Question

A random sample of 20 items is drawn from a population whose standard deviation is unknown. The sample mean is x¯x¯ = 930 and the sample standard deviation is s = 5. Use Appendix D to find the values of Student’s t.

(a) Construct an interval estimate of μ with 98% confidence. (Round your answers to 3 decimal places.)
  
The 98% confidence interval is from  to

(b) Construct an interval estimate of μ with 98% confidence, assuming that s = 10. (Round your answers to 3 decimal places.)
  
The 98% confidence interval is from  to

(c) Construct an interval estimate of μ with 98% confidence, assuming that s = 20. (Round your answers to 3 decimal places.)
  
The 98% confidence interval is from  to

Answer 1

a)

sample mean, xbar = 930
sample standard deviation, s = 5
sample size, n = 20
degrees of freedom, df = n - 1 = 19

Given CI level is 98%, hence α = 1 - 0.98 = 0.02
α/2 = 0.02/2 = 0.01, tc = t(α/2, df) = 2.539

ME = tc * s/sqrt(n)
ME = 2.539 * 5/sqrt(20)
ME = 2.839

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (930 - 2.539 * 5/sqrt(20) , 930 + 2.539 * 5/sqrt(20))
CI = (927.161 , 932.839)

The 98% Confidence interval is from 927.161 to 932.839

b)

sample mean, xbar = 930
sample standard deviation, s = 10
sample size, n = 20
degrees of freedom, df = n - 1 = 19

Given CI level is 98%, hence α = 1 - 0.98 = 0.02
α/2 = 0.02/2 = 0.01, tc = t(α/2, df) = 2.539

ME = tc * s/sqrt(n)
ME = 2.539 * 10/sqrt(20)
ME = 5.677

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (930 - 2.539 * 10/sqrt(20) , 930 + 2.539 * 10/sqrt(20))
CI = (924.323 , 935.677)

The 98% Confidence interval is from 924.323 to  935.677
c)

sample mean, xbar = 930
sample standard deviation, s = 20
sample size, n = 20
degrees of freedom, df = n - 1 = 19

Given CI level is 98%, hence α = 1 - 0.98 = 0.02
α/2 = 0.02/2 = 0.01, tc = t(α/2, df) = 2.539

ME = tc * s/sqrt(n)
ME = 2.539 * 20/sqrt(20)
ME = 11.355

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (930 - 2.539 * 20/sqrt(20) , 930 + 2.539 * 20/sqrt(20))
CI = (918.645 , 941.355)
The 98% Confidence interval is from 918.645 to 941.355