A survey of several 10 to 12 year olds recorded the following amounts spent on a trip to the mall: $23.30,$22.29,$21.23,$21.86,$21.57 Construct the 80% confidence interval for the average amount spent by 10 to 12 year olds on a trip to the mall. Assume the population is approximately normal.

(21.59, 22.51)

Step-by-step explanation:

Given the data:

23.30,22.29,21.23,21.86,21.57

We calculate the mean and standard deviation of the sample given :

Using calculator :

Mean (m) = 22.05

Standard deviation (s) = 0.799843735 = 0.7998

Sample size, n = 5

Zcritical at 80% = 1.28

To obtain the confidence interval, use the Relation :

m ± Zcrit*s /sqrt(n)

22.05 ± 1.28 * (0.7998/sqrt(5))

Lower boundary = 22.05 - 1.28 * (0.7998/sqrt(5)) = 22.05 - 1.28(0.3576814) = 21.592167808

= 21.59

Upper boundary = 22.05 + 1.28 * (0.7998/sqrt(5)) = 22.05 + 1.28(0.3576814) = 22.507832192

= 22.51

(21.59, 22.51)

this answer is !