Question

A company personnel officer wants to estimate the mean time between occurrences of personnel accidents that might provide the potential for liability lawsuits. A random sample of n=30 accidents from the company's records of the time x between an accident and the one proceding gave a sample mean of x-bar = 42.1 days and a standard deviation of s =19.6 days. Find a 90% confidence interval for the mean time between occurrences of personnel accidents possessing the potential for liability lawsuits.

Answer 1

Solution :

Given that,

= 42.1

s = 19.6

n = 30

Degrees of freedom = df = n - 1 = 30 - 1 = 29

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

t _/2,df = t_0.05,29 = 1.699

Margin of error = E = t_/2,df * (s /n)

= 1.699 * (19.6 / 30)

= 6.1

The 90% confidence interval estimate of the population mean is,

- E < < + E

42.1 - 6.1< < 42.1 + 6.1

36.0 < < 48.2

(36.0 , 48.2)