Question

A manufacturer of meter sticks selected at random a certain number of pieces and measured them with a precision ruler. The mean length is found to be 1.045 m, with a standard deviation of 10.5 mm. Assuming the lengths of the parts are normally distributed, what is the 90% confidence for the mean for each of the following sample izes: a. 10 pieces b. 25 pieces c. 100 pieces

Answer 1

a) If n=10<30 hence t statistic used

The formula for estimation is:

μ = M ± t(s_M)

where:

M = sample mean
t = t statistic determined by confidence level
s_M = standard error = √(s²/n)

Calculation

M = 1.045
t = 1.83
s_M = √(10.5²/10) = 3.32

μ = M ± t(s_M)
μ = 1.045 ± 1.83*3.32
μ = 1.045 ± 6.08665

90% CI [-5.04165, 7.13165].

b) If n=25 again t statistic is used

Calculation

M = 1.045
t = 1.71
s_M = √(10.5²/25) = 2.1

μ = M ± t(s_M)
μ = 1.045 ± 1.71*2.1
μ = 1.045 ± 3.59285

90% CI [-2.54785, 4.63785].

c) Since n=100>30 hence here we can use Z statistic insted if t ststistic

The formula for estimation will become now as:

μ = M ± Z(s_M)

where:

M = sample mean
Z = Z statistic determined by confidence level
s_M = standard error = √(s²/n)

M = 1.045
Z= 1.64
s_M = √(10.5²/100) = 1.05

μ = M ± Z(s_M)
μ = 1.045 ± 1.64*1.05
μ = 1.045 ± 1.7271

90% CI [-0.6821, 2.7721]