Question

The article "The Influence of Corrosion Inhibitor and Surface Abrasion on the Failure of Aluminum-Wired Twist-On Connections" (IEEE Trans. on Components, Hybrids, and Manuf. Tech., 1984: 20-25) reported data on potential drop measurements for one sample of connectors wired with alloy aluminum and another sample wired with EC aluminum. Does the accompanying SAS output suggest that the true average potential drop for alloy connections (type 1) is higher than that wfor EC connections (as stated in the article)? Carry out the appropriate test using a significance level of .01. In reaching your conclusion, what type of error might you have committed?

(Note: SAS reports the\(P\)-value for a two-tailed test.)

\(\begin{array}{*{20}{c}}{ Type }&{}&N&{ Mean }&{ Std Dev }&{ Std Error }\\1&{20}&{17.49900000}&{0.55012821}&{0.12301241}&{}\\2&{20}&{16.90000000}&{0.48998389}&{0.10956373}&{}\\{}&{ Variances }&T&{ DF }&{Prob > |T|}&{}\\{ Unequal }&{3.6362}&{37.5}&{0.0008}&{}&{}\\{}&{ Equal }&{3.6362}&{38.0}&{0.0008}&{}\end{array}\)

Short answer

Reject the null hypothesis.

Step-by-step solution

Testing the hypotheses

Let\({\mu _1}\)denotes the true average potential drop for alloy connections and\({\mu _2}\)denotes the true average for the EC connections.

The hypotheses of interest are\({H_0}:{\mu _1} - {\mu _2} = 0\)versus\({H_a}:{\mu _1} - {\mu _2} > 0\).

Testing this hypothesis, the results provided in the exercise, when the variances are unequal (see exercise output), is

\(t = 3.6362,df = 37.5\).

The pvalue for two tailed test is 0.0008.

The the p- value for the upper tailed test can be computed as

\(P = \frac{1}{2} \cdot 0.0008 = 0.0004\)

Since

\(\begin{array}{c}P = 0.0004\\ < 0.01\end{array}\)

reject the null hypothesis at 1% level of significance.

The type I error could have been made- rejecting the null hypothesis when it is true.