question archive Based on last week's report, the average number of processed documents per hour was 15
Based on last week's report, the average number of processed documents per hour was 15
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Based on last week's report, the average number of processed documents per hour was 15.11, with a standard deviation of 2.666. That is, one document was reviewed in 238.25 seconds. To be objective as much as possible, the manager spoke with an employee whose average was exactly 15 documents per hour. The employee claimed that if she was given a larger monitor, the processing time would be shorter.
They conducted an experiment with a large monitor and measured processing time. After reviewing 20 documents, the calculated average processing time per document was 190.58 seconds. The manager believes that a bigger monitor helped reduce the processing time for reviewing foreclosure documents. Conduct a hypothesis test using a 95% confidence level, which means that significance level a = 0.05.
Use the 5-step process, and explain each term or concept mentioned in each section in the following.
Step 1: Set Up Null and Alternative Hypotheses
Based on the request description, explain if a one-tailed or two-tailed test is needed. If a one-tailed test is needed, is it a left or right-tailed test? Please explain why one alternative is better than the other.
State both of the following hypotheses:
- Null hypothesis
- Alternative hypothesis
You will need the following information to progress to Step 2:
- Standard deviation: Explain what standard deviation is. Locate the calculated standard deviation in the assignment description, and enter here.
- Random variable: Explain what a random variable is. Locate it in the assignment description, and enter here.
- Test type: Compare and contrast t-test and z-test. Once done, determine which one is appropriate for the experiment given the fact that the sample size is less than 30.
Step 2: Decide the Level of Significance
Explain what the significance level is, and determine whether the one used in the assignment description is high, medium, or low. What does this significance level tell you about this test? Locate the level of significance in the given scenario, and list it in this step.
Significance level = ?
Determine the degree of freedom based on the number of reviewed documents in the new experiment (n = 20) and based on the formula Degree of freedom = n - 1.
Degree of freedom = ?
Critical value = (You will need to use the t-table and find the intersection point between the degree of freedom and the alpha value that is provided in the assignment description.)
Step 3: Calculate the Test Statistics
Calculate the test statistics based on the test type determined in Step 1.
If the determination was done correctly, you should use this formula to calculate the test statistics.
Test statistics = ?
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Answer:
Step 1
Hypotheses are:
H0:Mean reviewing time is 238.25 seconds.
H1:Mean reviewing time <238.25 seconds.
This is one tail test and left tailed since H1:μ<15.11
An alternative is determined according to the study. Here the investigator tested that large monitors reduce effect in reviewing time. That is, time is considered only in one direction which is less than 238.25 sec.
Step 2
α=0.05 which is medium.
n=20
Degrees of freedom(d.f.) based on t test =20-1
=19
Critical value for t-test =1.729 with 19 d.f.
Z table value =1.645
Step3
An appropriate test is the Z-test.
Test statistic =6.34
Step 4
Critical region is z<- zα
Here 6.34>-1.645
Step 5
Do not reject H0.
Conclusion: A larger monitor does not show any effect in reviewing time.
Step-by-step explanation
Given, the mean number of documents per hour processed is 15.11, and its standard deviation(SD) is 2.666.
From this information, we can say that processing time for reviewing is 238.25 sec.
Sample size =20
Sample mean= Average processing time
=190.58
That is, average number of processed documents =60*60/190.58
=18.8897 per hour
Hypotheses are:
H0:Mean of reviewing time is 238.25 seconds.
H1:mean of reviewing time <238.25 seconds.
That is,
H0:μ=15.11
H1:μ<15.11
This is one tail test and left tailed since H1:μ<15.11.
An alternative is determined based on the study. Here the investigator tested that large monitors reduce effect in reviewing time. That is, time is considered only in one direction which is less than 238.25 sec.
- Population SD =2.666(given)
- Let X be a number of documents processed per hour and assume that it follows normal X~N(15.11,2.6662).
- T-test is applied when population SD is unknown. The test statistic is:
-
Step 4
Critical region is z<- zα
Here, 6.34>-1.645
Step 5:
Do not reject H0.
Hence, larger monitors do not show any effect in reviewing time.
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