Question - Statistical Comparison of New Technique's Mean Time to Failure

The question is asking to perform a hypothesis test to compare the mean times for cracks to appear for two different techniques at a 5% significance level. We will use a t-test for comparing two independent means with the assumption that both populations are normally distributed, and we assume equal variances since \(s_{\bar{x}}\) is the same for both techniques.

Null hypothesis, \(H_0\): \(\mu_1 = \mu_2\) (The mean time to failure is the same for both techniques.)

Alternative hypothesis, \(H_1\): \(\mu_1 < \mu_2\) (The mean time to failure for the new technique is greater than the current technique.)

We calculate the test statistic using the formula:

Where \(\bar{x}_1\) is the mean for the current technique, \(\bar{x}_2\) is the mean for the new technique, \(s_{\bar{x}}\) is the standard deviation of the sample means, \(n_1\) is the sample size for the current technique, and \(n_2\) is the sample size for the new technique. Plugging the values from the table:

The degrees of freedom in this case is the smaller of \(n_1 - 1\) and \(n_2 - 1\), so we use \(n_2 - 1 = 99\).

Now we compare the calculated t-value to the critical t-value from the t-distribution table at the 5% significance level with 99 degrees of freedom. If our calculated t-value is less than the critical t-value we reject the null hypothesis.

(b) Based on the results of the statistical test, if the null hypothesis is rejected, we can say that the new technique significantly increases the mean time before components crack and need to be replaced. However, if the null hypothesis cannot be rejected, then we cannot support the statement that the new technique significantly increases the time before failure.