Answer to Question #214158 in Statistics and Probability for Anisha Radhika Kum

Question

Clients visiting a Chartered Accountant’s Firm on busy business day had to wait exactly 15 minutes less than an hour with a standard deviation of 10 minutes before they are able to see the Accountant. a) Compute the 90 percent confidence interval estimate for the average waiting time for all the clients visiting the office. Assume sample size is 36. b) Interpret your findings

Answer 1

15 minutes less than an hour is equivalent to 60-15=45min

mean =45min, sd=10min

sample mea, $\overline{X}=45$

population standard deviation $(\sigma)=10$

sample size (N) = 36

the critical value for $\alpha=0.1$ is $z_c=z_{1-\alpha /2}=1.645.$ The corresponding confidence interval is calculated as shown below

$CI=(\overline{X}-z_c\times \sigma / \sqrt n, \overline{X}+z_c\times \sigma/\sqrt n)$

$=(45-1.645\times 10/\sqrt{36}, 45+1.645\times 10/\sqrt {36})=(42.259,47.741)$

The 90% confidence interval for the population mean is $42.259 \lt \mu \lt47.741$

sample mean $\overline{X}=45$

sample standard deviation (s) =10

sample size (n) =36

The critical value for $\alpha=0.1$ and df=n-1=35 degrees of freedom is $t_c=z_{1-\alpha /2:n-1}=1.69$

its corresponding confidence interval is computed as shown below:

$CI=(X−tc​×s/\sqrt n ​,X+tc​×s/\sqrt n ​) =(45−1.69×10/\sqrt{36},45+1.69×10/\sqrt{36})=(42.184,47.816)$

Thus based on the data provided, the 90% confidence interval for the population mean is $42.184\lt \mu \lt 47.816,$ which indicates that we are 90% confident that the true population mean $\mu$ is contained by the interval (42.184,47.816).