Question 1206798: An important factor in selling a residential property is the number of people who look through the home. A sample of 11 homes recently sold in the Halifax, Nova Scotia, area revealed the mean number looking through each home was 17 and the standard deviation of the sample was 3 people. Develop a 90% confidence interval for the population mean. (Round the final answers to 2 decimal places.)
Confidence interval for the population mean is between
and
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
mu = population mean = unknown
sigma = population standard deviation = unknown
xbar = 17 = sample mean
s = 3 = sample standard deviation
n = 11 = sample size
Since we do not know the population standard deviation (sigma), and because n > 30 is not the case, we must use the T distribution.
df = degrees of freedom
df = n-1
df = 11-1
df = 10
At 90% confidence, and df = 10, the t critical value is roughly t =1.812
You can use a table such as this
https://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf
to determine the t critical value.
Highlight the row that has df = 10. Highlight the column that has "90%" at the bottom.
This row and column combo yields the value 1.812 which is approximate.
What this means is that P(-1.812 < t < 1.812) = 0.90 approximately when df = 10.
Another way to determine this t critical value is to use a stats calculator such as a TI84.
The specific function to use on a TI84 is called invT.
There are many other calculators that offer a similar feature.
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Here are the important values we need
t = 1.812 (approximate)
xbar = 17
s = 3
n = 11
Then,
E = margin of error
E = t*s/sqrt(n)
E = 1.812*3/sqrt(11)
E = 1.639016 approximately
L = lower bound of confidence interval
L = xbar - E
L = 17 - 1.639016
L = 15.360984
L = 15.36
U = upper bound of confidence interval
U = xbar + E
U = 17 + 1.639016
U = 18.639016
U = 18.64
The 90% confidence interval of the format (L, U) is roughly (15.36, 18.64)
This represents 15.36 < mu < 18.64 which is a more descriptive way of explaining what's going on (since it involves the parameter we're trying to estimate).
We are 90% confident that the population mean (mu) is somewhere between 15.36 and 18.64
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