How many boxes can a machine pack in 25 min if the machine packs boxes at a rate of 50 boxes per minute? Enter your answers into
2 answers:
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Answer:
1) the planning value for the population standard deviation is 10,000
2)
a) Margin of error E = 500, n = 1536.64 ≈ 1537
b) Margin of error E = 200, n = 9604
c) Margin of error E = 100, n = 38416
3)
As we can see, sample size corresponding to margin of error of $100 is too large and may not be feasible.
Hence, I will not recommend trying to obtain the $100 margin of error in the present case.
Step-by-step explanation:
Given the data in the question;
1) Planning Value for the population standard deviation will be;
⇒ ( 50,000 - 10,000 ) / 4
= 40,000 / 4
σ = 10,000
Hence, the planning value for the population standard deviation is 10,000
2) how large a sample should be taken if the desired margin of error is;
we know that, n = [ ( × σ ) / E ]²
given that confidence level = 95%, so = 1.96
Now,
a) Margin of error E = 500
n = [ ( × σ ) / E ]²
n = [ ( 1.96 × 10000 ) / 500 ]²
n = [ 19600 / 500 ]²
n = 1536.64 ≈ 1537
b) Margin of error E = 200
n = [ ( × σ ) / E ]²
n = [ ( 1.96 × 10000 ) / 200 ]²
n = [ 19600 / 200 ]²
n = 9604
c) Margin of error E = 100
n = [ ( × σ ) / E ]²
n = [ ( 1.96 × 10000 ) / 100 ]²
n = [ 19600 / 100 ]²
n = 38416
3) Would you recommend trying to obtain the $100 margin of error?
As we can see, sample size corresponding to margin of error of $100 is too large and may not be feasible.
Hence, I will not recommend trying to obtain the $100 margin of error in the present case.