Answer to Question #339250 in Statistics and Probability for Eduardo Valle

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 $(0.7799, 0.8534).$