Question

Research by Steelcase found the average worker get interrupted every 11 minutes and takes 23 minutes to get back on task. From a random sample of 200 workers, 168 said they are interrupted every 11 minutes by email, texts, alerts, etc. Find the 90% confidence interval of the population proportion of workers who are interrupted every 11 minutes.

Answer 1

Solution :

Given that,

n = 200

x = 168

= x / n = 168 / 200 = 0.840

1 - = 1 - 0.840 = 0.160

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

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

= 1.645 * (((0.840 * 0.160) / 200)

= 0.043

A 90% confidence interval for population proportion p is ,

- E < P < + E

0.840 - 0.043 < p < 0.840 + 0.043

0.797 < p < 0.893

(0.797 , 0.893)