Example 1: Suppose you have a die and suspect that it is biased towards the number three, and so run an experiment in which you throw the die 10 times and count that the number three comes up 4 times. Determine whether the die is biased.
Define x = the number of times the number three occurs in 10 trials. This random variable has the binomial distribution where π is the population parameter corresponding to the probability of success on any trial. We use the following null and alternative hypotheses:
H0: π ≤ 1/6; i.e. the die is not biased towards the number 3
H1: π > 1/6
Setting α = .05, we have
P(x ≥ 4) = 1–BINOM.DIST(3, 10, 1/6, TRUE) = 0.069728 > 0.05 = α.
and so we cannot reject the null hypothesis that the die is not biased towards the number 3 with 95% confidence.
Example 2: We suspect that a coin is biased towards heads. When we toss the coin 9 times, how many heads need to come up before we are confident that the coin is biased towards heads?
We use the following null and alternative hypotheses:
H0: π ≤ .5
H1: π > .5
Using a confidence level of 95% (i.e. α = .05), we calculate
BINOM.INV(n, p, 1–α) = BINOM.INV(9, .5, .95) = 7
which means that if 8 or more heads come up then we are 95% confident that the coin is biased towards heads, and so can reject the null hypothesis.
We confirm this conclusion by noting that P(x ≥ 8) = 1–BINOM.DIST(7, 9, .5, TRUE) = 0.01953 < 0.05 = α, while P(x ≥ 7) = 1–BINOM.DIST(6, 9, .5, TRUE) = .08984 > .05.
Example 3: Historically a factory has been able to produce a very specialized nano-technology component with 35% reliability, i.e. 35% of the components passed its quality assurance requirements. They have now changed their manufacturing process and hope that this has improved the reliability. To test this, they took a sample of 24 components produced using the new process and found that 13 components passed the quality assurance test. Does this show a significant improvement over the old process?
We use a one-tailed test with null and alternative hypotheses:
H0: p ≤ .35
H1: p > .35
p-value = 1–BINOM.DIST(12, 24, .35, TRUE) = .04225 < .05 = α
and so conclude with 95% confidence that the new process shows a significant improvement.