Question 1174988: The high temperatures (in degrees Fahrenheit) of a random sample of 10 small towns are:
98.8
97
97.1
98.3
97.5
98.5
99.1
96.3
97.9
96.7
Assume high temperatures are normally distributed. Based on this data, find the 95% confidence interval of the mean high temperature of towns. Enter your answer as an open-interval (i.e., parentheses) accurate to two decimal places (because the sample data are reported accurate to one decimal place).
95% C.I. =
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Answer: (97.04, 98.40)
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Work Shown:
xbar = sample mean
xbar = 97.72
This is found by adding up all of the numbers given in the table, then dividing by 10 since there are 10 values in this data table.
So this means n = 10 is the sample size.
s = sample standard deviation
s = 0.946103 which is approximate
While it is possible to find this by hand, it's a bit of a pain with this many numbers. I recommend using either a spreadsheet or some other similar form of technology to quickly compute this value.
Because n is not greater than 30, and because we don't know the population standard deviation (sigma), this means we must use a T distribution (instead of a Z distribution)
At 95% confidence and with degrees of freedom = n-1 = 10-1 = 9, the critical t value is roughly t = 2.262
Use a table or calculator to determine this.
The lower bound (L) of the confidence interval is
L = xbar - t*s/sqrt(n)
L = 97.72 - 2.262*0.946103/sqrt(10)
L = 97.72 - 0.676754
L = 97.043246
L = 97.04
The upper bound (U) of the confidence interval is
U = xbar + t*s/sqrt(n)
U = 97.72 + 2.262*0.946103/sqrt(10)
U = 97.72 + 0.676754
U = 98.396754
U = 98.40
The 95% confidence interval that estimates the population mean mu is approximately (97.04, 98.40)
This is the same as writing 97.04 < mu < 98.40
We're 95% confident that mu is somewhere between 97.04 and 98.40
mu = population mean of high temperatures (in degrees Fahrenheit) of towns.
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