Question

A survey of 87 randomly selected homeowners finds that they spend a mean of $65 per month on home maintenance. Assume that the population standard deviation is $13 per month.

Construct and interpret a 97% confidence interval for the mean amount of money spent per month on home maintenance by all homeowners.

Answer 1

Given that,

mean=65,

standard deviation=13

random sample size=87

97% Confidence Interval for Mean Amount of Money Spent per month on Home Maintenance by all Homeowners

97% Confidence Interval means α = 0.03.

97% C.I =

alpha = 0.03, alpha/ 2= 0.015= 2.17

So Z value when alpha/ 2= 0.015 is given from the normal distribution table, z = 2.17

97% C.I = = 65+/- 2.17 * 13/sqrt(87)

= 65 + 3.02and 67 - 3.02 = 61.98and 68.02

So 97% C.I. is 61.98-68.02 for the Mean Amount of Money Spent per month on Home Maintenance by all Homeowners.