H0 (null hypothesis): μ1 = μ2 = μ3 = μ4 = μ5 (all the population means are equal) H1 (null hypothesis): at least one population mean is different from the rest Where μ1, μ2, μ3, μ4, μ5 are average daily returns (on the stock you are assigned) on MON, TUE, WED, THR, and FRI respectively. Please explain your answe
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H0 (null hypothesis): μ1 = μ2 = μ3 = μ4 = μ5 (all the population means are equal)
H1 (null hypothesis): at least one population
Where μ1, μ2, μ3, μ4, μ5 are average daily returns (on the stock you are assigned) on MON, TUE, WED, THR, and FRI respectively. Please explain your answer.
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- Suppose a researcher studied sleeping patterns in second-, fourth-, and sixth-grade children, and there were 7 students in each group, showing sleep periods in minutes below. Grade 2nd 4th 6th 580 525 562 570 575 532 594 497 565 543 478 537 552 517 525 506 455 493 537 532 490 Mean minutes participants slept 562.57 527.00 505.43 Suppose the researcher found significant ANOVA test result: the observed F statistic is 7.14 at alpha= .05 level, and the researcher made the decision to reject the null hypothesis. Finish up the ANOVA test with your interpretation in terms of the research question (you do not need to perform the test, just provide interpretation based the information available). Your interpretation: Given MSerror= 41, conduct the Tukey’s HSD post-hoc test to determine which groups differed significantly from one another. Include your calculation work, and show each comparison and state the outcome of each…Hamstrings Athletes who had suffered hamstringinjuries were randomly assigned to one of two exerciseprograms. Those who engaged in static stretchingreturned to sports activity in a mean of 15.2 days fasterthan those assigned to a program of agility and trunkstabilization exercises. (Journal of Orthopaedic & SportsPhysical Therapy 34 [March 2004]: 3)a) Explain why it was important to assign the athletes tothe two different treatments randomly.b) There was no control group consisting of athleteswho did not participate in a special exercise program.Explain the advantage of including such a group.c) How might blinding have been used?d) One group returned to sports activity in a mean of37.4 days 1SD = 27.6 days2 and the other in a meanof 22.2 days 1SD = 8.3 days2. Do you think thisdifference is statistically significant? Explain.The data processing department of the Arizona Bank has five data entry clerks. Each working day their supervisor verifies the accuracy of a random sample of 250 records. A record containing one or more errors is considered defective and must be redone. The results of the last 30 samples are shown in the table. All were checked to make sure that none was out of control. Sample Number of DefectiveRecords Sample Number of DefectiveRecords Sample Number of DefectiveRecords Sample Number of DefectiveRecords 1 2 3 4 5 6 7 8 7 5 19 10 11 8 12 9 9 10 11 12 13 14 15 16 6 13 18 5 16 4 11 8 17 18 19 20 21 22 23 12 4 6 11 17 12 6 24 25 26 27 28 29 30 Total 7 13 10 14 6 11 9 300 a. Based on these historical data, set up a p-chart using z = 3.b. Samples for the next 4 days showed the following: Sample Number of Defective Records Tues Wed Thurs Fri 17 15 22 21 What is the supervisor’s assessment of the data-entry process likely to be?
- Suppose that the average waiting time for a patient at a physician's office is just over 29 minutes. In order to address the issue of long patient wait times, some physicians' offices are using wait-tracking systems to notify patients of expected wait times. Patients can adjust their arrival times based on this information and spend less time in waiting rooms. The following data show wait times (minutes) for a sample of patients at offices that do not have a wait-tracking system and wait times for a sample of patients at offices with a wait-tracking system. Without Wait-Tracking System With Wait-TrackingSystem 24 14 60 9 10 33 22 10 34 18 47 30 12 12 12 17 27 9 30 3 (a) Considering only offices without a wait-tracking system, what is the z-score for the 10th patient in the sample (wait time = 30 minutes)? If required, round your intermediate calculations and final answer to two decimal places. z-score = (b) Considering only offices…Suppose that the average waiting time for a patient at a physician's office is just over 29 minutes. In order to address the issue of long patient wait times, some physicians' offices are using wait-tracking systems to notify patients of expected wait times. Patients can adjust their arrival times based on this information and spend less time in waiting rooms. The following data show wait times (minutes) for a sample of patients at offices that do not have a wait-tracking system and wait times for a sample of patients at offices with a wait-tracking system. Without Wait-Tracking System With Wait-TrackingSystem 20 8 56 12 12 11 35 16 26 32 48 36 16 8 10 9 27 15 36 5 (a) Considering only offices without a wait-tracking system, what is the z-score for the 10th patient in the sample (wait time = 36 minutes)? If required, round your intermediate calculations and final answer to two decimal places. z-score = (b) Considering only offices…A bank with a branch located in a commercial district of a city has the business objective of developing an improved process for serving customers during the noon-to-1 p.m. lunch period. Management decides to first study the waiting time in the current process. The waiting time is defined as the time that elapses from when the customer enters the line until he or she reaches the teller window. Data are collected from a random sample of 35 customers. Suppose that another branch, located in a residential area, is also concerned with improving the process of serving customers in the noon-to-1 p.m. lunch period. Data are collected from a random sample of 40 customers. The obtained data from the two branches are stored. At the 5% significance level, is there evidence of a difference in the mean waiting time between the two branches? What is/are implications of the result?
