Question

An experiment was performed to compare the wear of two different laminated materials. Twelve pieces of material 1 were tested by exposing each piece to a machine measuring wear. Ten pieces of material 2 were similarly tested. In each case, the depth of wear was observed. The samples of material 1 gave an average (coded) wear of 85 units with a sample standard deviation of 4, while the samples of material 2 gave an average of 81 with a sample standard deviation of 5. Can we conclude at the 0.05 level of significance that the abrasive wear of material 1 exceeds that of material 2 by more than two units? Assume the populations to be approximately normal with equal variances pdf

Answer 1

At 0.05 level of significance, the abrasive wear of material 1 exceeds that of material 2 by more than 2 units

Explanation:

We hypothesize that mean difference between abrasive wear of material 1 and material 2 is greater than 2.

So we write the null hypothesis
`H_0 : \mu_1 - \mu_2 >2`,

and the alternative hypothesis
`H_1: \mu_1 - \mu_2 \leq 2`.

We will find the T-score as well as the p-value. If the p-value is less than the level of significance, we will reject the null hypothesis, i.e. we will conclude that the abrasive wear of material 1 is less than that of material 2. Otherwise, we will accept the null hypothesis.

Since the variance is unknown and assumed to be equal, we will use the pooled variance

`s_p^2 = ((n_1-1)s_1^2 + (n_2-1)s_2^2)/(n_1+n_2-2) = 20.05`,

where
`n_1 = 12, n_2 =10, s_1 =4, s_2 = 5`.

The mean of material 1 and material 2 are
`\mu_1 =85, \mu_2=81` respectively and mean difference
`d` is equal to 4. The hypothesize difference
`d_0` is equal to 2.

To find the T-score, we use the following formula

`T = \frac{d - d_0}{\sqrt{(s_p^2)/(n_1) + (s_p^2)/(n_2) }}`

Substituting all the values into the T-score formula gives us
`T = 1.04`, and the respective p-value is equal to 0.31. This means we have enough statistical evidence not to reject the null hypothesis, and at 5% significance level, the abrasive wear of material 1 exceeds that of material 2 by more than 2 units.