Question

The body temperatures in degrees Fahrenheit of a sample of adults in one small town are: 97.1 99.3 99.9 99.5 97.5 96.4 99.1 96.6 99.4 98.7 97.6 98.1 Assume body temperatures of adults are normally distributed. Based on this data, find the 90% confidence interval of the mean body temperature of adults in the town. Enter your answer as an open-interval (i.e., parentheses) accurate to 3 decimal places. Assume the data is from a normally distributed population. 90% C.1. =

Answer 1

type all data in excel

to find sample mean = AVERAGE (choose the array)

sample standard deviation = SQRT(VAR.S(choose the array))

from the codes we get ,

sample mean = x bar = 98.26667

sample SD = s =1.210059

n=12

here population SD is unknown so we use t confidence interval

degrees of freedom = 11

alpha = 0.1

Based on the provided information, the critical t-value for alpha = 0.1 and df = 11 degrees of freedom is t_c = 1.796

The 90% confidence for the population mean mu is computed using the following expression

CI =( X bar - [ {t_c * s}/{sqrtn}] , X bar + [ {t_c * s}/{sqrtn}] )

Therefore, based on the information provided, the 90 % confidence for the population mean mu is

CI = (98.26667 - { 1.796 *1.210059}/{sqrt 12} , 98.26667 +{ 1.796*1.210059}/{sqrt{ 12} )

CI=(98.26667−(12​1.796×1.210059)​ , 98.26667+(12​1.796×1.210059​))

= (98.26667 - 0.627, 98.26667 + 0.627)

=(98.26667−0.627,98.26667+0.627)

= (97.639, 98.894)

=(97.639,98.894)

which completes the calculation.

so 90% confidence interval = (97.639,98.894)

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