Question

A simple random sample from a population with a normal distribution of 100 body temperatures has a mean of 98.40 and s=0.68 degree F. Construct a 90% confidence interval.

Answer 1

Solution :

Given that,

= 98.40

s = 0.68

n = 100

Degrees of freedom = df = n - 1 = 100 - 1 = 99

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

t _/2,df = t_0.05,99 = 1.660

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

= 1.660 * (0.68 / 100)

= 0.11

The 90% confidence interval estimate of the population mean is,

- E < < + E

98.40 - 1.11 < < 98.40 + 1.11

97.29 < < 99.51

(97.29 , 99.51)