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If the p-value is greater than 0.05, what are the possible values of X?

(i) 4 (ii) 5 (iii) 6

Possible MCQ options: (a) (i) only (b) (i) and (ii) only. (c) (i), (ii) and (iii). (d)None of the above, X must be greater than 7

The answer is (a). Why is this so? How do you come to that conclusion?

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    $\begingroup$ Welcome to CV. If this question relates to a class exercise, please see stats.stackexchange.com/tags/self-study/info and add the tag to modify the question accordingly. $\endgroup$
    – Pitouille
    Nov 17 '21 at 7:35
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    $\begingroup$ There isn't enough information to answer because it isn't specified whether the test is 1-tailed or 2-tailed. (Based on the supposed correct answer, the question must be assuming that the test was 1-tailed.) $\endgroup$
    – fblundun
    Nov 17 '21 at 8:21
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    $\begingroup$ For a one-sided test: Using R as a calculator, code round(.5^(4:10) ,3) returns P-values $0.062,0.031,0.016,0.008,$ $0.004,0.002,0.001,$ which correspond to $4,5,6,…$ tosses. For a 2-sided test, you need to double these P-values. // Yet another poorly written multiple choice question. No wonder you're puzzled. $\endgroup$
    – BruceET
    Nov 17 '21 at 9:24
  • $\begingroup$ It’s homework, I wouldn’t be overthinking the answer. $\endgroup$
    – Aksakal
    Nov 17 '21 at 16:16
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This comment is for the future, not for your immediate frequentist statistics-laden question. To obtain direct evidence for fairness of a coin you need to define the state of prior knowledge and to define fairness, e.g, the probability of heads $\theta$ being between 0.49 and 0.51. Then acquire a lot of data and compute the Bayesian posterior probability that $\theta \in [0.49, 0.51]$.

The state of prior knowledge considers such things as the following:

  • Was the coin selected at random from a bag of coins? If so how was the bag selected?
  • If the coin was intentionally made to be unfair, would the person who did this be crafty enough to avoid detection by not making $\theta$ to be outside the interval $[0.4, 0.6]$? If so the prior distribution might be uniform on $[0.4, 0.6]$.
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Because the problem does not provide enough information, I will show four relevant exact binomial tests in R.

For such small numbers of tosses, I would not use normal approximations.


Right-sided alternative.

Four tosses. right-sided $(H_0: p=.5$ against: $H_a: p > .5).$ Getting four Heads in four tosses does not quite lead to rejection at the 5% level.

binom.test(4, 4, alt="greater")

        Exact binomial test

data:  4 and 4
number of successes = 4, number of trials = 4, 
  p-value = 0.0625
alternative hypothesis: 
 true probability of success is greater than 0.5
95 percent confidence interval:
 0.4728708 1.0000000
sample estimates:
probability of success 
                     1

Five Heads out of five tosses does lead to rejection at the 5% level, against the right-sided alternative.

binom.test(5, 5, alt="greater")$p.value
[1] 0.03125

Two-sided alternative.

Five tosses. two-sided $(H_0: p=.5$ against: $H_a: p \ne .5).$ Five Heads do not quite lead to rejection at the 5% level.

binom.test(5, 5, alt="two")

        Exact binomial test

data:  5 and 5
number of successes = 5, number of trials = 5,
 p-value = 0.0625
alternative hypothesis: 
 true probability of success is not equal to 0.5
95 percent confidence interval:
 0.4781762 1.0000000
 sample estimates:
                 1

Getting six Heads in six tosses does lead to rejection at the 5% level.

binom.test(6, 6, alt="two")$p.val
[1] 0.03125
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