Question

A systems analyst tests a new algorithm designed to work faster than the currently-used algorithm. Each algorithm is applied to a group of 52 sample problems. The new algorithm completes the sample problems with a mean time of 16.34 hours. The current algorithm completes the sample problems with a mean time of 16.93 hours. The standard deviation is found to be 3.913 hours for the new algorithm, and 3.6243.624 hours for the current algorithm. Conduct a hypothesis test at the 0.01 level of significance of the claim that the new algorithm has a lower mean completion time than the current algorithm. Let μ1μ1 be the true mean completion time for the new algorithm and μ2 be the true mean completion time for the current algorithm.

do we reject or do we fail to reject?

Answer 1

Answer : Let X₁ denote the time taken to complete the sample problem by the new algorithm and X₂ denotes the time taken by the currently used algorithm.

Then, we are given,

n=52 16.34 hours   16.93 hours   

3.913 hours 3.624 hours

Let us now construct hypothesis for the given problem.

Since, we are given the sample size(n) to be 52( which is large ), hence we apply Normal or z-test.

Ho(Null Hypothesis): , that is, there is no significant difference between the average time taken by the new algorithm as well as the current algorithm while solving sample problems.

H1(Alternative Hypothesis): (Left-tailed test), the average time taken by the new algorithm is less than the time taken by the current algorithms in solving problems.

Test Statistic:

Under Ho, the test statistic (for large samples) is given by:

Here n₁=n₂=n(52) ,so on solving we get:

Z = -0.7977

Critical Region: For a one-tailed test, the critical value of Z at 1% level of significance is 2.3263. the critical region for left tailed test thus consists of all values of Z <= -2.3263.

Conclusion: Since the calculated value of Z(-0.7977) is more than the critical value(-2.3263), thus the null hypothesis is not rejected at 1% level of significance, hence, we conclude that the new algorithm does not has the lower mean completion time than the current algorithm.