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Use the numbering format in our answers (1a, 1b, 2a, 2b,etc) that you see below. You can copy and paste the example below and then modify.
Item 1: One- tail (upper): Label 1a, 1b, 1c etc. (use format below)
1a. Preparation: Explain the problem you would test for (make up numbers) and state the hypotheses in both numeric and narrative form. Explain why this would be a upper tailed test. At this time you have not collected any data in your pretend scenario. State what course of action your management team will make if (1) Ho is rejected.
1b: Sampling: explain exactly what is the population of interest and the basic procedures you will take to select a sample from the population. Don’t be vague, describe the mechanics of selecting the data and the sample size (n).
1c. The data: pretend you now have the sample; State the test statistic and the rejection value for alpha at .05. Just make these up but be sure they are consistent with either rejecting nor not rejecting Ha. Be sure to use the words, “fail to reject” or reject Ho and a one sentence of the management action you will take.
Do no calculations or list any formulas. Justify why this would be worth testing so stay away from frivolous examples. Include you sampling methodology and management decision you will make if you can accept Ha (see the paragraph above Item 1).
Item 2: One-tail (lower) Use the same format as above, 2a, 2b, 2c.
Item 3: Two- tail: Use the same format as above, 3a, 3b, 3c. Explain clearly why this must be a 2T test (see module3, lesson4, and textbook). I suggest base it on a quality control problem. If you care about something be too little or too much do NOT use 2T! Do no calculations or list any formulas. Justify why this would be worth testing so stay away from frivolous examples. Include you sampling methodology and management decision you will make if you can accept Ha (see the paragraph above Item 1). Many students do poorly on this because they do not simply think of a QC scenario. Caution: often students do poorly on item 3 because they are not using a QC example.
Answer three more questions:
Item 4: In a short answer (if possible) explain why as the preceding forms show, the equality part of the expression (either >=, <=, or =) always appears in the null hypothesis. Why is this true?
Item 5: What do/may Type I and Type II errors have to do with the type of test you might conduct (upper or lower tail) or the selection of the level of significance?
Item 6: Explain why the phrase “we accept the null hypothesis (Ho) is always is an incorrect statement.”
example another student did
It has come to my attention that the desire for more flavors has been brought up by our customers on numerous occasions. Currently only offering 5 main flavors and a special flavor of the week, we have been seriously considering expanding our menu. In order to do this, however, we would need to raise our prices in order to compensate for the possible loss we could experience in offering more variety. We currently offer our cupcakes at a base price of $12 a cupcake and have not experienced an issue in moving the product off our shelves. We would like to expand our menu to 15 flavors and a flavor of the day. We expect this change to affect the budgets for personnel, perishable supplies, and equipment. With all of that taken into consideration, not knowing exactly how much sales will increase from this change, we would need to sell our cupcakes for at least $14.50. We would like to conduct a short survey with the current customers at each of our 5 locations and the patrons who conduct business within the shopping district around the locations. We would like to see if our current customers and possible future customers would be willing to pay $14.50 for customized cupcake if offered more variety. If the null hypothesis is rejected we will expand our menu to add more variety to our customized cupcakes. Since we are wanting to acquire more customers with this transition, we would like to set our alpha to .05 so to minimize the probability of a Type 1 error. The hypothesis for the one-tailed test are as follows:
Hypothesis (numeric expression)
Ho: u=$14.50
Ha: u>$14.50
Narrative
Ho: population mean is equal to $14.50 the maximum dollar amount customers are willing to pay for customized cupcakes with more variety.
Ha: population mean is greater than $14.50 the maximum dollar amount customers are willing to pay for customized cupcakes with more variety.
This is a one-tailed test upper since we are interested in a population mean being greater than a value of $14.50 which would then be to the right half of the distribution and the far right would be the rejection region, alpha. If we reject the null hypothesis, then we would like to expand our menu to add more variety.
1b: We will be conducting a survey of all the customers at each of our 5 establishments and surrounding locations. The survey within the establishment will be set up by the exit/entrance of our facilities. We will place a printed survey with the following questions next to a drop box for them to drop the survey when it is complete.
As a ‘thank you’ for participating in our survey, customers who fill out the survey in our establishment will receive a free topping of their choice on their next purchase. The questions are as follows: 1) How often do you visit any of our ‘Snacker’s Cupcakes and more’ locations? 2.How would you rate our current flavors?(1-best, 5-worst) 3. How much would you pay for a customized cupcake with more variety? 4. What is your favorite flavor cupcake? (please include icing and any toppings). To collect the opinions of patrons in the surrounding locations we will drop off a stack of 20 printed coupons good for 50% off their next purchase with a survey on the back at local establishments who are willing to post them in a place for their customers and employees to see. When they come in to redeem their coupon we will be able to collect the survey. The costs of the survey will be around $150 for each establishment, including the loss of profits from the coupons. The survey will be collected for the entire month of November to capture a large number of possible participants. The minimum sample size will be 350. Our population of interest are the patrons who conduct business at or around our 5 establishments. Possible bias include non-response, incomplete, or unreadable surveys collected.
