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Felicia observed getting “6 spots” 3 times out of 6 rolls of a six-sided die. Eve suggested that the die is biased towards “6 spots”. In order to quantitatively evaluate Eve’s suggestion, Felicia set up the null hypothesis that P(“6 spots”) = 1/6. Which of the following choices has exactly those events that are equally or more extreme than the observed event?

Getting 6 spots A) 0,1,2 times B) 0,1,2,3 times C) 3,4,5,6 times D) 4,5,6 times

Using binomial distribution formula, I calculated that when "6 spots" is obtained 4,5 or 6 times per 6 rolls, then the null hypothesis would be rejected and the die is biased. Am i right that the answer is D?

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  • $\begingroup$ I don't understand your comment about rejecting the null hypothesis, and I'd say that $C$ is the correct answer. easy to check that $3,4,5,6$ occurrences of $6$ are all equally or more improbable than getting $3$ occurrences (assuming the null hypothesis). $\endgroup$
    – lulu
    Feb 15, 2017 at 15:20
  • $\begingroup$ I'll reject the null hypothesis if the calculated p-value < 0.05 $\endgroup$
    – Loki123
    Feb 15, 2017 at 15:56
  • $\begingroup$ That doesn't appear anywhere in your question, nor does it have anything to do with the multiple choice question. More to the point, doing a statistical experiment with six trials is never going to be reliable...and, of course, you would need to do the experiment again with fresh trials, after clearly declaring what you plan to measure. $\endgroup$
    – lulu
    Feb 15, 2017 at 16:48

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