Question

Sleep apnoea is a sleep disorder that causes a person to stop breathing momentarily and then awaken briefly. A study of 742 Victorian commercial truck drivers found that 63 of them suffered from sleep apnoea. Use the survey results to estimate, with 90% confidence, the proportion of Victorian commercial truck drivers who suffer from this sleeping disorder. Report the lower bound of the interval only, giving your answer as a percentage to two decimal places.

Answer 1

Solution :

Given that,

n = 742

x = 63

= x / n = 63 /742 = 0.085

1 - = 1 - 0.085 = 0.915

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

= 0.10

Z = Z0.10= 1.282

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.282 * (((0.085 * 0.915) / 742) = 0.0131

A 90 % confidence interval for population proportion p is ,

- E

0.085 - 0.0131

0.0719

answer= 7.19%