Answer:
The 99% confidence intervals are
(98.4346 , 98.7654)
C) This suggests that the mean body temperature could very possibly be 98.6degrees°F.
Step-by-step explanation:
Given data a data set includes 106 body temperatures of healthy adult humans having a mean of 98.7degreesF and a standard deviation of 0.66degreesF
sample size n = 106
Mean of the population ' μ' = 98.7degreesF
standard deviation of population 'σ' = 0.66degreesF
Mean of the sample x⁻ = 98.6degreesF
Level of significance ∝ = 0.99 0r 0.01
Confidence limits :-
The values [tex](x^{-} - 2.58\frac{S.D}{\sqrt{n} } ,x^{-} +2.58\frac{S.D}{\sqrt{n} } )[/tex] are called 99% of confidence limits
For the mean of the Population corresponding to the given sample.
[tex](98.6 - 2.58\frac{0.66}{\sqrt{106} } ,98.6 +2.58\frac{0.66}{\sqrt{106} } )[/tex]
on calculation, we get
(98.6 - 0.1654,98.6 + 0.1654)
The 99% confidence intervals are
(98.4346 , 98.7654)
C) This suggests that the mean body temperature could very possibly be 98.6degrees°F.
