Premium Essay

Bus 308 Week 3 Anova and Paired T-Test

In: Business and Management

Submitted By eileneleavell
Words 1045
Pages 5
BUS 308 Week 3 ANOVA and Paired T-test

Click Link Below To Buy: http://hwcampus.com/shop/bus-308-week-3-anova-paired-t-test/ 1Week 3 ANOVA and Paired T-test

At this point we know the following about male and female salaries.
a. Male and female overall average salaries are not equal in the population.
b. Male and female overall average compas are equal in the population, but males are a bit more spread out.
c. The male and female salary range are almost the same, as is their age and service.
d. Average performance ratings per gender are equal.
Let's look at some other factors that might influence pay - education(degree) and performance ratings.

1 Last week, we found that average performance ratings do not differ between males and females in the population.
Now we need to see if they differ among the grades. Is the average performace rating the same for all grades?
(Assume variances are equal across the grades for this ANOVA.) You can use these columns to place grade Perf Ratings if desired.
A B C D E F
Null Hypothesis:
Alt. Hypothesis:
Place B17 in Outcome range box.

Interpretation:
What is the p-value:
Is P-value < 0.05?
Do we REJ or Not reject the null?
If the null hypothesis was rejected, what is the effect size value (eta squared):
Meaning of effect size measure:

What does that decision mean in terms of our equal pay question:

2 While it appears that average salaries per each grade differ, we need to test this assumption.
Is the average salary the same for each of the grade levels? (Assume equal variance, and use the analysis toolpak function ANOVA.)
Use the input table to the right to list salaries under each grade level.

Null Hypothesis: If desired, place salaries per grade in these columns
Alt. Hypothesis: A B C D E F

Place B55 in Outcome range box.

What is the p-value:
Is…...

Similar Documents

Premium Essay

T-Test

...attempt to match eachother on any variable. 2. t = –3.15 describes the difference between women and men for what variable in this study? Is this value significant? Provide a rationale for your answer. t=-3.15 describes the differences between women and men for the mental health variable. It has the smallest p value, at p-0.002 and that is considered significant because the p=0.002 is smaller than alpha that was set at 0.05 for this study. Small p values indicate significant findings. 3. Is t = –1.99 significant? Provide a rationale for your answer. Discuss the meaning of this result in this study. T=-1.99 describes the difference in health functioning between men and women after an MI. t=-1.99 is significant because it’s p value (0.049), is still smaller than alpha that was set at 0.05 for this study. The meaning of this result in this study is that the quality of life measures in regards to health functioning show a significant result. This finding indicates that women had a mean of 17.9 with an sd of 4.1, and men had a mean of 19.3, with an sd of 4.6. Results support the study and findings that, “women rate lower levels in physical and psychological dimensions of quality of life” 4. Examine the t ratios in Table VI. Which t ratio indicates the largest difference between the males and females post MI in this study? Is this t ratio significant? Provide a rationale for your answer. 5. Consider t = –2.50 and t = –2.54. Which t ratio has the smaller p value?......

Words: 363 - Pages: 2

Premium Essay

Bus 308 Week 3 Assignments

...7.11 A) By using the Central Limit Theorem with the sample being greater than 30 we can assume that the distribution and shape of x will be normally distributed. B) Mean = 20 (no equations as it was given) 4/ 64 = 4/8 = 0.5 Standard Deviation C) Z = (21-20) / (4/ 64) = (1/0.5) = 2 P = 1-.09772 = .0228 D) Z = (19.385 – 20) / (4/ 64) = -0.615/0.5 = -1.23 P = 0.1093 7.30 A) Z = (0.32 - 0.3)/√(0.3(1 - 0.3)/1011) = 1.3877 P = (p’ > 0.32) = P (z > 1.3877) = 0.0826 B) If we set the a to anything below 0.01 the claims can be refuted but on the other hand if we set it above than the claims would be true so the claims can neither be refuted or true . 8.8 A) N = 100 X-bar = 5.46 s = 2.47 % = 95 Standard Error = σ/√n = 0.2470 z score = 1.9600 Width = 1.9600 * 0.2470 = 0.4841 Lower Limit = 5.46 – 0.4841 = 4.9759 Upper Limit = 5.46 + 0.4841 = 5.9441 Confidence interval of 95% [4.976, 5.944] N = 100 X-bar = 5.46 s = 2.47 % = 99 Standard Error = σ/√n = 0.2470 z- score = 2.5758 Width = 2.5758 * 0.2470 = 0.6362 Lower Limit = 5.46 – 0.6362 = 4.8238 Upper Limit = 5.46 + 0.6362 = 6.0962 Confidence interval of 99% [4.824, 6.096] B) Since the entire confidence interval is less than 6 the manager can confident that 95% is less than 6 minutes. C) Because the confidence interval at 99% is over 6 the manager cannot be 99% confident that it’s less than 6 minutes. D) If we choose α = 0.05, 95% is......

