Question

The body temperatures in degrees Fahrenheit of a sample of 6 adults in one small town are:

97.9

99.6

98.6

98.5

99.5

97.7

Assume body temperatures of adults are normally distributed. Based on this data, find the 98% confidence interval of the mean body temperature of adults in the town. Enter your answer as an open-interval (i.e., parentheses) accurate to twp decimal places (because the sample data are reported accurate to one decimal place).

98% C.I. =

Answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places.

Answer 1

Solution :

Given that 97.9,99.6,98.6,98.5,99.5,97.7

=> Mean x = sum of terms/number of terms

= 591.8/6

= 98.6333

=> standard deviation s = 0.7891

=> n = 6 , df = n - 1 = 5

=> The critical value for 98% confidence level at df = 5 can be found using the t-table. That is
t(0.02/2,5) = 3.365

=> The 98% confidence interval of the mean is

=> x +/- t*s/sqrt(n)

=> 98.6333 +/- 3.365*0.7891/sqrt(6)

=> (97.5493 , 99.7173)

=> (97.55 , 99.72) (rounded)