An initial time study resulted in an average observed time of 2.2 minutes per cycle, and a standard deviation of 0.3 minutes per cycle. The performance rating was 1.20. What sample size, including the 20 observations in the initial study, would be necessary to have a confidence of 95.44 percent that the observed time was within 4 percent of the true value?

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Answer:

The sample size required is 225.

Step-by-step explanation:

The confidence interval for population mean (μ) is:

[tex]\bar x\pm z_{\alpha /2}\times \frac{\sigma}{\sqrt{n}}[/tex]

The margin of error is:

[tex]MOE= z_{\alpha /2}\times \frac{\sigma}{\sqrt{n}}[/tex]

Given:

[tex]\bar x=2.2\\\sigma=0.3[/tex]

The margin of error is, MOE = 0.04

The confidence level is 95.44%.

The critical value of z for 95.44% confidence interval is:

[tex]P(-2\leq Z\leq 2)=0.9544[/tex]

So, [tex]z_{\alpha /2}=2[/tex].

Determine the sample size as follows:

[tex]MOE= z_{\alpha /2}\times \frac{\sigma}{\sqrt{n}}\\0.04=2\times\frac{0.30}{\sqrt{n}}\\ n=(15)^{2}\\=225[/tex]

Thus, the sample size required is 225.

You can find the margin of error and then use it to find the needed confidence interval.

The sample size needed for specified condition is 225

How to find the margin of error and confidence interval for population mean?

The margin of error is given by

[tex]MOE = Z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}[/tex]

And the confidence interval for the population mean [tex]\mu[/tex] is

[tex]CI = \overline{x} \pm MOE\\\\CI = \overline{x} \pm Z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}[/tex]

where we have

[tex]\overline {x} = \text{sample mean}\\n = \text{sample size}\\\alpha = \text{level of significance}\\\sigma = \text{population standard deviation}[/tex]

Since we have:

[tex]\overline {x} = 2.2\\\alpha = 100 - 95.44\% = 4.66\% = 0.0466 \\\sigma = 0.3[/tex]

Critical value of z at level of significance 0.0466 is

[tex]Z_{\alpha/2} = Z_{0.0233} = \pm 1.99[/tex] (two tailed) (from critical value calculator)

The margin of error is 0.04 since the observed time was within 4% = 0.04 width(0.02 on left and 0.02 on right)

Thus, using the MOE (margin of error) formula to get sample size:

[tex]MOE = Z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}\\\\0.04= \pm 1.99 \times \dfrac{0.3}{\sqrt{n}}\\n = (\dfrac{0.597}{0.04})^2 = (14.925)^2 \approx 225[/tex]

Thus,

The sample size needed for specified condition is 225

Learn more about confidence interval here:

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