Question

A survey of several 88 to 1010 year olds recorded the following amounts spent on a trip to the mall:

$19.02,$21.15,$22.96,$27.58$⁢19.02,$⁢21.15,$⁢22.96,$⁢27.58

Construct the 99%99% confidence interval for the average amount spent by 88 to 1010 year olds on a trip to the mall. Assume the population is approximately normal.

Copy Data

Step 4 of 4:

Construct the 99%99% confidence interval. Round your answer to two decimal places.

Answer 1

I think the values have been copied twice. I am putting down what I assume is the correct set of values down below.If you have further doubts please leave them in the comments section

19.02

21.15

22.96

27.58

n =4
  
Σ-2 n 1 Σ22.68 S 3.6435

Assuming the population follows a normal dist and the sample is random and independent

99% confidence interval for average amount spent by 88 to 101 yr olds is

( we use t-dist since population s.d. is unknown)

ST (tn-1,a/2 tn-1,a/2-

Where

=1-0.990.01

(values can be found using t-dist tables online or excel function 'tinv')

t3,0.01/27.4533

Substituting the values

(22.68 7.4533 3.6435 22.68 7.4533 3.6435 V4

Final Ans: (9.10, 36.26)

The width of the confidence interval is high because the variation is high when the sample is too small.