MATH SOLVE

2 months ago

Q:
# A data set includes 103 body temperatures of healthy adult humans having a mean of 98.3degreesF and a standard deviation of 0.73degreesF. Construct a 99% confidence interval estimate of the mean body temperature of all healthy humans. What does the sample suggest about the use of 98.6degreesF as the mean body temperature?

Accepted Solution

A:

Answer:CI = (98.11 , 98.49)The value of 98.6°F suggests that this is significantly higherStep-by-step explanation:Data provided in the question:sample size, n = 103Mean temperature, μ = 98.3°Standard deviation, σ = 0.73Degrees of freedom, df = n - 1 = 102Now,For Confidence level of 99%, and df = 102, the t-value = 2.62 [from the standard t table]Therefore,CI = [tex](Mean - \frac{t\times\sigma}{\sqrt{n}},Mean + \frac{t\times\sigma}{\sqrt{n}})[/tex]Thus, Lower limit of CI = [tex](Mean - \frac{t\times\sigma}{\sqrt{n}})[/tex]orLower limit of CI = [tex](98.3 - \frac{2.62\times0.73}{\sqrt{103}})[/tex]orLower limit of CI = 98.11and,Upper limit of CI = [tex](Mean + \frac{t\times\sigma}{\sqrt{n}})[/tex]orUpper limit of CI = [tex](98.3 + \frac{2.62\times0.73}{\sqrt{103}})[/tex]orUpper limit of CI = 98.49Hence,CI = (98.11 , 98.49)The value of 98.6°F suggests that this is significantly higher and the mean temperature could very possibly be 98.6°F