Question

1.     Work Interruptions Survey found that out of 200 workers, 168 said they were interrupted three or more times an hour by phone messages, faxes, etc. Find the 90% confidence interval of the population proportion of workers who are interrupted three or more times an hour.

Answer 1

The sample proportion is computed as follows, based on the sample size $N = 200$  and the number of favorable cases $X = 168:$

$$\hat{p}=\dfrac{X}{N}=\dfrac{168}{200}=0.84$$

The critical value for $\alpha = 0.1$ is $z_c = z_{1-\alpha/2} = 1.6449.$

The corresponding confidence interval is computed as shown below:

$$CI(Proportion)=(\hat{p}-z_c\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}},$$

$$\hat{p}+z_c\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}})$$

$$=(0.84-1.6449\sqrt{\dfrac{0.84(1-0.84))}{200}},$$

$$0.84+1.6449\sqrt{\dfrac{0.84(1-0.84))}{200}})$$

$$=(0.79736,0.88264)$$

Therefore, based on the data provided, the 90% confidence interval for the population proportion is $0.79736 < p < 0.88264,$ which indicates that we are 90% confident that the true population proportion $p$ is contained by the interval $(0.79736, 0.88264).$