Let X be the value of the population. σ = 0.8, a) b) c) x = 5.1 n = 38, σ σx = = 0.8/√38 = 0.1298 n x ± (1)σ x = 5.1 ± 0.1298 = (4.97, 5.23)
(correct to 2 decimal places)
Confidence level the interval is 68.26%
3.
Let X be the value of the population. a) b) c) d) x ± Z 0.2 σ x x ± Z 0.1 σ x x ± Z 0.05 σ x x ± Z 0.02 σ x = = = = x ± 0.84σ x x ± 1.28σ x x ± 1.64σ x x ± 2.05σ x
4.
Let X be the waiting time of the customer. a) b) c) σ = 5/2 = 2.5, x = 25 x ± Z0.025 σ = 25 ± 1.96 (2.5) = (20.10, 29.90) minutes σ = 5/3 = 1.6667, x = 15 x ± Z0.025 σ = 15 ± 1.96 (1.6667) = (11.73, 18.27) minutes Confidence interval is used to estimate a population parameter.
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Business Administration Discipline
Quantitative Techniques 2 Chapter 3
Tutorial Exercise 4 (Suggested Solution) (cont.)
5.
Let X be the value of the population. s = 6.1, a) x = 20.9, N = 360, n = 35
σx =
σ
n ˆ x = 1.0311 σ
≅
6.1 s = n 35
b)
96% C.I for μ: = x ± Zα/2 σ x = x ± Z0.02 (6.1/√35) = 20.9 ± 2.05 (1.0311) (correct to 2 decimal places) = (18.79, 23.01)
6.
Let X be the tire pressure. s = 2.1, a) b) c) n = 62,
ˆ σ = s = 2.1
x = 24
ˆ σx =
ˆ σ n
=
s n
=
2.1 62
= 0.2667
95% C.I for μ: = x ± Zα/2 σ x = x ± Z0.025 (2.1/√62) = 24 ± 1.96 (0.2667) (correct to 2 decimal places) = (23.48, 24.52)
