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 46 sample problems. The new algorithm completes the sample problems with a mean time of 18.78 hours. The current algorithm completes the sample problems with a mean time of 19.06 hours. The standard deviation is found to be 5.614 hours for the new algorithm, and 5.012 hours for the current algorithm. Conduct a hypothesis test at the 0.1 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.A. Reject null hypothesis
B. Fail to reject null hypothesis

Answer 1

Answer:

B. Fail to reject null hypothesis .

Step-by-step explanation:

We are given that a systems analyst tests a new algorithm designed to work faster than the currently-used algorithm.

Also, $\mu_1$ = true mean completion time for the new algorithm

        $\mu_2$ = true mean completion time for the current algorithm

Null Hypothesis, $H_0$ : $\mu_1 = \mu_2$  {Both new and current algorithm has same

                                                            completion time}

Alternate Hypothesis, $H_1$ : $\mu_1 < \mu_2$ {New algorithm has lower mean

                                                           completion time than current algorithm}

The test statistics we use here will be :

              $\frac{(X_1bar - X_2bar)- (\mu_1 - \mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ follows $t_n__1 + n_2-2$

where, $X_1bar$ = 18.78 hours   and      $X_2bar$ = 19.06 hours

              $s_1$ = 5.614 hours         and      $s_2$ = 5.012 hours

              $n_1$ = 46                        and      $n_2$ = 46

              $s_p = \sqrt{\frac{(n_1-1)s_1^{2} + (n_2-1)s_2^{2} }{n_1 + n_2 - 2} }$ = 5.321

Here, we use t test statistics because we know nothing about population standard deviations.

     Test statistics = $\frac{(18.78 - 19.06)- 0}{5.321\sqrt{\frac{1}{46}+\frac{1}{46} } }$ follows $t_9_0$

                             = -0.2524

At 0.1 or 10% level of significance t table gives a critical value between -1.296 and -1.289 at 90 degree of freedom. Since our test statistics is more than the critical table value of t as -0.2524 > -1.296 to -1.289 so we have insufficient evidence to reject null hypothesis.

Therefore, we conclude that new algorithm has same mean completion time with that of current algorithm.