Question

A simple random sample of 30 households was selected from a particular neighborhood. The number of cars for each household is shown below. Construct the 98% confidence interval estimate of the population mean.

2 0 1 2 3 2 1 0 1 4

1 3 2 0 1 1 2 3 1 2

1 0 0 5 0 2 2 1 0 2

A) 1.0 cars < μ < 2.0 cars

B) 0.9 cars < μ < 2.1 cars

C) 1.3 cars < μ < 1.7 cars

D) 1.5 cars < μ < 2.0 cars

Answer 1

Solution:

Sample size n = 30

2 0 1 2 3 2 1 0 1 4 1 3 2 0 1 1 2 3 1 2 1 0 0 5 0 2 2 1 0 2

Using calculator or Excel ,

Sample mean = 1.5

Sample standard deviation = s = 1.253

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 = 0.01

Also, d.f = n - 1 = 30 - 1 = 29

    =    =  _0.01,29 = 2.462

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

The margin of error is given by

E =  /2,d.f. * (s / n)

= 2.462 * (1.253 / 30)

= 0.56322054984

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

( - E ) <   <  ( + E)

(1.5 - 0.56322054984)   <   <  (1.5 + 0.56322054984)

0.93677945016 <   <  2.06322054984

0.9 <   < 2.1

Answer : 0.9 cars  <   < 2.1 cars