Words: 400 - Pages: 2

Premium Essay

Bus 370 Week 3 Assignment

...Bus 370: Organizational Development Angel Lozada Week 3 Dr. Barbara-Leigh Tonelli August 13, 2012 Introduction       Organization Development (OD) increases the use of human resources in order to improve administrative efficiency. Its goals focus on making an organization more effective and to provide individuals the best opportunity to meet or exceed their potential. Brown (2011, p.4) states “employee involvement and commitment are the true keys to successful changes in continued operations of any company”. With change comes employee resistance, if this is mishandled it can prevent a change from successfully taking place. Implementing OD specialists and principles will make sure continued operation of any business. This paper will analyze the challenges facing a retail dry goods store using OD principle and how OD principles will be applied to ensure the continued operation of this business.   Analysis of the Changes     An owner of a 100 employee retail dry goods store, which is located in a town where the major employer is an automotive support industry that has just announced they are relocating their facility to Mexico, must implement a change within their organization. A challenge facing this business is coming up with a strategic plan to incorporate automotive support into the retail dry......

Words: 927 - Pages: 4

Premium Essay

T-Test

...T-test: Salaries of Female and Male Human Resource Managers Christy Newman Argosy University T-test: Salaries of Female and Male Human Resource Managers An independent-samples t-test using the raw measurement data presented was completed to check the difference, if any, of the salaries between female and male human resource managers, t(18) = -0.408, p = 0.688, but no significant difference was found (Female Mean = 62.2; Male Mean = 63.7). The preparation for a statistical analysis test began with a well-developed, clear research question: What is the difference, if any, between the salaries of female and male HR Directors? Next, the null and alternative hypotheses must be defined. The null H₀ is that the salaries of female human resource managers = salaries of male human resource managers. The alternative Ha is that salaries of female human resource managers ≠ salaries of male human resource managers. The next step in the process is to determine the appropriate statistical test and sampling distribution. Since σ is unknown and the number of salaries being tested is less than 100, the t-test will be used. Because we are comparing data from two different groups, the two-sample t-test will be used. We do not know if the variance is different or equal, so we will use the two-tailed t-test for two-sample assuming unequal variances. Subsequently, we have to choose the Type 1 Error rate. For this t-test, α = 0.05, which is the standard rate used by statisticians for......

Words: 804 - Pages: 4

Premium Essay

T-Test and Anova

...BUS 310 Notes regarding Two-Sample t-Tests and ANOVAs In Chapter 9, we learned how to conduct a t test of a hypothesis when we were testing the mean of a single sample group against some pre-determined value (i.e., the 21.6 gallons of milk consumption as the national average). This week, in Chapter 10, we will see how to test hypotheses that involve more than one sample group—such as testing to see if males are significantly taller than females. If we have two groups, then the technique that we will use will still be a t test. If we have more than two groups, then we will have to use a different test called Analysis of Variance (ANOVA, for short). The good news is that the decision rules for hypothesis testing that we learned last week are still exactly the same: Set #1: If the absolute value (ignore any negative sign) of the test statistic is greater than or equal to the critical value, then you reject the null. If the absolute value of the test statistic is less than the critical value, you do not reject the null. Set #2: If the p value is less than or equal to α, reject the null. If the p value is greater than α, do not reject the null. (Remember that we must either reject or not reject the null—we never accept the null.) In order to conduct these tests, we will need to use the data analysis feature of Excel, which probably is not installed for you, but that’s OK, because it’s available and pretty simple to install—just follow these steps: ...

Words: 2114 - Pages: 9

Premium Essay

Bus 308 Ashford Week 3 Quiz

...This document of BUS 308 Week 3 Quiz includes answers to the next questions: 1. Question : There is little difference between the values of ta/2 and Za/2 when the sample size is small. size is large. mean is small. mean is large. standard deviation is small. 2. Question : If the population proportion is .4 with a sample size of 20, then is this sample large enough so that the sampling distribution of is a normal distribution. True False 3. Question : For non-normal populations, as the sample size (n) _________, the distribution of sample means approaches a/an __________ distribution. decreases, uniform increases, normal decreases, normal increases, uniform increases, exponential 4. Question : As the sample size ________the variation of the sampling distribution of ___________. decreases, decreases increases, remains the same decreases, remains the same increases, decreases 5. Question : Assuming the same value of a, as the sample size increases, the value of ta/2 approaches the value of Za/2. True False 6. Question : If the sampled population has mean 48 and standard deviation 16, then the mean and the standard deviation for the sampling distribution of for n = 16 are 4 and 1. 12 and 4. 48 and 4. 48 and 1. 48 and 16. True False 8. Question : As our sample standard deviation increases when all other parts of the confidence interval stay the same, then the confidence interval will become: wider narrower remain the......