1c. Throughout the month of November we collected a total of 516 surveys. The main question of interest from our survey was question number 3 ‘How much would you be willing to pay for a customized cupcake with more variety?’ The alpha value for a upper one-tailed test at .05 is z=1.645. We will then reject if our test statistic is z>1.645. Since our computed test statistic from our actual data is z=2.105 (sample mean of our data is $15.25, the mean response on how much a customer would be willing to pay for more variety), we reject the null hypothesis because our test statistic (z=2.105) is greater than the z score of our alpha level of .05 (z=1.645). We therefore accept the alternative hypothesis and recommend to management (see part 1a) that we expand our menu and offer more variety to our customers.
A presentation from the Manager of Operations at Target to the Board of Directors
2a. With the recent activity in our economy, it has been apparent for several months that our sales have been down drastically compared to the last five years. With this inevitable continual decrease in sales, I see it necessary to decrease the amount of merchandise on hand and the personnel employed at 17 of our domestic locations in order to counter the effects of the loss of sales we will experience. We would like to conduct an analysis of the sales volume for the next 60 days to determine if a 20% reduction in inventory volume and a 15% reduction in personnel costs will be necessary. For this test, I have designated personnel at each of the 17 domestic locations experiencing a current decrease in sales to analyze and record the daily sales for the months of June and July. The mean daily sales from 2006 until 2011 were $8,925,000 combined. If the mean daily sales drops below $7,000,000 we will have no choice but to adjust our inventory and personnel to fit our new budget. We will set an alpha level of .05 in order to decrease the chance of making a Type 1 error. This will also be considered a one-tailed test. Since we are only concerned if our sales drop it will be a lower tailed test with our rejection region being below the set alpha level of .05.
Hypothesis (numeric expression)
Ho: u=$7,000,000
Ha: u<$7,000,000
Narrative
Ho: sales mean is equal to $7,000,000; the minimum mean daily sales amount between June and July of the 17 locations, combined.
Ha: population mean is less than $7,000,000; the minimum mean daily sales amount between June and July of the 17 locations, combined.
This is a one-tailed test lower since we are interested in a sales mean less than $7,000,000. The left tail, below .05, would be the rejection region, alpha. If we reject the null hypothesis then we will be forced to decrease our volume of inventory and personnel costs.
2b. To conduct this analysis, the designees at each of the 17 locations will be monitoring and recording the daily sales their store achieves. This amount is in the daily report that is already pulled on a nightly basis after closing the store so no extra cost will be accrued for this process. We will also be implementing more promotions in the community in order to try to generate more revenue to decrease the chances of down-sizing, causing people to be laid off indefinitely. At the end of the two months, we will combine the means from the 17 locations for the month of June and also for the month of July. If the combined mean for both June and July is below $7,000,000 we will send store managers their notifications on August 15th of what actions will be taking place effective September 1st. One bias we must keep in mind while conducting this analysis is the time we will be analyzing the data. We experience a higher volume of guests during the summer months due to families visiting from out of town. This could produce more sales than what could be expected in the coming months. However, we will not be adjusting the data to compensate for this because we expect some of our regular customers to travel out of town as well and their sales will not be considered in the collections of data. We feel this will even itself out.
2c. Throughout the months of June and July we conducted an analysis of the daily sales at 17 of our domestic locations. The test statistic for the alpha value for a lower one-tailed test at .05 is z=-1.645. We will then reject our null hypothesis if our test statistic is z<-1.645. Since our computed test statistic from our actual data for June is z=-1.857 and July is z=-1.952 (June mean daily sales = $6,589,250 and July mean daily sales =$5,985,157), we reject the null hypothesis because our test statistics (z=-1.857 and z=-1.952) are both less than the z score of our alpha level of .05 (z=-1.645). We, therefore, accept the alternative hypothesis and inform the managers of each location the actions that must take place. They must decrease inventory volume by 20% and personnel costs by 15% in order to counter the decrease in sales.
