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How many work cycles should be timed to estimate the average cycle time to within 5 percent of the sample mean with a confidence of 99.0 percent if a pilot study yielded these times (minutes): 4.0, 4.5, 5.0, 4.5, 4.5, and 4.0

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Answer:

Following are thew solution to this question:

Step-by-step explanation:

Please find the complete question in the attached file.

Using formula:

[tex]\text{Margin of error} = \frac{(z \times \sigma) }{\sqrt{n}}[/tex]

n = the number of work cycles for 95 confidence interval

z-score = 1.96

So, [tex]\frac{(z \times \sigma)}{\sqrt{n}}[/tex] = [tex]\ Mean \times 0.02[/tex]  

[tex]\frac{(1.96 \times 0.376)}{\sqrt{n}} = 4.416666667 \times 0.02 \\\\\\[AVERAGE(4,4.5,5,4.5,4.5,4) = 4.416666667][/tex]

[tex]\sqrt{n} = \frac{(1.96 \times 0.376)}{(4.416666667 \times 0.02)} \\\\[/tex]

     [tex]= 8.342943396\\\\[/tex]

[tex]n = 69.6047045[/tex]

   [tex]= 70[/tex]

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Using the sample size formula, the number of samples required is 919.

  • Standard deviation, σ = 0.376

From the data given :

  • Mean = (4 + 4.5 + 5 + 4.5 + 4.5 + 4) / 6 = 4.417
  • Standard deviation = 0.376

Using the sample size relation thus:

  • [tex] n = (\frac{T_{critical}σ}{E})^{2}[/tex]

  • E = Margin of Error = 5% = 0.05
  • Tcritical at 99%, df = 5 = 4.032 (T distribution table)

Substituting the values into the equation :

[tex] n = (\frac{4.032 \times 0.376 }{0.05})^{2}[/tex]

[tex] n = (\frac{1.516032}{0.05})^{2} = 919.34[/tex]

Hence, the sample size required is 919.

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