# To test whether a coin is fair, Jane tossed it X times and observed that all outcomes were heads

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?

• 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. Nov 17 '21 at 7:35
• 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.) Nov 17 '21 at 8:21
• 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. Nov 17 '21 at 9:24
• It’s homework, I wouldn’t be overthinking the answer. Nov 17 '21 at 16:16

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 $$[0.4, 0.6]$$.

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