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 53 sample problems. The new algorithm completes the sample problems with a mean time of 19.15 hours. The current algorithm completes the sample problems with a mean time of 22.01 hours. The standard deviation is found to be 5.784 hours for the new algorithm, and 4.293 hours for the current algorithm. Conduct a hypothesis test at the 0.05 level of significance of the claim that the new algorithm has a lower mean completion time than the current algorithm. Let μ1 be the true mean completion time for the new algorithm and μ2 be the true mean completion time for the current algorithm. Step 1 of 4 : State the null and alternative hypotheses for the test.

Answer 1

Explanation:

This is a test of 2 independent groups. The population standard deviations are not known. it is a two-tailed test. Let μ1 be the true mean completion time for the new algorithm and μ2 be the true mean completion time for the current algorithm.

The random variable is μ1 - μ2 = difference in the mean completion time between the new algorithm and the current algorithm.

We would set up the hypothesis.

The null hypothesis is

H0 : μ1 ≤ μ2 H0 : μ1 - μ2 ≤ 0

The alternative hypothesis is

H1 : μ1 > μ2 H1 : μ1 - μ2 > 0

Since sample standard deviation is known, we would determine the test statistic by using the t test. The formula is

(μ1 - μ2)/√(s1²/n1 + s2²/n2)

From the information given,

μ1 = 19.15

μ2 = 22.01

s1 = 5.784

s2 = 4.293

n1 = 53

n2 = 53

t = (19.15 - 22.01)/√(5.784²/53 + 4.293²/53)

t = - 2.89

The formula for determining the degree of freedom is

df = [s1²/n1 + s2²/n2]²/(1/n1 - 1)(s1²/n1)² + (1/n2 - 1)(s2²/n2)²

df = [5.784²/53 + 4.293²/53]²/[(1/53 - 1)(5.784²/53)² + (1/53 - 1)(4.293²/53)²] = 0.958/0.0099876314

df = 100

We would determine the probability value from the t test calculator. It becomes

p value = 0.0023

Since alpha, 0.05 > than the p value, 0.0023, then we would reject the null hypothesis.