Question

for sample of 29 operating rooms taken in a hospital study,the mean noise level was 42.7 decibel,and the standard deviation was 6.5.find the 98% confidence interval of the true mean of noise level in the operating rooms.assum the variable is normally distributed.use a graphical calculate and round the answer to at least one decimal place.

Answer 1

n= 29,   = 42.7, s= 6.5

c= 98%

using graphical calculator ( TI-84 ) we get confidence interval as follows

( 39.722 , 45.678 )

Manually we get answer as follows

formula for confidence interval is

Where tc is the t critical value for c=95% with df=n-1 = 29-1 = 28

using t table we get critical value as

tc = 2.048

39.722 <    <  45.678

( 39.722 , 45.678 )