Question

The body temperatures in degrees Fahrenheit of a sample of adults in one small town are:

99.2

98.8

96.4

99.9

96.3

97.4

99.3

98.4

98.1

97.9

98.7

98.5

96.6

Assume body temperatures of adults are normally distributed. Based on this data, find the 95% 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.

95% C.I. = ________

Answer 1

ANSWER:

Given data,

99.2

98.8

96.4

99.9

96.3

97.4

99.3

98.4

98.1

97.9

98.7

98.5

96.6

Sample mean = = x / n = (99.2+98.8+96.4+99.9+96.3+97.4+99.3+98.4+98.1+97.9+98.7+98.5+96.6)/13 = 1275.5/13 = 98.1154

Sample standard deviation = s = sqrt( ( x -)^2/ (n-1))

= sqrt(((99.2-98.1154)^2+(98.8-98.1154)^2+(96.4-98.1154)^2+(99.9-98.1154)^2+(96.3-98.1154)^2+(97.4-98.1154)^2+(99.3-98.1154)^2+(98.4-98.1154)^2+(98.1-98.1154)^2+(97.9-98.1154)^2+(98.7-98.1154)^2+(98.5-98.1154)^2+(96.6-98.1154)^2)/(13-1)))

= 1.1509

c = 95% =95/100 = 0.95

= 1-c = 1-0.95 = 0.05

/2 = 0.05/2 = 0.025

df = n-1 = 13-1 = 12

Critical value :

t_/2,df = t_0.025,12 = 2.179

95% Confidence interval

t_/2,df * (s/sqrt(n))

98.1154 2.179 * (1.1509/sqrt(13))

98.1154 0.69554

95% CI = ( 98.1154 - 0.69554 , 98.1154 + 0.69554 )

95% CI = ( 97.41986 , 98.81094 )

95% CI = ( 97.419 , 98.811 ) (Rounded to three decimal places)

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