Question

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

98.3 96.7 98.6 98.2 97.7 99.5 99.4 98

Assume body temperatures of adults are normally distributed. Based on this data, find the 80% 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.

80% C.I. =

Answer 1

Solution:

x	x2
98.3	9662.89
96.7	9350.89
98.6	9721.96
98.2	9643.24
97.7	9545.29
99.5	9900.25
99.4	9880.36
98	9604
x=786.4	x2=77308.88

The sample mean is

Mean   = (x / n) )

=98.3+96.7+98.6+98.2+97.7+99.5+99.4+988

=786.4 /8

=98.3

Mean   = 98.3

The sample standard is S

  S = ( x² ) - (( x)² / n ) n -1
= (77308.88-(786.4)²8 )7

=(77308.88-77303.12 / 7)

=5.76 / 7

=0.8229

=0.9071

The sample standard is =0.907

Degrees of freedom = df = n - 1 = 25 - 1 = 24

At 80% confidence level the t is ,

= 1 - 80% = 1 - 0.80 = 0.20

/ 2 = 0.20 / 2 = 0.10

t _/2,df = t_0.10,7 =1.415

Margin of error = E = t_/2,df * (s /n)

= 1.415 * (0.907 / 8)

= 0.454

Margin of error = 0.454

The 80% confidence interval estimate of the population mean is,

- E < < + E

98.3 - 0.454 < < 98.3 + 0.454

97.846 < < 98.754

(97.846 , 98.754)