Question

During a study about “dry” and “wet” drilling in rock, six holes were drilled, three corresponding to each process. In a dry hole compressed air down the drill flushes the cutting and the drive hammer, whereas in a wet hole one forces water. With significance level a = 0.05, n = m = 80, x_bar = 805.53sec and y_bar = 943.82sec, (sn)2 = 23768.09, and (sm)2 = 15327, is wet or dry drilling faster?

Answer 1

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u_Dry> u_Wet
Alternative hypothesis: u_Dry < u_Wet

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the mean difference between sample means is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = sqrt[(s₁²/n₁) + (s₂²/n₂)]
SE = 22.1063
DF = 158

t = [ (x₁ - x₂) - d ] / SE

t = - 6.26

where s₁ is the standard deviation of sample 1, s₂ is the standard deviation of sample 2, n₁ is thesize of sample 1, n₂ is the size of sample 2, x₁ is the mean of sample 1, x₂ is the mean of sample 2, d is the hypothesized difference between population means, and SE is the standard error.

The observed difference in sample means produced a t statistic of -6.26.

Therefore, the P-value in this analysis is less than 0.001.

Interpret results. Since the P-value (almost 0) is less than the significance level (0.05), we have to reject the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that wet drilling is faster than dry drilling.