An industrial engineer collected data on the labor time required to produce an order of automobile mufflers using a heavy stamping machine. The data on times (hours) for n = 52 orders of different parts 2.15 2.27 0.99 0.63 2.45 1.30 2.63 2.20 0.99 1.00 1.05 3.44 0.49 0.93 2.52 1.05 1.39 1.22 3.17 0.85 1.18 2.27 1.52 0.48 1.33 4.20 1.37 2.70 0.63 1.13 3.81 0.20 1.08 2.92 2.87 2.62 1.03 2.76 0.97 0.78 4.68 5.20 1.90 0.55 1.00 2.95 0.45 0.70 2.43 3.65 4.55 0.33 has x = 1.865 hours and s2 = 1.5623 so s = 1.250 hours. What can one assert with 95% confidence about the maximum error if x = 1.865 hours is used as a point estimate of the true population mean labor time required to run the heavy stamping machine?
To determine the maximum error (also known as the margin of error) for a 95% confidence interval when using the sample mean (( \bar{x} )) as a point estimate of the true population mean (( \mu )), we can use the formula for the margin of error in the context of a t-distribution, given that the sample size is relatively small (n = 52) and the population standard deviation is unknown.
The formula for the margin of error (E) is:
[ E = t_{\alpha/2} \cdot \frac{s}{\sqrt{n}} ]
where:
Given:
First, we need to find the critical value ( t_{\alpha/2} ) for a 95% confidence level with ( n-1 = 51 ) degrees of freedom. Using a t-table or a calculator, we find that ( t_{0.025, 51} \approx 2.009 ).
Now, we can calculate the margin of error:
[ E = 2.009 \cdot \frac{1.250}{\sqrt{52}} ]
[ E = 2.009 \cdot \frac{1.250}{7.211} ]
[ E = 2.009 \cdot 0.1733 ]
[ E \approx 0.348 ]
Therefore, with 95% confidence, the maximum error when using ( \bar{x} = 1.865 )