Question

A sample of 100 lab rats run a complex maze and the time it takes each rat to complete the maze is recorded. It is shown that, on average, rats complete the maze in 2.55 minutes. From many replications of the same study over many decades, the standard deviation for all rats to complete the maze has been found to be 0.75 minutes.

a.) With 95% confidence, calculate the expected average population mean of maze completion times.

b.) If you were to calculate a 99% confidence interval, what Z value would you use? How would this change your estimate from part a?

Answer 1

a) we have  

Sample size n = 100

Sample mean xbar = 2.55

Standard deviation = 0.75

We have to find 95% confidence interval for population mean of maze completion time

xbar - Z_{​​​​​​a/2}*(/√n) < < xbar + Z_a/2*(/√n)

For c = 0.95 , a = 0.05

Z_{​​​​​​a/2} = Z_{​​​​​​0.025} = 1.96

2.55 - 1.96*(0.75/√100) < < 2.55 + 1.96 *(0.75/√100)

2.55 - 0.147 < < 2.55 + 0.147

2.403 < < 2.697

This is 95% confidence interval for average population mean of maze completion times.

b) if we have confidence level c = 0.99 , a = 0.01

Then Z_{​​​​​0.005} = 2.58

Here in part b for 99% confidence Z value is larger than part a)

So 99% confidence interval is wider than 95% confidence interval