Question

Sleep apnoea is a sleep disorder that causes a person to stop breathing momentarily and then awaken briefly. A study of 418 Victorian commercial truck drivers found that 36 of them suffered from sleep apnoea. Use the survey results to estimate, with 92% confidence, the proportion of Victorian commercial truck drivers who suffer from this sleeping disorder. Report the upper bound of the interval only, giving your answer as a percentage to two decimal places.

Answer 1

Solution:

In case of proportions, confidence interval can be found as:

CI = p^ -/+ z-score*(p^*(1 - p^)/sample size^(1/2)) ; where p^ is sample proportion. In given case, p^ = 36/418 = 0.0861

With 92% confidence, required z-score is calculated for alpha (1 - 92%)/2 = 0.04, and this value is achieved for z value of 1.75 (exact value is 0.0401, so a close approximation)

Then, CI = 0.0861 -/+ 1.75*(0.0861*0.9139/20.445)

CI = 0.0861 -/+ 1.75*0.00385

CI = 0.0861 -/+ 0.0067

Upper bound can be found using plus sign part of CI, so upper bound value = 0.0861 + 0.0067 = 0.0928 or 9.28% (in percentage terms, on round off to 2 decimal places).