Question

A meteorologist in Jenkins county is interested in estimating the average annual rainfall of the region. The annual rainfall is assumed to follow a normal distribution. She takes a sample of last 10 years and obtains an alterage annual rainfall of 39.5 inches and a standard deviation of 3.8 inches. Estimate a 95% confidence interval for the population mean. 39.5 +/-

Answer 1

Solution :

Given that,

Point estimate = sample mean = $\bar x$ = 39.5

sample standard deviation = s = 3.8

sample size = n = 10

Degrees of freedom = df = n - 1 = 10 - 1 = 9

At 95% confidence level the t is ,

$\alpha$ = 1 - 95% = 1 - 0.95 = 0.05

$\alpha$ / 2 = 0.05 / 2 = 0.025

t$\alpha$ /2,df = t0.025,9 = 2.262

At 95% confidence level of  $\mu$ is

$\bar x$$\pm$  t$\alpha$/2,df * (s /$\sqrt$n)

39.5 $\pm$ 2.262 * (3.8 / $\sqrt$ 10)

39.5 $\pm$ 2.7