Chabra industries is considering a new computer-assisted program to train maintenance employees to do machine repairs.
In order to fully evaluate the program, the director of manufacturing requested an estimate of the population mean time
required for maintenance employees to complete the computer assisted training. Use 7.64 days as a planning value for the
population standard deviation. (Round your answers up to the nearest whole number.)
(a) Assuming confidence, what sample size would be required to obtain a margin of error of 0.5 days?
(b) If the precision statement was made with confidence, what sample size would be required to obtain a margin of
error of 3 days?

Question

Chabra industries is considering a new computer-assisted program to train maintenance employees to do machine repairs.
In order to fully evaluate the program, the director of manufacturing requested an estimate of the population mean time
required for maintenance employees to complete the computer assisted training. Use 7.64 days as a planning value for the
population standard deviation. (Round your answers up to the nearest whole number.)
(a) Assuming

confidence, what sample size would be required to obtain a margin of error of 0.5 days?
(b) If the precision statement was made with

confidence, what sample size would be required to obtain a margin of
error of 3 days?

UpStudy AI · Accepted Answer

**(a)** For a $$95\%$$ confidence level, the sample size $$n$$ is determined by the formula

$$
n = \left( \frac{z_{\alpha/2} \sigma}{E} \right)^2,
$$

where  
$$z_{\alpha/2}$$ is the critical value,  
$$\sigma = 7.64$$ days is the planning value for the standard deviation, and  
$$E = 0.5$$ days is the desired margin of error.

For $$95\%$$ confidence, the critical value is $$z_{0.025} \approx 1.96$$. Substituting these values gives

$$
n = \left( \frac{1.96 \times 7.64}{0.5} \right)^2.
$$

First, calculate the numerator:

$$
1.96 \times 7.64 \approx 14.97.
$$

Then, the expression becomes

$$
n = \left( \frac{14.97}{0.5} \right)^2 = (29.94)^2 \approx 896.4.
$$

Since $$n$$ must be rounded up to the nearest whole number, we have

$$
n = 897.
$$

**(b)** For a $$90\%$$ confidence level, using the same formula

$$
n = \left( \frac{z_{\alpha/2} \sigma}{E} \right)^2,
$$

with $$ \sigma = 7.64$$ days and now $$E = 3$$ days. For $$90\%$$ confidence, the critical value is $$z_{0.05} \approx 1.645$$. Substituting these values gives

$$
n = \left( \frac{1.645 \times 7.64}{3} \right)^2.
$$

First, calculate the numerator:

$$
1.645 \times 7.64 \approx 12.57.
$$

Then, the expression becomes

$$
n = \left( \frac{12.57}{3} \right)^2 \approx (4.19)^2 \approx 17.56.
$$

Rounding up to the nearest whole number, we have

$$
n = 18.
$$
**(a)** To achieve a $$95\%$$ confidence level with a margin of error of 0.5 days, the required sample size is **897**. **(b)** For a $$90\%$$ confidence level and a margin of error of 3 days, the necessary sample size is **18**.

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