Question

Refer to the Scheer Industries example in Section 8.2. Use 6.83 days as a planning value for the population standard deviation.

Answer 1

Answer:

a)Sample size  would be required to obtain a margin of error of 1 days is

n = 179

b) sample size would be required to obtain a margin of error of 2.5 days is n = 20

Step-by-step explanation:

step(i):-

Given Population standard deviation  = 6.83 days

a) Given margin of error = 1 day

The margin of error is determined by

$M.E = \frac{Z_{\alpha } S.D }{\sqrt{n} }$

Step(ii):-

Given 95 % of confidence level

Now the critical value Z₀.₀₅ = 1.96

$1 = \frac{1.96 X 6.83 }{\sqrt{n} }$

√n = 13.38

Squaring on both sides, we get

n =  179.206

b)

step(i):-

a) Given margin of error = 2.5 day

The margin of error is determined by

$M.E = \frac{Z_{\alpha } S.D }{\sqrt{n} }$

Step(ii):-

Given 90 % of confidence level

Now the critical value Z₀.₁₀ = 1.645

$2.5= \frac{1.645 X 6.83 }{\sqrt{n} }$

Cross multiplication , we get

$\sqrt{n} = \frac{1.645 X 6.83 }{2.5 }$

√n   = 4.494

Squaring on both sides, we get

n = 20.19

Final answer:-

a)Sample size  would be required to obtain a margin of error of 1 days is

n = 179

b) sample size would be required to obtain a margin of error of 2.5 days is n = 20