Skip to main content

Test of proportion: drug test failure rate

Short answer

For H0:p=0.058 versus H1:p<0.058, p-hat=58/1520=0.0382, z=(0.0382-0.058)/0.006=-3.31. Critical z at 0.05 is -1.645. Reject H0; failure rate is now lower, suggesting fewer job applicants use drugs.

In 1990, 5.8% job applicants who were tested for drugs failed the test. At the 0.05 significant level, the test claim that the failure rate is now lower if a simple random sample of 1520 current job applicants results in 58 failure. Does the result suggest that fewer job applicants now use drugs?

This is a left-tailed test of a single proportion (large sample).

n = 1520, x = 58, p̂ = 58/1520 = 0.0382,  p₀ = 0.058, q₀ = 0.942
  1. H0: p = 0.058 (failure rate is still 5.8%) H1: p < 0.058 (failure rate is now lower)
  2. Level of significance: α = 0.05
  3. Test statistic:
SE = √(p₀q₀/n) = √(0.058×0.942/1520) = 0.00600
z  = (p̂ − p₀)/SE = (0.0382 − 0.058)/0.00600 = -3.310
  1. Critical value: z(0.05, left-tail) = −1.645. Reject H0 if z < −1.645.
  2. Decision: z = -3.310 < −1.645 (p-value = 0.0005), so reject H0.

Conclusion: at the 5% level the failure rate is now lower than 5.8% (sample rate 3.82%). The result suggests that fewer job applicants now use drugs.

✓ Answer checked by Claude Sonnet 5.5

Related questions

All Probability and Statistics chapter 7 questions →