---
title: "Test of proportion: drug test failure rate"
subject: "Probability and Statistics"
chapter: 7
key: "semester_5/probability_and_statistics:job-applicants-drug-test-failure-rate"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/job-applicants-drug-test-failure-rate
site: BCT Engineering
---

# Test of proportion: drug test failure rate

Probability and Statistics · Chapter 7: Inference concerning proportion

- Asked in: 2076 Asoj, 5 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/job-applicants-drug-test-failure-rate
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-7

## Question

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?

## 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.

## Answer

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

```text
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:

```text
SE = √(p₀q₀/n) = √(0.058×0.942/1520) = 0.00600
z  = (p̂ − p₀)/SE = (0.0382 − 0.058)/0.00600 = -3.310
```

4. Critical value: z(0.05, left-tail) = −1.645. Reject H0 if z < −1.645.
5. 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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The claim is that the failure rate is now lower. What is H1?
   A. p > 0.058
   B. p ≠ 0.058
   C. p = 0.058
   D. p < 0.058 (left-tailed)
   Answer: D. A lower rate gives H1: p < 0.058. (Key point: Direction)

2. What is p̂?
   A. 58/1520 = 0.0382
   B. 0.058
   C. 1520/58
   D. 0.942
   Answer: A. p̂ = x/n = 58/1520. (Key point: Sample proportion)

3. What is the critical value?
   A. +1.645
   B. −1.645
   C. −2.326
   D. −1.96
   Answer: B. Left-tailed at 0.05 gives −1.645. (Key point: Critical value)

4. z = −3.310. What is concluded?
   A. Rate is higher
   B. Fail to reject H0
   C. Reject H0; the failure rate is now lower
   D. No change
   Answer: C. −3.310 < −1.645, so the rate is significantly below 5.8%. (Key point: Decision)

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