Question

An engineer wants to test two different ways to transfer large data volumes over the internet. The first method resulted in an average time of 4.7 seconds with a calculated standard deviation of 0.3 seconds, after 40 tests. The second method resulted in an average time of 4.5 seconds with a calculated standard deviation of 0.25 seconds, after 40 tests also. He needs to prove that the second method is significantly faster than the first. State the hypotheses that the engineer should test. Which test should be used to test the hypotheses?

Answer 1

The provided sample means are shown below: X1 = 4.7 X2 = 4.5 Also, the provided sample standard deviations are: $1 = 0.3 82 = 0.25 and the sample sizes are n1 = 40 and n2 = 40. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: Mi sue Ha: Mi > 2 This corresponds to a right-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.05, and the degrees of freedom are df = 75.543. In fact, the degrees of freedom are computed as follows, assuming that the population variances are unequal: Hence, it is found that the critical value for this right-tailed test is te = 1.665, for a = 0.05 and df = 75.543

The rejection region for this right-tailed test is R={t:t > 1.665). (3) Test Statistics Since it is assumed that the population variances are unequal, the t-statistic is computed as follows: X1 - X2 t=2 Vt 4.7 – 4.5 = 3.239 10:32 +0.25 (4) Decision about the null hypothesis Since it is observed that t = 3.239 >te = 1.665, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p = 0.0009, and since p = 0.0009 < 0.05, it is concluded that the null hypothesis is rejected. (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that population mean uj is greater than um, at the 0.05 significance level.

Hence, there is enough evidence to conclude that second method is significantly faster than the first at 5% level of significance.