Question

In a sample of 82 hip surgeries of a certain type, the average surgery time (x) was 126.5 minutes with a sample standard deviation (s) of 21.6 minutes. Follow the steps below to construct a 99 % confidence interval for the mean surgery time (μ) for this procedure. Assume all assumptions have been satisfied. What would the point estimate and critical value be?

Answer 1

Solution :

Given that,

n = 82

   = 126.5

s = 21.6

Note that, Population standard deviation() is unknown..So we use t distribution.

Our aim is to construct 99% confidence interval.   

c = 0.99

= 1 - c = 1 - 0.99 = 0.01

  /2 = 0.01 2 = 0.005

Also, d.f = n - 1 = 81

    =    =  0.005,81 = 2.638

( use t table or t calculator to find this value..)

The margin of error is given by

E =  /2,d.f. * ( / n )

= 2.638* (21.6 / 82 )

= 6.292

Now , confidence interval for mean() is given by:

( - E ) <   <  ( + E)

( 126.5 - 6.292 )   <   <  ( 126.5 + 6.292)

120.208 <   < 132.792

Required.99% confidence interval is ( 120.208 , 132.792 )

The point estimate = 126.5

The critical value = 2.638