Art Snakeoil, the tool manufacturer’s sales representative, now wishes to introduce a new line of pulse tools to the Purchasing Department. He indicates that historically his company has determined that the mean output torque of his line of pistol grip pulse tools is x = 42.2 Newton-Meters (Nm)

Question

Art Snakeoil, the tool manufacturer’s sales representative, now wishes to introduce a new line of pulse tools to the Purchasing Department. He indicates that historically his company has determined that the mean output torque of his line of pistol grip pulse tools is $\overline{x} = 42.2$ Newton-Meters (Nm) with a standard deviation σ = 5.0 Nm. However, Art Snakeoil can provide no data on the population mean.

The Purchasing Department requests that Engineering provide information on the statistical range of torque output for these pulse tools, using, in this case, a level of significance of α = 0.01 and a sample size of n = 32.

Accepted Answer

Approach:
1. [latex]H_0: μ ≠ μ_0[/latex] (Note that this indicates a two-tailed test.)
2. [latex]H_a: μ = μ_0[/latex]
3. α = 0.01 (Level of significance)
4. n is large (n = 32)
σ is known σ = 5.0

[latex]\overline{ x} = 42.2[/latex]

μ is unknown
5. n = 32
6. Find for [latex]α = 0.01→α/2 = 0.005 = Z_{0.005} = 2.576[/latex]
From Table 4.2 we find:
[latex]Z_{0.005} = 2.576[/latex]
The range of tool output will be determined by the relationship:

[latex]\overline{ x} − Z_{α/2} imes \sigma /\sqrt{n}\lt\mu \lt \overline{ x} +Z_{α/2} imes \sigma /\sqrt{n}[/latex]

The general form of the sketch for use above is given by:

And we find

[latex]42.2-\left(2.576 ight)\frac{5.0}{\sqrt{32} }\lt \mu \lt 42.2-\left(2.576 ight)\frac{5.0}{\sqrt{32} }[/latex]

or

[latex]39.92\lt \mu \lt 44.48[/latex]

Sketching this data yields:

That is, based on the data provided the mean will range between 39.92 and 44.48 Nm, as shown in the sketch.
Therefore, at a confidence level of 0.01 we can anticipate the torque output of this group of pistol grip pulse tools to range from 39.92 to 44.48 Nm. Stated differently, we can say we are 99% confident that the unknown population mean output torque lies within the interval of 39.92–44.48 Nm.

TABLE 4.2
$z_{P}$ Corresponding to P for the Normal Curve, z is the Standard Normal Variable [3]

p	0	0.01	0.02	0.03	0.04	0.05	0.06	0.07	0.08	0.09
0	–	−2.33	−2.05	−1.88	−1.75	−1.64	−1.55	−1.48	−1.41	−1.34
0.1	−1.28	−1.23	−1.18	−1.13	−1.08	−1.04	−0.99	−0.95	−0.92	−0.88
0.2	−0.84	−0.81	−0.77	−0.74	−0.71	−0.67	−0.64	−0.61	−0.58	−0.55
0.3	−0.52	−0.50	−0.47	−0.44	−0.41	−0.39	−0.36	−0.33	−0.31	−0.28
0.4	−0.25	−0.23	−0.20	−0.18	−0.15	−0.13	−0.10	−0.08	−0.05	−0.03
0.5	0	0.03	0.05	0.08	0.1	0.18	0.15	0.18	0.2	0.23
0.6	0.25	0.28	0.31	0.33	0.36	0.39	0.41	0.44	0.47	0.5
0.7	0.52	0.55	0.58	0.61	0.64	0.67	0.71	0.74	0.77	0.81
0.8	0.84	0.88	0.92	0.95	0.99	1.04	1.08	1.13	1.18	1.23
0.9	1.28	1.34	1.41	1.48	1.55	1.64	1.75	1.88	2.05	2.33

Question 4.DS.7: Art Snakeoil, the tool manufacturer’s sales representative, ......

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