416 words - 2 pages

Business Administration Discipline

Quantitative Techniques 2 Chapter 3

Tutorial Exercise 4 (Suggested Solution) 1. Let X be the value of the population.

∑ x = 129.4 ∑ x

n 1 s2 = n −1 x=

2

= 2027.86

n=9

∑ x = 129.4/9 = 14.38 (thousands)

(correct to 2 decimal places)

(∑ x

2

− nx 2 =

)

1 (2027.86 − (9)(14.3778) 2 ) = 20.92 (thousands) 8

ˆ σ 2 ≅ s2 = 20.92

(correct to 2 decimal places)

a)

b) 2.

ˆ μ ≅ x = 14.3778,

96% C.I. ...view middle of the document...

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.

Page 1 of 2

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)

Page 2 of 2

Find the perfect research document on any subject