Question

A sample of 106 healthy adults have a mean body temperature of 98.2, and a standard deviation of 0.62. Construct a 99% confidence interval estimate of the mean body temperature of all healthy humans. Group of answer choices ( 98.04, 98.36 ) ( 97.54, 97.95 ) ( 97.84, 98.12 ) ( 97.82, 98. 15 ) ( 97.95, 98.15 )

Answer 1

Answer:
`(98.04,\ 98.36)`

Explanation:

Given : Sample size of healthy adults: n= 106

Degree of freedom = df =n-1 = 105

Sample mean body temperature :
`\overline{x}=98.2`

Sample standard deviation :
`s= 0.62`

Significance level ;
`\alpha= 1-0.99=0.01`

∵ population standard deviation is unknown , so we use t- critical value.

Confidence interval for the population mean :

`\overline{x}\pm t_(\alpha/2, df)(s)/(√(n))`

Using t-distribution table , we have

Critical value for df = 105 and
`\alpha=0.01`=
`t_(\alpha/2, df)=t_(0.005 , 105)=2.623`

A 99% confidence interval estimate of the mean body temperature of all healthy humans will be :

`98.2\pm (2.623)(0.62)/(√(106))`

`98.2\pm (2.623)(0.0602197234662)`

`98.2\pm (0.157956334652)`

`\approx98.2\pm 0.16`

`(98.2-0.16,\ 98.2+0.16)`

`(98.04,\ 98.36)`

Hence, a 99% confidence interval estimate of the mean body temperature of all healthy humans =
`(98.04,\ 98.36)`