Question 1202152: The body temperatures in degrees Fahrenheit of a sample of adults in one small town are: 97.1 98.1 98 97.7 97.4 99.3 96.8
Assume body temperatures of adults are normally distributed. Based on this data, find the 90% confidence interval of the mean body temperature of adults in the town. Enter your answer as an open-interval (i.e., parentheses) accurate to 3 decimal places. Assume the data is from a normally distributed population.
90% C.I. =
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Answer: (97.169, 98.374)
That is the condensed form of 97.169 < mu < 98.374
Work Shown:
mu = μ = Greek letter representing population mean
In this problem's context, mu is the population mean body temperature in degrees Fahrenheit.
The goal is to estimate mu using a confidence interval.
Given data set = {97.1,98.1,98,97.7,97.4,99.3,96.8}
n = 7 = sample size
xbar = sample mean
xbar = (add up the values)/(number of values)
xbar = (97.1+98.1+98+97.7+97.4+99.3+96.8)/(7)
xbar = 97.77143 approximately
Use a calculator or spreadsheet to determine the sample standard deviation is approximately s = 0.81999
df = degrees of freedom = n-1 = 7-1 = 6
Because the population standard deviation (sigma) is not known, and because n > 30 isn't the case, we must use the T distribution.
Refer to this T table
https://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf
Locate the row labeled df = 6
Locate the column labeled "90% confidence". The confidence labels are at the bottom.
The approximate value t = 1.943 is at this row and column intersection. It is the t critical value.
Specialized stats calculators such as this one
https://www.omnicalculator.com/statistics/critical-value
can find the t critical value.
Make sure to do a two-tailed test.
Also set the significance level to 0.10 (recall that alpha = 1-C where C is the confidence level)
Compute the margin of error
E = t*s/sqrt(n)
E = 1.943*0.81999/sqrt(7)
E = 0.60219
That result is approximate.
Then,
L = lower boundary
L = xbar - E
L = 97.77143 - 0.60219
L = 97.16924
L = 97.169
and
U = upper boundary
U = xbar + E
U = 97.77143 + 0.60219
U = 98.37362
U = 98.374
The 90% confidence interval in the format L < mu < U is roughly 97.169 < mu < 98.374
That is then condensed to the format (L, U) so we get (97.169, 98.374) as the final answer.
We are 90% confident that the population mean temperature is somewhere between 97.169 degrees Fahrenheit and 98.374 degrees Fahrenheit.
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