- A Deputy Registrar at a certainty university conducted a Chi-Square test of association to establish whether or not employee grade and level of absenteeism were associated. Table 2 below shows part of the results which were obtained. Table 2: Cross tabulation of employee grade and level of absenteeism Contingency Tables Grade LAbsence 1 2 3 Total low Observed 8 1 0 9 Expected 2.90 3.48 2.61 9.00 high Observed 2 11 9 22 Expected 7.10 8.52 6.39 22.00 Total Observed 10 12 9 31 Expected 10.00 12.00 9.00 31.00 2.1 State the appropriate measurement scales for the variables. 2.2 State the null and alternative hypotheses. 2.3 Calculate the expected frequency corresponding to a Grade 3 academic with a high level of absenteeism. 2.4 Find the Chi-Square critical value and test statistic.…A Deputy Registrar at a certainty university conducted a Chi-Square test of association to establish whether or not employee grade and level of absenteeism were associated. Table 2 below shows part of the results which were obtained. Table 2: Cross tabulation of employee grade and level of absenteeism Contingency Tables Grade LAbsence 1 2 3 Total low Observed 8 1 0 9 Expected 2.90 3.48 2.61 9.00 high Observed 2 11 9 22 Expected 7.10 8.52 6.39 22.00 Total Observed 10 12 9 31 Expected 10.00 12.00 9.00 31.00 2.2 State the null and alternative hypotheses.A Deputy Registrar at a certainty university conducted a Chi-Square test of association to establish whether or not employee grade and level of absenteeism were associated. Table 2 below shows part of the results which were obtained. Table 2: Cross tabulation of employee grade and level of absenteeism Contingency Tables Grade LAbsence 1 2 3 Total low Observed 8 1 0 9 Expected 2.90 3.48 2.61 9.00 high Observed 2 11 9 22 Expected 7.10 8.52 6.39 22.00 Total Observed 10 12 9 31 Expected 10.00 12.00 9.00 31.00 2.3 Calculate the expected frequency corresponding to a Grade 3 academic with a high level of absenteeism.
- A Deputy Registrar at a certainty university conducted a Chi-Square test of association to establish whether or not employee grade and level of absenteeism were associated. Table 2 below shows part of the results which were obtained. Table 2: Cross tabulation of employee grade and level of absenteeism Contingency Tables Grade LAbsence 1 2 3 Total low Observed 8 1 0 9 Expected 2.90 3.48 2.61 9.00 high Observed 2 11 9 22 Expected 7.10 8.52 6.39 22.00 Total Observed 10 12 9 31 Expected 10.00 12.00 9.00 31.00 2.4 Find the Chi-Square critical value and test statistic. Hence, conduct a Chi-Square test of association at 5% level of significance clearly stating your decision and conclusion.A Deputy Registrar at a certainty university conducted a Chi-Square test of association to establish whether or not employee grade and level of absenteeism were associated. Table 2 below shows part of the results which were obtained. Table 2: Cross tabulation of employee grade and level of absenteeism Contingency Tables Grade LAbsence 1 2 3 Total low Observed 8 1 0 9 Expected 2.90 3.48 2.61 9.00 high Observed 2 11 9 22 Expected 7.10 8.52 6.39 22.00 Total Observed 10 12 9 31 Expected 10.00 12.00 9.00 31.00 2.1 State the appropriate measurement scales for the variables.The Occupational Safety and Health Administration (OSHA) mandates certain regulations that have to be adopted by corporations. Prior to the implementation of the OSHA program, a company found that for a sample of 40 randomly selected months, the mean employee time lost due to job-related accidents was 45 hours. After implementation of the OSHA program, for a random sample of 45 months, the mean employee time lost due to job-related accidents was 39 hours. It can be assumed that the variability of time lost due to accidents is about the same before and after implementation of the OSHA program (with a standard deviation being 3.5 hours). Find a 90% confidence interval for the difference in the mean time lost due to accidents. Test the hypothesis that implementation of the OSHA program has reduced the mean employee lost time. Use a level of significance of 0.10.
