Question

A random sample of 25 items is drawn from a population whose standard deviation is unknown. The sample mean (x-bar) is 850. Construct a confidence interval with 95% confidence for the different values of s. Do not forget to use the t distribution in this case. a. Assume s = 15. b. Assume s = 30. c. Assume s = 60. d. Describe how the confidence interval changes as s increases.

Answer 1

a)

sample mean, xbar = 850
sample standard deviation, s = 15
sample size, n = 25
degrees of freedom, df = n - 1 = 24

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 2.064

ME = tc * s/sqrt(n)
ME = 2.064 * 15/sqrt(25)
ME = 6.192

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (850 - 2.064 * 15/sqrt(25) , 850 + 2.064 * 15/sqrt(25))
CI = (843.81 , 856.19)

b)

sample mean, xbar = 850
sample standard deviation, s = 30
sample size, n = 25
degrees of freedom, df = n - 1 = 24

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 2.064

ME = tc * s/sqrt(n)
ME = 2.064 * 30/sqrt(25)
ME = 12.384

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (850 - 2.064 * 30/sqrt(25) , 850 + 2.064 * 30/sqrt(25))
CI = (837.62 , 862.38)

c)

sample mean, xbar = 850
sample standard deviation, s = 60
sample size, n = 25
degrees of freedom, df = n - 1 = 24

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 2.064

ME = tc * s/sqrt(n)
ME = 2.064 * 60/sqrt(25)
ME = 24.768

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (850 - 2.064 * 60/sqrt(25) , 850 + 2.064 * 60/sqrt(25))
CI = (825.23 , 874.77)

d)

As s increases width of interval increases