Question

Work Interruptions Survey found that out of 300 workers, 245 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 $N = 300$ and the number of favorable cases $X = 245:$

$$\hat{p}=\dfrac{X}{N}=\dfrac{245}{300}=\dfrac{49}{60}$$

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:

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

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

$$=(\dfrac{49}{60}-1.6449\sqrt{\dfrac{\dfrac{49}{60}(1-\dfrac{49}{60})}{300}},$$

$$\dfrac{49}{60}+1.6449\sqrt{\dfrac{\dfrac{49}{60}(1-\dfrac{49}{60})}{300}})$$

$$=(0.7799, 0.8534)$$

Therefore, based on the data provided, the 90% confidence interval for the population proportion is $0.7799< p < 0.8534,$ which indicates that we are 90% confident that the true population proportion $p$ is contained by the interval $(0.7799, 0.8534).$