Question

Q1. Assume a normal distribution with a known variance. Calculate the Lower Confidence Level (LCL) and Upper Confidence Level (UCL) for each of the following:

a. X-Bar = 50; n = 64; σ = 40; α = 0.05

LCL ==>

UCL ==>

b. X-Bar = 85; n = 225; σ2 = 400; α = 0.01

LCL ==>

UCL ==>

c. X-Bar = 510; n = 485; σ = 50; α = 0.10

LCL ==>

UCL ==>

Answer 1

Solution :

Given that,

a) Point estimate = sample mean = = 50

Population standard deviation = = 40

Sample size = n = 64

= 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

Margin of error = E = Z_/2* ( /n)

= 1.96 * (40 / 64)

= 9.8

- E < < + E

50 - 9.8 < < 50 + 9.8

40.2 < < 59.8

(40.2 , 59.8)

LCL ==> 40.2

UCL ==> 59.8

b) Point estimate = sample mean = = 85

variance = σ^{2 = 400}

Population standard deviation = = 20

Sample size = n = 225

= 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

Margin of error = E = Z_/2* ( /n)

= 2.576 * ( 20/ 225)

= 3.4

- E < < + E

85 - 3.4 < < 85 + 3.4

81.6 < < 88.4

(81.6 , 88.4)

LCL ==> 81.6

UCL ==> 88.4

c)Point estimate = sample mean = = 510

Population standard deviation = = 50

Sample size = n = 485

, = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

Margin of error = E = Z_/2* ( /n)

= 1.645 * ( 50 / 485)

= 0.2

- E < < + E

510 - 0.2 < < 510 + 0.2

509.8 < < 510.2

(509.8 , 510.2)

LCL ==> 509.8

UCL ==>510.2