Answer to Question #346759 in Statistics and Probability for Bethsheba Kiap

Question

A government agency was charged by the legislature with estimating the length of time it takes citizens to fill out various forms. Two hundred randomly selected adults were timed as they filled out a particular form. The times required had mean 12.8 minutes with standard deviation 1.7 minutes. Construct a 90% confidence interval for the mean time taken for all adults to fill out this form.

Answer 1

The critical value for $\alpha = 0.10, df=n-1=199$ degrees of freedom is $t_c​=z_{1−α/2;n−1}= 1.652547$

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{x}-t_c\times\dfrac{s}{\sqrt{n}}, \bar{x}+t_c\times\dfrac{s}{\sqrt{n}})$$ $$=(12.8- 1.652547\times\dfrac{1.7}{\sqrt{200}},$$ $$12.8+ 1.652547\times\dfrac{1.7}{\sqrt{200}})$$

$$=(12.601, 12.999)$$

Therefore, based on the data provided, the 90% confidence interval for the population mean is $12.601 < \mu < 12.999,$ which indicates that we are 90% confident that the true population mean $\mu$ is contained by the interval $(12.601, 12.999).$