Problem 2) Your friend wants to play a game with you involving a die, which she

Question

Problem 2)

Your friend wants to play a game with you involving a die, which she claims is fair. Assume that her claim is correct.

a) Consider a sample proportion of even faces in 100 flips. What is the mean, standard deviation and distribution of this sample proportion?

b) How would your answers to a) changed if the sample proportion were based on 25 flips of a fair die? Explain

c) What is the probability that the sample proportion of even faces in 100 flips of a die will be between 0.45 and 0.55? Suppose that in 100 flips of die, 38 resulted in an even face.

d) Construct a 98% confidence interval for the probability of a die coming up even face and interpret it .Based on your confidence interval is your friend’s claim correct? If it isn’t how is the die biased?

e) What is the margin of error of the confidence interval in a)?

f) What sample size would you need to obtain a 98% confidence interval for the probability of a die coming up even face with the margin of error of not more that 0.05?

Explanation / Answer

(a) In a sample proportion of even faces in a fair die. Assuming the die has 6 faces and 3 of them are even faces then the probabilty of getting an even face is 0.5 in every flip.

So in 100 flips there will be 50 flips of even faces, the mean would therefore be mean of 2 ,4,ad 6 = 4 and the std. deviation = 2, the distribution followed would be a normal distribution

(b) Our answer will not change if the sample proprtion were based on 25 flips as the die is an fair die and the mean, std. deviation and the distribution would all remain the same

(c) Now if 38 result in an even face out of 100 flips then P0= 0.38 and Pbar = 0.45

z1=(0.45-0.38)/SQRT(0.38*(1-0.38)/100) = 1.44 and z2 = (0.55-0.38)/SQRT(0.38*(1-0.38)/100) = 3.5

The probabiltiy that the sample proprtionof even faces lie between 0.45 and 0.55 would be the area under the curve on the std. normal distribution between z1 and z2 which can found by referring the corresponding z dist table

(d) 98% C.I means an alpha = 0.02, so the value of z has to be found such that the area to the right of z equals alpha/2 = 0.02/2 = 0.01. This ican be found by looking up the corresponding z dist table. Let us assume it is z

So the interval will be 0.5 +- ( z * Sqrt(0.5*(1-0.5)/100) ). So the sample proportion of even faces would lie anywhere between these figures in reality

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