College

A data set includes 108 body temperatures of healthy adult humans, having a mean of 98.2 and a standard deviation of 0.64. Construct a 99% confidence interval estimate of the mean body temperature of all healthy humans. What does the sample suggest about the use of 98.6 as the mean body temperature?

Answer :

The 99% confidence interval for mean temperature of human is :

( 98.04f, 98.40f )

A data set includes 108 body temperatures of healthy adult humans having a mean of 98.2 and a standard deviation of 0.64 .

Now, We have to construct a 99% confidence interval estimate of the mean body temperature of all healthy humans.

We use these formula for upper limit and lower limit:

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

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

We have the information from question:

x = sample mean = 98.2

s = standard deviation = 0.64

n = sample size = 108.

[tex]Z_\frac{\alpha }{2}[/tex] = critical value for a 2 tailed test performed at a 1% level of significance ( 100% - 99%) = 2.58

Now, Substitute the parameter,

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.

The 99% confidence interval for mean temperature of human is :

( 98.04f, 98.40f )

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