Asample 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 )

Step-by-step explanation:

Given : Sample size of healthy adults: n= 106

Degree of freedom = df =n-1 = 105

Sample mean body temperature  :

Sample standard deviation :

Significance level ;

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

Confidence interval for the population mean :

Using t-distribution table , we have

Critical value for df = 105 and =

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

Hence, a 99% confidence interval estimate of the mean body temperature of all healthy humans =

gmt is 4 hrs ahead of edt so just add four hours to 1: 14: 20 (minutes and seconds stay the same)

gmt is 5: 14: 20

hope this !

~just a girl in love with shawn mendes

40 i don’t know lololol