An important factor in selling a residential property is the number of people who look through the home. A sample of 20 homes re
cently sold in the Buffalo, New York, area revealed the mean number looking through each home was 22 and the standard deviation of the sample was 7 people. Develop a 98 percent confidence interval for the population mean.
1 answer:
You might be interested in
Part A
Given info:
- xbar = sample mean = 4.90 ppm
- s = sample standard deviation = 1.12 ppm
- n = 23 = sample size
Because n > 30 is not true, and we don't know the population standard deviation (sigma), this means we must use a T distribution.
The degrees of freedom here are n-1 = 23-1 = 22.
At 90% confidence and the degrees of freedom mentioned, the t critical value is roughly t = 1.717
Use a T distribution table or calculator to determine this. If you don't have a calculator for the task, then you can search out "inverse T calculator" and there are tons of free options to pick from.
The margin of error E is
E = t*s/sqrt(n)
E = 1.717*1.12/sqrt(23)
E = 0.400982
This is approximate and accurate to 6 decimal places.
The confidence interval is going to be xbar plus or minus that E value
L = lower bound = xbar - E = 4.90 - 0.400982 = 4.499018 = 4.50
U = upper bound = xbar + E = 4.90 + 0.400982 = 5.300982 = 5.30
The confidence interval in the format of (L, U) is (4.50, 5.30)
You could also express it as the format L < mu < U and it would be 4.50 < mu < 5.30; however, I'll stick to the first method.
<h3>Answer: (4.50, 5.30)</h3>=====================================================
Part B
Since we know sigma = 2.6 is the population standard deviation, we can use a Z distribution now.
At 90% confidence, the z critical value is roughly 1.645; use a table or calculator to determine this.
Round this up to the nearest integer to get 20326. For min sample size problems, <u>always</u> round up.
<h3>Answer: 20326</h3>