One-Tailed Binomial Test and Sample Size
A company claims that 30% of customers prefer a new design. A researcher suspects the true proportion is lower and surveys 40 customers, of whom 8 prefer the new design. The researcher models the number who prefer the design as X ~ B(40, p) and plans a one-tailed test at the 5% significance level. A colleague says that because 8 is below the expected 12, the claim is clearly false and no formal test is needed. The researcher must state hypotheses, calculate a p-value, and write a conclusion in context. They must also explain whether the conclusion would change at the 10% level or if a two-tailed test were used. No calculator is available for cumulative binomial probabilities, so values may be quoted to 3 significant figures.
1. State the hypotheses and carry out the one-tailed test at the 5% level. Give the p-value to 3 s.f. and conclude in context.
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H0: p=0.30; H1: p<0.30, where p is the population proportion who prefer the new design. Under H0, X~B(40,0.30). The observed value is 8.
The p-value is P(X<=8). Summing binomial probabilities from r=0 to r=8 gives 0.112 (3 s.f.). Since 0.112>0.05, do not reject H0.
There is insufficient evidence at the 5% level to support the researcher's suspicion that the true proportion is lower than 30%. This is not proof that the claim is true; it means the sample is consistent with p=0.30.
2. The company says the difference is too small to matter. Assess this using the p-value and the sample size, and state whether the conclusion changes at the 10% level.
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The p-value of 0.112 means that, if p=0.30, about 11% of samples of size 40 would produce 8 or fewer preferring the design. That is not a rare event at the 10% level either (0.112>0.10), so the conclusion does not change.
The difference of 4 customers is based on only 40 observations, so it is not strong evidence. A larger sample would give a more precise test. At the 10% level the same decision remains: do not reject H0. A two-tailed test would use a larger p-value (about 0.229 for this observation), so it would also not reject at 5%.
3. Suppose a second survey of 80 customers finds 16 who prefer the design. Use a normal approximation with continuity correction to estimate the p-value for the same one-tailed test, and compare with the first test.
Read the worked answer
For n=80, under H0 the mean is np=80(0.30)=24 and the variance is np(1-p)=80(0.30)(0.70)=16.8, so sd=sqrt(16.8)=4.099.
With continuity correction, P(X<=16) is approximated by P(Z<=(16.5-24)/4.099)=P(Z<=-1.830). This gives 0.0336 (3 s.f.).
Since 0.0336<0.05, reject H0 at the 5% level; there is evidence that p<0.30. This is not inconsistent with the first test: the second sample is larger, so the same sample proportion (16/80=0.20 versus 8/40=0.20) gives stronger evidence. The exact binomial p-value would be somewhat different, but the normal approximation with continuity correction is reasonable because np and n(1-p) are both large.