Question

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

84.6

81.1

95.7

65.5

90.8

80.6

Find the 90% 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).

90% C.I. =  

Answer 1

Solution:

Given a sample of size n = 6

df = n - 1 = 5

First we find sample mean and sample SD s using calculator.

= 83.05

s = 10.395143096658

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

Our aim is to construct 90% confidence interval.   

c = 0.90

= 1- c = 1- 0.90 = 0.10

  /2 = 0.10 2 = 0.05

Also, d.f = n - 1 = 6 - 1 = 5  

    =    =  _0.05,5 = 2.015

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

The margin of error is given by

E =  /2,d.f. * ( / n)

=  2.015 * (10.395143096658/ 6 )

= 8.55

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

( - E ) <   <  ( + E)

(83.05 - 8.55)   <   <  (83.05 + 8.55)

74.50 <   < 91.60

Required 90% confidence interval is (74.50  , 91.60 )