A presentation from the Branch Managers of Bank of America in New Hampshire to the Board of Executives of Operations and Human Resources Manager
3a. With only five months left in the year and more than 65% of our budgeted goal to be met, we feel it is necessary to implement an incentive program to motivate our employees to work harder towards their individual goals. In order to encourage our employees to achieve a higher volume of loan dollars, we will decide on an appropriate incentive program when they achieve a certain level of success. We would like to conduct an anonymous survey of a randomized list of employees from the branches in New Hampshire. The survey will help determine what levels of success will warrant an incentive. If the mean dollar amount of loans achieved last year was more than the goal they are set to achieve this year, the incentive will be designer gifts such as purses and watches. If the mean dollar amount of loans achieved last year was lower than the goal they are set to achieve this year, they will receive additional paid vacation or cash compensation. The goal they are set to achieve is $1,200,000 per person. We will set an alpha level of .05 in order to decrease the chance of making a Type 1 error. This will also be considered a two-tailed test. Since we are measuring the possibility of the achievement on either side of the spectrumb we will consider both the upper and lower possibilities.
Hypothesis (numeric expression)
Ho: u=$1,200,000
Ha: u> $1,200,000 or u<$1,200,000
Narrative
Ho: mean of the employees total disbursed loan dollars from last year equals the goal the employees are set to achieve this year. ($1,200,000)
Ha: mean of the employees total disbursed loan dollars last year is greater than $1,200,000 or less than $1,200,000.
This is a two-tailed test since we are interested in a the possibility of mean being less than or greater than $1,200,000, which would then be to the left and right sides of the distribution. The extreme left tail and extreme right tail will count as rejection regions defined by our alpha level of .05. If we reject the null hypothesis, then we will determine which incentive to implement based on the tail the test statistic falls in.
3b. To conduct this survey, a randomized, computer-generated list of 250 branch employees from the New Hampshire locations that were here in 2012 will be emailed. This email will contain four questions; 1. How long have you been employed with Bank of America. 2. Which department do you work in. 3. What was your total dollar amount of loans disbursed for 2012. 4. Would a cash or gift compensation motivate you to achieve more? The results from this email will be gathered From June 1st through the 10th in order to gather information from part time workers and those who may be on leave between those dates. Possible bias include non-response, incomplete, or misread/typo’s in the responses. The employee will simply reply to the email. The data from the emails will be compiled and calculated by the team of branch managers from New Hampshire. Once the data is collected, we can then determine if an incentive is necessary, and if so, which type. This survey will take each employee about two minutes to complete, not incurring any costs to the company. To create the survey and calculate the data received, it will take about 4 hours. Since the management team is exempt and receive salary, not hourly, wages, this will not cost the company any extra funds.
3c. The results were compiled and calculated. The test statistic for the alpha value for a two-tailed test at 0.05 is z=-1.960 and z=1.960. (Since this is a two-tailed test, we must divide alpha by two to set the rejection region for the upper and lower tails of the distribution.) From the surveys collected, the mean dollar amount of loans disbursed for 2012 was $1,054,962. The test statistic that we calculate from the data is z=-2.901. We will then reject our null hypothesis since our calculated statistic is within the rejection region z<-1.960. Since the mean was below $1,200,000.00, the incentive that will be implemented is cash or vacation compensation. This will be implemented Starting July 1st. Any employee to achieve $130,000 a month in loan dollars disbursed will qualify for the incentive. By achieving the incentive every month (July through December) they will meet their individual goal for the year. The vacation time will start with 8 hours and will increase by the percentage of the excess met each month. For example, if the employee achieves 20% above the incentive, they receive 20% more vacation time. If the employee doesn’t want the vacation time, they may choose to take the cash compensation.
Item 4: In a short answer (if possible) explain why as the preceding forms show, the equality part of the expression (either >=, <=, or =) always appears in the null hypothesis. Why is this true?
The use of an equality statement is used with the null hypothesis due to the fact that there has to be a generally accepted value or “status quo” to test against. This Ho is meant to represent the mean of all the sample means or sample proportions. The mean of anything is a well-defined number, not a range of numbers.
Item 5: What do/may Type I and Type II errors have to do with the type of test you might conduct (upper or lower tail) or the selection of the level of significance?
A type I error occurs when you reject the null hypothesis even though it is true. A type II error occurs when you accept the null hypothesis even though it is false. Whichever sided test you conduct, upper or lower tail or two-tailed, an alpha (type I) or beta (type II) can be conducted. In the three examples above, and typically in statistics, a type I error will be calculated because it is more likely to occur. The main importance is selecting your alpha level when conducting the type I error test. If you do not choose a high enough alpha level, you are more likely to experience a type II error and fail to reject something that is false. If you do not choose a low enough alpha level, you are more likely to experience a type I error and reject something that is true.
Item 6: Explain why the phrase “we accept the null hypothesis (Ho) is always is an incorrect statement.”
Simply because it is nearly impossible in most scenarios to run statistics on an entire population. Only when you know your population can you accept anything. If my population size is ten, I can say I accept all day long. In most cases, unfortunately, it is not possible to do this in a z-distribution. When running a type I test, we are not trying to prove anything is correct, we are trying to prove that it is incorrect. You either reject or fail to reject; Never accept unless you know your entire population.
