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?

Answer: (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.