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Answer :
The 95% confidence interval of the mean body temperature of adults in the town is (97.584, 98.876)°F (open-interval notation), accurate to 3 decimal places.
What is the 95% confidence interval of the mean body temperature of adults in the town?
To find the 95% confidence interval (C.I.) of the mean body temperature, we need to use the formula:
C.I. = x ± z*(σ/√n)
Where:
x is the sample mean
σ is the population standard deviation (unknown, but we can estimate it using the sample standard deviation)
n is the sample size
z is the critical value for the desired confidence level (95% in this case)
First, we need to calculate the sample mean, sample standard deviation, and sample size:
x = (98 + 97.9 + 99 + 97.3 + 98.9 + 99.6 + 97.8 + 99.8 + 99.1 + 98.4 + 98.7 + 97.6 + 96.5) / 13 = 98.23
s = √((1/(n-1)) * Σ(xi - x)²)
s = 1.339
n = 13
Next, we need to find the critical value, z, for a 95% confidence level. We can look this up in a standard normal distribution table or use a calculator. For a 95% confidence level, z = 1.96.
Now we can plug in the values into the formula to get the 95% confidence interval:
C.I. = 98.23 ± 1.96*(1.339/√13) = (97.584, 98.876)
Learn more about confidence intervals at: https://brainly.com/question/17097944
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