An ordinary die is “fair” or “balanced” if each face has an equal chance of land

Question

An ordinary die is “fair” or “balanced” if each face has an equal chance of landing on top when the die is rolled. Thus the proportion of times a three is observed in a large number of tosses is expected to be close to 1/6 or 0.16. Suppose a die is rolled 240 times and shows three on top 36 times, for a sample proportion of 0.15. Find the probability that a fair die would produce a proportion of 0.15 or less. You may assume that the normal distribution applies. Suppose the sample proportion 0.15 came from rolling the die 2,400 times instead of only 240 times. Rework part (a) under these circumstances. Give an interpretation of the result in part (c). How strong is the evidence that the die is not fair?

Explanation / Answer

Assume, r.v X denote number 3 showing on top. The mean of 1-proportion sampling distribution is phat=0.16, standard deviation of sampling distribution, SE=sqrt[phat(1-phat)/n]=sqrt[0.16(1-0.16)/240]=0.0237. Substitute values in Z score formula, Z=(X-xbar)/s, where, X is raw score, xbar is sample mean, and s is sample standard deviation.

P(X<=0.15)=P[Z<(0.15-0.16)/0.0237]=P(Z<-0.42)=0.3372

Under second condition, applying same rule as before,

P(X<=0.15)=P[Z<(0.15-0.16)/0.0075]=P(Z<-1.33)=0.0918

The probability that obtaining 0.15 or less is much lesser for rolling the dice for 2400 times instead of 240 times.

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