Question

Each algorithm is applied to a group of 52 sample problems. The new algorithm completes the sample problems with a mean time of 24.04 hours. The current algorithm completes the sample problems with a mean time of 24.52 hours. The standard deviation is found to be 5.769 hours for the new algorithm, and 5.556 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.

Answer 1

Answer:

Therefore we fail to reject the Null hypothesis $H_0$

Because the is no sufficient evidence to support the claim of the alternative hypothesis $H_a$

Step-by-step explanation:

From the question we are told that

Sample size $n_1 =52$

Mean time 1 $\=x_1=24.04\ hours$

Mean time 2 $\=x_2=24.52\ hours$

Standard deviation $\partial _1=5.769$

Standard deviation $\partial _2=5.556$

Level of significance 0.05

The null hypothesis is $H_o: \mu_1=\mu_2$

Alternative hypothesis $H_a: \mu_1$

Generally the test statistics is mathematically represented as

      $Z= \frac{\=x_1-\=x_2}{\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}}$

     $Z= \frac{24.04-24.52}{\sqrt{\frac{5.769^2}{52}+\frac{5.5562^2}{52}}}$

     $Z= -0.4321583379$

     $Z= -0.43$

From Table the Z-critical value at $\alpha =0.05\ is\ -1.6449$

Therefore

Zcal> Zcritical

(-0.43)=(-1.645)

Generally we fail to reject the null hypothesis $H_0$ at $\alpha = 0.05$

Therefore we fail to reject the$H_0$

Because the is no sufficient evidence to support the claim of $H_a$