Question 4.DS.10: Two separate assembly lines in the same manufacturing plant ......
Two separate assembly lines in the same manufacturing plant are utilized to apply 18 Nm torque to threaded fasteners. The torque process can be assumed to be normal, with standard deviations of σ_{1} = 0.019 and σ_{2} = 0.021. Upper management and the legal office insist that both assembly lines must torque to the same value, whether or not this torque value is 18 Nm. Due to a number of serious fastener failures occurring in the plant’s product, an outside expert is retained to evaluate the torque process reliability. He takes random samples from each line and evaluates the torque on each line to find the following:
Assembly Line 1: \overline{x}_{1} = 18.259; \sigma _{1} = 0.019; n_{1} = 32
Assembly Line 2: \overline{x}_{2} = 18.250; \sigma _{2} = 0.021; n_{2} = 32
He follows the format below when performing the statistical experiment:
a. State the appropriate hypothesis for this experiment.
b. Test the hypothesis using α = 0.01. State the conclusions.
c. Determine the 99% confidence interval on the difference in mean torque level for the two assembly lines.
Approach:
1. H_{0}: μ_{1} = μ_{2} H_a: μ_{1} ≠ μ_{2} (Indicates a Two-Tailed test)
2. α = 0.01
3. \overline{x}_{1} = 18.259 \overline{x}_{2} = 18.250 From the given data
σ_{1} = 0.019 σ_{2} = 0.021 σ_{1}, σ_{2} known
n_{1} = 32, n_{2} = 32 n large Α use Z statistic
From Table 4.2 find for Z_{a /2} = Z_{0.05} = 1.645
Hence, since z_{0} = 1.798 is outside the bound of 1.645, we reject H_{0}. Hence, we note that there is a statistical difference between the two assembly lines, contrary to the insistence of upper
management and the corporate attorneys. The 99% confidence interval on the difference on the torque level for the lines is given by
\left(\overline{x}_{1}-\overline{x}_{2}\right) -Z_{a/2}\sqrt{\frac{\sigma ^{2}_{1} }{n_{1}}+\frac{\sigma ^{2}_{2}}{n_{2}}} \leq \left(\mu _{1}-\mu _{2}\right) \leq+\left(\overline{x}_{1}-\overline{x}_{2}\right) -Z_{a/2}\sqrt{\frac{\sigma ^{2}_{1} }{n_{1}}+\frac{\sigma ^{2}_{2}}{n_{2}}}Substituting for values then yields
\left(18.259-18.250\right)-1645\sqrt{\frac{0.019^{2}}{32}+\frac{0.021^{2}}{32} }\leq \left(\mu _{1}-\mu _{2}\right) \leq \left(18.259-18.250\right)+1645\sqrt{\frac{0.019^{2}}{32}+\frac{0.021^{2}}{32} }\\0.0008 \leq \left(\mu _{1}-\mu _{2}\right) \leq 0.0174Hence, at a confidence level of 0.01, we can anticipate that the sample combined mean will fall between 0.0008 and 0.0174. Stated otherwise, we are 99% confident that the combined mean difference will range from 0.0008 to 0.0174.
| TABLE 4.2 | ||||||||||
| z_{P} Corresponding to P for the Normal Curve, z is the Standard Normal Variable [3] | ||||||||||
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| p | 0.00 | 0.01 | 0.02 | 0.03 | 0.04 | 0.05 | 0.06 | 0.07 | 0.08 | 0.09 |
| 0.00 | – | −2.33 | −2.05 | −1.88 | −1.75 | −1.64 | −1.55 | −1.48 | −1.41 | −1.34 |
| 0.10 | −1.28 | −1.23 | −1.18 | −1.13 | −1.08 | −1.04 | −0.99 | −0.95 | −0.92 | −0.88 |
| 0.20 | −0.84 | −0.81 | −0.77 | −0.74 | −0.71 | −0.67 | −0.64 | −0.61 | −0.58 | −0.55 |
| 0.30 | −0.52 | −0.50 | −0.47 | −0.44 | −0.41 | −0.39 | −0.36 | −0.33 | −0.31 | −0.28 |
| 0.40 | −0.25 | −0.23 | −0.20 | −0.18 | −0.15 | −0.13 | −0.10 | −0.08 | −0.05 | −0.03 |
| 0.50 | 0.00 | 0.03 | 0.05 | 0.08 | 0.10 | 0.18 | 0.15 | 0.18 | 0.20 | 0.23 |
| 0.60 | 0.25 | 0.28 | 0.31 | 0.33 | 0.36 | 0.39 | 0.41 | 0.44 | 0.47 | 0.50 |
| 0.70 | 0.52 | 0.55 | 0.58 | 0.61 | 0.64 | 0.67 | 0.71 | 0.74 | 0.77 | 0.81 |
| 0.80 | 0.84 | 0.88 | 0.92 | 0.95 | 0.99 | 1.04 | 1.08 | 1.13 | 1.18 | 1.23 |
| 0.90 | 1.28 | 1.34 | 1.41 | 1.48 | 1.55 | 1.64 | 1.75 | 1.88 | 2.05 | 2.33 |
