SOLUTION: ou want to rent an unfurnished one-bedroom apartment in Boston next year. The mean monthly rent for a simple random sample of 32 apartments advertised in the local newspaper is $1

Question 1196688: ou want to rent an unfurnished one-bedroom apartment in Boston next year. The mean monthly rent for a simple random sample of 32 apartments advertised in the local newspaper is $1 400. Assume that the standard deviation is known to be $220.
Find a 99% confidence interval for the mean monthly rent for unfurnished one-bedroom apartments available for rent in this community.
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

At 99% confidence, the z critical value is roughly z = 2.576
Use a table like this
https://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf
to get that value. Look at the bottom row labeled "Z" and above the 99% confidence level.

We use a Z distribution instead of a T distribution since
a) sigma is given
b) n > 30

Given info:
n = 32 = sample size
xbar = 1400 = sample mean
sigma = 220 = population standard deviation

Side note: We often won't know what sigma is since it's a population parameter. The goal of statistics is to estimate parameter. Use the sample standard deviation (variable 's') to estimate sigma, if sigma is unknown.

E = margin of error
E = z*sigma/sqrt(n)
E = 2.576*220/sqrt(32)
E = 100.18288875851

L = lower boundary of confidence interval
L = xbar - E
L = 1400 - 100.18288875851
L = 1299.8171112415
L = 1299.82

U = upper boundary of confidence interval
U = xbar + E
U = 1400 + 100.18288875851
U = 1500.1828887585
U = 1500.18

Answer: (1299.82, 1500.18) which is in the format (L, U)
This is equivalent to saying 1299.82 < mu < 1500.18 in the format L < mu < U.
Your answers may slightly vary depending on how you round the decimal values.

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