Question

When σ is unknown and the sample is of size n ≥ 30, there are two methods for computing confidence intervals for μ. Method 1: Use the Student's t distribution with d.f. = n − 1. This is the method used in the text. It is widely employed in statistical studies. Also, most statistical software packages use this method. Method 2: When n ≥ 30, use the sample standard deviation s as an estimate for σ, and then use the standard normal distribution. This method is based on the fact that for large samples, s is a fairly good approximation for σ. Also, for large n, the critical values for the Student's t distribution approach those of the standard normal distribution. Consider a random sample of size n = 41, with sample mean x = 44.9 and sample standard deviation s = 6.5.

(a) Compute 90%, 95%, and 99% confidence intervals for μ using Method 1 with a Student's t distribution. Round endpoints to two digits after the decimal.

90% 95% 99%

lower limit 43.19 42.85 42.16   

upper limit   46.61 46.95 47.65

(b) Compute 90%, 95%, and 99% confidence intervals for μ using Method 2 with the standard normal distribution. Use s as an estimate for σ. Round endpoints to two digits after the decimal.

90% 95% 99%

lower limit 43.23 42.91 42.28

upper limit 46.57 46.89 47.52

(c) Compare intervals for the two methods. Would you say that confidence intervals using a Student's t distribution are more conservative in the sense that they tend to be longer than intervals based on the standard normal distribution?

No. The respective intervals based on the t distribution are shorter.

No. The respective intervals based on the t distribution are longer.

Yes. The respective intervals based on the t distribution are shorter.

Yes. The respective intervals based on the t distribution are longer. (This answer is correct)

(d) Now consider a sample size of 71. Compute 90%, 95%, and 99% confidence intervals for μ using Method 1 with a Student's t distribution. Round endpoints to two digits after the decimal.

90% 95% 99%

lower limit

upper limit

(e) Compute 90%, 95%, and 99% confidence intervals for μ using Method 2 with the standard normal distribution. Use s as an estimate for σ. Round endpoints to two digits after the decimal.

90% 95% 99%
lower limit

upper limit

Answer 1

d) sample Mean = 44.9
t critical= 1.67
sM = √(6.5^2/71) = 0.77

μ = M ± t(sM)
μ = 44.9 ± 1.67*0.77
μ = 44.9 ± 1.286
90% CI [43.61, 46.19].

You can be 90% confident that the population mean (μ) falls between 43.61 and 46.19.

M = 44.9
t critical = 1.99
sM = √(6.5^2/71) = 0.77

μ = M ± t(sM)
μ = 44.9 ± 1.99*0.77
μ = 44.9 ± 1.53
95% CI [43.37, 46.43].

You can be 95% confident that the population mean (μ) falls between 43.37 and 46.43.

M = 44.9
t critical = 2.65
sM = √(6.5^2/71) = 0.77

μ = M ± t(sM)
μ = 44.9 ± 2.65*0.77
μ = 44.9 ± 2.0405
99% CI [42.86, 46.94].

You can be 99% confident that the population mean (μ) falls between 42.86 and 46.94.

e) Sample Mean = 44.9
Z critical= 1.64
s_M = √(6.5²/71) = 0.77
μ = M ± Z(s_M)
μ = 44.9 ± 1.64*0.77
μ = 44.9 ± 1.263

90% CI [43.64, 46.16].

You can be 90% confident that the population mean (μ) falls between 43.64 and 46.16.

Z Critical = 1.96
sM = √(6.5^2/71) = 0.77

μ = M ± Z(sM)
μ = 44.9 ± 1.96*0.77
μ = 44.9 ± 1.512
95% CI [43.39, 46.41].

You can be 95% confident that the population mean (μ) falls between 43.39 and 46.41

Z critical = 2.58
sM = √(6.5^2/71) = 0.77

μ = M ± Z(sM)
μ = 44.9 ± 2.58*0.77
μ = 44.9 ± 1.987
99% CI [42.91, 46.89].

You can be 99% confident that the population mean (μ) falls between 42.91 and 46.89.