Question

2. The mean replacement time for a random sample of 18 washing machines is $.6 years and the standard deviation is 1.5 years. Construct a 98% confidence interval for the time of all washing machins of this typs.(use distribution for analysis).

Answer 1

Given that, sample size ( n ) = 18

sample standard deviation ( s ) = 1.5 years

Degrees of freedom = 18 - 1 = 17

confidence level = 0.98

=> significance level $(\alpha)$ = 1 - 0.98 = 0.02

/2 0.02/1- 0.01

1-a/2- 10.01 0.99

Using Excel we get, chi-square critical values,

CHỊÎNV (0.01.17) 33.409

X2 CHIINV(0.99. 17) 6.408

The 98% confidence interval for population standard deviation is,

$\sqrt {\frac {(n-1)* s^2}{\chi_L^2} } < \sigma < \sqrt {\frac {(n-1)* s^2}{\chi_R^2} }$

$\sqrt {\frac {(18-1)* (1.5)^2}{33.409 } } < \sigma < \sqrt {\frac {(18-1)* (1.5)^2}{6.408} }$

1.07 < σ < 2.44

Answer: 98% confidence interval is, (1.07, 2.44) years.