for an analysis of the case of finding the probability of a given, fixed person having the same birthday as any of the remaining ( [N - 1] ) people.Hence total chances are: [22+21+20+cdots+1 = 253] , so comparing every person to all of the others allows 253 distinct chances (combinations): in a group of 23 people there are [ extstyle {23 choose 2} = rac{23 cdot 22}{2} = 253] distinct possible combinations of pairing.

Presuming all birthdays are equally probable,[5] the probability of a given birthday for a person chosen from the entire population at random is 1/365 (ignoring February 29). Although the number of pairings in a group of 23 people is not statistically equivalent to 253 pairs chosen independently, the birthday problem becomes less surprising if a group is thought of in terms of the number of possible pairs, rather than as the number of individuals.As far as abstraction proof Let A be the statement "Everybody in the set [mathcal{S}] has a unique birthday" (so P(A') is what we are actually looking for). By definition, P(A) is the fraction of injective functions out of all possible functions (i.e., the probability of the birthday function being one that assigns only one person to each birthdate.

2. If we combine the studies medication A had 306 successes out of 400.

B had 317 successes out of 400.

So if we want to decide which had a better "overall" success rate it would be B.

This seems counter-intuitive with the individual test data and is an example of what is called in statistics "Simpson's Paradox."

3.His recommendation very simply was to armor plate the unhit areas that the returning planes had in common. When he surveyed the undersides of these planes, he noticed that there were a few spots that all of them had in common that had no bullet holes. And he had to assume that the ones that hadn't returned had bullet holes in those locations. They were in the English Channel someplace. Reader Mike Cox wrote us that the mathematician was the famous statistician Abraham Wald and gave the following references for Wald's work: "A Method of Estimating Plane Vulnerability Based on Damage of Survivors, Abraham Wald, Center for Naval Analyses, 423, July 1980. and "Abraham Wald's Work on Aircraft Survivability, Marc Mangel;Fancisco J. Samaniego, Journal of the American Statistical Association,Vol. 79, No. 386, pp. 259-267

Asumes that we have the following data for planes participating in combat:

(1) The total number N of planes participating in combat.

(2) For any integer i (i = 0,1,2,...) the number of planes that received exactly hits but have not been downed, i.e., have returned from combat.

The data in (2) is an example of missing data since the corresponding information is not known for the planes that do not return.

Wald assumes that the probability that a plane will be shot down does not depend on the number of previous non-destructive hits. This allows him to compare the vulnerability of these areas. He illustrating this by the example.

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