A data includes 108 body temperature od healthy adult humans having a mean of 98.2 f and a standard deviation of .64f. Contruct a 99% confidence interval estimu of the mean body temperature of all healthy humans. What dose the sample suggest about the use of 98.6 f as the mean body temperature?

( 98.04f, 98.40f ), it is not accepted

Step-by-step explanation: the question is suggesting we construct a 99% confidence interval for mean body temperature.

The formulae for Constructing a 99% confidence interval is given below as

u = x + Zα/2 × (s/√n) For the upper limit

u = x - Zα/2 × (s/√n) For the lower limit

Where u = population mean

x = sample mean = 98.2

s = sample standard deviation = 0.64

n = sample size = 108.

Zα/2 = critical value for a 2 tailed test performed at a 1% level of significance ( 100% - 99%) = 2.58

We are making use of our z test because sample size is greater than 30 ( n = 108).

By substituting the parameters, we have that

For upper limit

u = x + Zα/2 × (s/√n)

u = 98.2 + 2.58 ( 0.64/√108)

u = 98.2 + 2.58 ( 0.0616)

u = 98.2 + 0.1589

u = 98.4

For lower limit

u = x - Zα/2 × (s/√n)

u = 98.2 - 2.58 ( 0.64/√108)

u = 98.2 - 2.58 ( 0.0616)

u = 98.2 - 0.1589

u = 98.04.

Hence the 99% confidence interval for mean temperature of human is ( 98.04f, 98.40f )

Using 98.6f as the average mean body temperature is wrong because it is out of the confidence interval calculated above.

Our confidence interval states that at 99% confidence interval, an average human temperature should be at least 98.04f and at most 98.40f.

98.6f is out of the range.

your answer is d 2,500

step-by-step explanation: