Question

You must estimate the mean temperature (in degrees Fahrenheit) with the following sample temperatures:

62.4
102.8
79.9
52.7
45.5
70.6
76.2
102.3
62.4

Find the 98% confidence interval. Enter your answer as an open-interval (i.e., parentheses) accurate to two decimal places (because the sample data are reported accurate to one decimal place).

98% C.I. =

Answer should be obtained without any preliminary rounding.

Answer 1

Solution :

We are given a data of sample size n = 9

62.4,102.8,79.9,52.7,45.5,70.6,76.2,102.3,62.4

Using this, first we find sample mean() and sample standard deviation(s).

=   

= (62.4 + 102.8.......+ 62.4)/9

= 72.75

Now ,

s=   

Using given data, find Xi - for each term.Take square for each.Then we can easily find s.

s = 20.01

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

Our aim is to construct 98% confidence interval.

   c = 0.98

= 1 - c = 1 - 0.98 = 0.02

  /2 = 0.02 2 = 0.01

Also, d.f = n - 1 = 9 - 1 = 8

    =    =  _0.01,8 = 2.896

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

The margin of error is given by

E =  /2,d.f. * ( / n )

= 2.896 * ( 20.01/ 9 )

= 19.319

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

( - E ) <   <  ( + E)

( 72.75 - 19.319 )  <   <  ( 72.75 + 19.319 )

53.43 <   < 92.07

Required 98% confidence interval is ( 53.43 , 92.07 )