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 72 sample problems. The new algorithm completes the sample problems with a mean time of 24.68 hours. The current algorithm completes the sample problems with a mean time of 27.14 hours. Assume the population standard deviation for the new algorithm is 4.299 hours, while the current algorithm has a population standard deviation of 4.903 hours. 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.

Step 2 of 4: Compute the value of the test statistic. Round your answer to two decimal places.

Step 3 of 4: Determine the decision rule for rejecting the null hypothesis H0. Round the numerical portion of your answer to three decimal places.

Step 4 of 4: Make the decision for the hypothesis test.

A. Reject null hypothesis

B. Fail to reject null hypothesis

Answer 1

1) Null hypothesis:
`\mu_(1) \geq \mu_(2)`

Alternative hypothesis:
`\mu_(1) < \mu_(2)`

2)
`z=\frac{24.68-27.14}{\sqrt{(4.299^2)/(72)+(4.903^2)/(72)}}}=-3.201`

3) For this case we are conducting a left tialed test so we need to find a quantile on the normal standard distribution that accumulates 0.05 of the area on the left tail and on this case is:

`z_(crit)= -1.64`

And our rjection zone for the null hypothesis would be Z< -1.64

4) A. Reject null hypothesis

Explanation:

Data given and notation

`\bar X_(N)=24.68` represent the mean for the sample with new algorithm

`\bar X_(C)=27.14` represent the mean for the sample with the current algorithm

`\sigma_(N)=4.299` represent the population standard deviation for the sample for the new algorithm

`\sigma_(C)=4.903` represent the population standard deviation for the sample for the current algorithm

`n_(N)=72` sample size selected for the New algorithm

`n_(C)=72` sample size selected for the Current algorithm

`\alpha=0.05` represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

`p_v` represent the p value for the test (variable of interest)

Part 1: State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if new algorithm has a lower mean completion time than the current algorithm

Null hypothesis:
`\mu_(1) \geq \mu_(2)`

Alternative hypothesis:
`\mu_(1) < \mu_(2)`

If we analyze the size for the samples both are higher than 30 and we know the population deviations so for this case is better apply a z test to compare means, and the statistic is given by:

`z=\frac{\bar X_(1)-\bar X_(2)}{\sqrt{(s^2_(1))/(n_(1))+(s^2_(2))/(n_(2))}}` (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Part 2: Calculate the statistic

We can replace in formula (1) the info given like this:

`z=\frac{24.68-27.14}{\sqrt{(4.299^2)/(72)+(4.903^2)/(72)}}}=-3.201`

Part 3: Decision rule and P-value

For this case we are conducting a left tialed test so we need to find a quantile on the normal standard distribution that accumulates 0.05 of the area on the left tail and on this case is:

`z_(crit)= -1.64`

And our rjection zone for the null hypothesis would be Z< -1.64

Since is a left side test the p value would be:

`p_v =P(Z<-3.201)=0.00069`

Part 4: Conclusion

As we can see our calculated value is on the rejection zone so then the best conclusion would be:

A. Reject null hypothesis