Question

A data set includes 106 body temperatures of healthy adult humans having a mean of 98.7degreesF and a standard deviation of 0.66degreesF. 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?
What is the confidence interval estimate of the population mean?

?(Round to three decimal places as? needed.)

What does this suggest about the use of 98.6degrees°F as the mean body? temperature?

A. This suggests that the mean body temperature could be higher than 98.6degrees°F.
B. This suggests that the mean body temperature could be lower than 98.6degrees°F.
C. This suggests that the mean body temperature could very possibly be 98.6degrees°F.

Answer 1

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 $(x^{-} - 2.58\frac{S.D}{\sqrt{n} } ,x^{-} +2.58\frac{S.D}{\sqrt{n} } )$ are called 99% of confidence limits

For the mean of the Population corresponding to the given sample.

$(98.6 - 2.58\frac{0.66}{\sqrt{106} } ,98.6 +2.58\frac{0.66}{\sqrt{106} } )$

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.