Question

a data set includes 110 body temperatures of healthy adult humans having a mean of 98.1F and a standard deviation of 0.64F. Construct a 99% confidence interval estimate of the mean body temperature of all healthy humans

Answer 1

Explanation:

Answer 2

The 99% confidence interval is
`97.94 < \mu < 98.26`

Explanation:

From the question we are told that

The sample size is n = 110

The sample mean is
`\= x = 98.1 \ F`

The standard deviation is
`\sigma = 0.64 \ F`

Given that the confidence level is 99% the level of significance i mathematically evaluated as

`\alpha = 100 - 99`

`\alpha = 1\%`

`\alpha = 0.01`

Next we obtain the critical value of
`(\alpha )/(2)` from the normal distribution, the values is

`Z_{(\alpha )/(2) } = Z_{(0.01 )/(2) } = 2.58`

Generally the margin of error is mathematically represented as

`E = Z_{(\alpha )/(2) } * (\sigma)/(√(n) )`

substituting values

`E = 2.58 * ( 0.64)/(√(110) )`

`E = 0.1574`

Generally the 99% confidence interval is mathematically represented as

`\= x - E < \mu < \= x + E`

substituting values

`98.1 - 0.1574 < \mu < 98.1 + 0.1574`

`97.94 < \mu < 98.26`