AI HL · 13.4
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Testing a proportion: the binomial test
13.4.15
Screenshot and video to come
Screenshot and video to come
A website says that 30% of the people who visit its page sign up for the newsletter. After the page is redesigned, 17 of a random sample of 40 visitors sign up. The owner thinks the proportion who sign up has increased. (a) Write down suitable hypotheses. (b) Find the p-value and test at the 5% significance level.
Have a go first. Then open the working one line at a time.
Let X be the number of visitors, out of 40, who sign up. Let p be the probability that a visitor signs up. X ~ B(40, p).
The hypotheses are about p, the probability of success, not about X.
H0: p = 0.3, H1: p > 0.3
The owner thinks the proportion has increased, so H1 is upper-tailed. Discrete tests in this course are one-tailed.
Assume H0 is true: X ~ B(40, 0.3)
The test works with the distribution you would get if H0 were true.
P(X ≥ 17) = 1 − P(X ≤ 16) = 0.0633
From the calculator, binomial cumulative probability: P(X ≤ 16). Take it from 1 for the upper tail.
0.0633 > 0.05, so do not reject H0.
The p-value is the probability of a result at least this extreme if H0 is true.
There is insufficient evidence, at the 5% significance level, that the proportion of visitors who sign up has increased.
Write it in the words of the question.
A seed company says that 85% of its seeds germinate. A gardener plants 30 seeds, and 21 germinate. She thinks the true proportion is less than 85%. Test her belief at the 5% significance level.
Have a go first. Then open the working one line at a time.
Let X be the number of seeds, out of 30, that germinate. H0: p = 0.85, H1: p < 0.85
She thinks the proportion is lower, so the tail is lower.
X ~ B(30, 0.85)
Use the distribution you get when H0 is true.
P(X ≤ 21) = 0.0278
A lower tail needs no subtraction: the calculator's cumulative probability is the p-value.
0.0278 < 0.05, so reject H0.
Compare p with 0.05.
There is sufficient evidence, at the 5% significance level, that the proportion of seeds that germinate is less than 85%.
Say what it means for the seeds.
A council says that 64% of households in the county recycle glass. A councillor thinks that fewer households recycle glass in her ward. She will ask a random sample of 60 households. Write down hypotheses and find the critical region at the 5% significance level.
Have a go first. Then open the working one line at a time.
Let X be the number of households, out of 60, that recycle glass. H0: p = 0.64, H1: p < 0.64
Fewer households: lower tail. Under H0, X ~ B(60, 0.64).
Mean = 60 × 0.64 = 38.4. The critical region is below this.
A lower critical region is made of values well below the mean.
From the calculator, a table of P(X ≤ k) near the 5% mark: P(X ≤ 30) = 0.0182, P(X ≤ 31) = 0.0334, P(X ≤ 32) = 0.0580
Build a column of k values and a column of cumulative probabilities. Look for where the values cross 0.05.
P(X ≤ 32) = 0.0580 > 0.05; P(X ≤ 31) = 0.0334 < 0.05
Check the value either side of the crossing. An inverse binomial can be one out.
Critical region: X ≤ 31
31 is the largest value whose cumulative probability is at most 0.05. If the sample has 31 or fewer households recycling glass, reject H0.
A garden centre says that 20% of its sweet pea seeds give a red flower. A customer thinks the proportion is higher. She grows 80 seeds. Write down hypotheses and find the critical region at the 10% significance level.
Have a go first. Then open the working one line at a time.
Let X be the number of seeds, out of 80, that give a red flower. H0: p = 0.2, H1: p > 0.2
Under H0, X ~ B(80, 0.2). Mean = 16, so the region is above 16.
P(X ≥ c) = 1 − P(X ≤ c − 1)
For an upper region, subtract the cumulative probability of the value just below c.
From the calculator, cumulative table: P(X ≤ 20) = 0.893, P(X ≤ 21) = 0.934
Read down the table to where 1 − P(X ≤ c − 1) drops below 0.10.
P(X ≥ 21) = 0.107 > 0.1; P(X ≥ 22) = 0.0660 < 0.1
The smallest c with P(X ≥ c) at most 0.10.
Critical region: X ≥ 22
22 or more red flowers gives sufficient evidence that the proportion is higher.
A coin is spun 20 times to test H0: p = 0.5 against H1: p < 0.5, where p is the probability of heads, at the 5% significance level. The graph shows the probabilities for X ~ B(20, 0.5). (a) Find the critical region. (b) Find the actual probability of rejecting H0 when it is true. (c) Explain why this is not 5%.
Have a go first. Then open the working one line at a time.
P(X ≤ 6) = 0.0577 > 0.05; P(X ≤ 5) = 0.0207 < 0.05. Critical region: X ≤ 5
Test adjacent values, as before.
P(reject H0 | H0 true) = P(X ≤ 5) = 0.0207
If H0 is true, the chance of landing in the critical region is 2.07%, not 5%.
Adding the next value takes the total to 0.0577, which is over 5%.
A discrete variable jumps from one probability to the next. Each value you add moves the total by a whole bar of the graph, so it cannot stop exactly at 0.05.
The test is stricter than asked: 2.07% instead of 5%.
The region is chosen so that its probability is at most the significance level. It is never over.