Question

Suppose that I take a sample of 10 observations from a Normal distribution whose true mean ('mu') and standard deviation ('sigma') are unknown. If x-bar and s are the mean and standard deviation of the sample, which expression gives a 95% confidence interval for mu?

x-bar plus and minus 1.96*sigma/square-root(10)

x-bar plus and minus 1.96*sigma/square-root(10)

x-bar plus and minus 2.262*s/square-root(10)

x-bar plus and minus 2.228*s/square-root(10)

x-bar plus and minus 2.262*s/square-root(9)

Answer 1

Solution :

Given that,

Point estimate = sample mean =

sample standard deviation = s

sample size = n = 10

Degrees of freedom = df = n - 1 = 9

t _/2,df = 2.262

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

The 95% confidence interval estimate of the population mean is,

  t_/2,df * (s /n) = 2.262 * (s /10)