Words: 373 - Pages: 2

Premium Essay

Bus 308 Ashford Week 5 Final Part I

...This document of BUS 308 Week 3 Quiz includes answers to the next questions: 1. Question : There is little difference between the values of ta/2 and Za/2 when the sample size is small. size is large. mean is small. mean is large. standard deviation is small. 2. Question : If the population proportion is .4 with a sample size of 20, then is this sample large enough so that the sampling distribution of is a normal distribution. True False 3. Question : For non-normal populations, as the sample size (n) _________, the distribution of sample means approaches a/an __________ distribution. decreases, uniform increases, normal decreases, normal increases, uniform increases, exponential 4. Question : As the sample size ________the variation of the sampling distribution of ___________. decreases, decreases increases, remains the same decreases, remains the same increases, decreases 5. Question : Assuming the same value of a, as the sample size increases, the value of ta/2 approaches the value of Za/2. True False 6. Question : If the sampled population has mean 48 and standard deviation 16, then the mean and the standard deviation for the sampling distribution of for n = 16 are 4 and 1. 12 and 4. 48 and 4. 48 and 1. 48 and 16. True False 8. Question : As our sample standard deviation increases when all other parts of the confidence interval stay the same, then the confidence interval will become: wider narrower remain the......

Words: 370 - Pages: 2

Free Essay

Anova and Nonparametric Tests

...ANOVA and Parametric tests Looking at the concepts in the book it is interesting with regard to placing them into the work world. The most recent concepts were ANOVA and nonparametric tests. ANOVA, also known as analysis of variance is a concept that allows you to “compare more than two means simultaneously and how to trace sources of variation to potential explanatory factors”. One of the biggest take aways from this concept is that the ANOVA tests can take on many factors or “treatments. This can be very beneficial when dealing with many factors. Although this does not mean that it is the norm to have many factors. Most researchers focus on a limited amount of factors. Another big lesson is with nonparametric tests. Working for a company that requires that we learn how comfortable our users are with their technology it is very important to use ordinal data for informative decision making. In our data sets there is a large majority of information that does not come in with a normal distribution. This is where it comes into play that a parametric test can aid due to the ability to examine information without normal distribution. One of the largest lessons in terms of these concepts is the data size that it takes to utilize these tests. Unfortunately as with any questionnaire my department usually gets only a small amount back. When working with parametric testing a small sample size would only go to hurt the result of the testing. When working with these tests it is possible...

Words: 344 - Pages: 2

Free Essay

Bus 475 Week 3 Dq

...BUS 475/DQ Week 3 Question 1 What are the different strategies and what is different about them? The main strategies commonly referred to as Generic and Grand. Generic strategies are used by companies to gain ground in a competitive market. This type has subcategories, which are cost leadership, differentiation, and focus. Firms that elect to use generic strategies, may offer cheaper prices for the same service as their competitors. For instance, when purchases a plane ticket, they can go with Southwest Airlines, which is a no frills company and get a lower price. Southwest opts to omit the bells and whistles (unlike Virgin Air) to offer low prices and gain a competitive edge within the airline industry. Generic strategies can be used in all industries and on any product type. Whereas Grand strategies are elaborate and over the top and change the course of business, because their focus is long term. This type has subgroups, which are market growth, product development, turnaround, liquidation. Firms that focus on market growth, want a low risk strategy and will growing market interest on a product already being marketed. The opposite of market growth is product development, where firms focus on research and development. Pharmaceutical companies are heavily into research and development. Turnaround strategy is just what it is, firms whose product is performing poorly, will switch how they present their product to generate interest. Liquidation is a desperate......

Words: 578 - Pages: 3

Premium Essay

Bus 308 Week 2 a++ 100% Accurate and Most Economical Answer Can Be Download from Here

...BUS 308 Week 2 Download 100% accurate A++ and most economical answer from here http://www.homeworkmarket.com/content/bus-308-week-2-5173059-2 http://www.homeworkmarket.com/content/bus-308-week-2-5173059-2 Week 2 Testing means In questions 2 and 3, be sure to include the null and alternate hypotheses you will be testing. In the first 3 questions use alpha = 0.05 in making your decisions on rejecting or not rejecting the null hypothesis. 1 Below are 2 one-sample t-tests comparing male and female average salaries to the overall sample mean. (Note: a one-sample t-test in Excel can be performed by selecting the 2-sample unequal variance t-test and making the second variable = Ho value -- see column S) Based on our sample, how do you interpret the results and what do these results suggest about the population means for male and female average salaries? Males Females Ho: Mean salary = 45 Ho: Mean salary = 45 Ha: Mean salary =/= 45 Ha: Mean salary =/= 45 Note: While the results both below are actually from Excel's t-Test: Two-Sample Assuming Unequal Variances, having no variance in the Ho variable makes the calculations default to the one-sample t-test outcome - we are tricking Excel into doing a one sample test for us. Male Ho Female Ho ......

Words: 264 - Pages: 2

Premium Essay

Bus 308 Week 2 Quiz

...1. Question : The one-sample t-test differs from the z-test in which way? @Answer found in section 4.3 The One-sample t-Test, in Statistics for Managers Student Answer: There are no parameter values involved in a t-test. The t-test is more sensitive to minor differences between sample and population. With the t-test one can be confident of the normality of the data. The t-test requires no parameter standard error of the mean. Points Received: 0 of 1 Comments: Question 2. Question : If a certifying agency raises the requirements for real estate agents, what sort of decision error is the agency protecting against? Student Answer: Type I Type II Type III Type IV Instructor Explanation: Answer found in section 3.4 Statistical Significance, in Statistics for Managers Points Received: 0 of 1 Comments: Question 3. Question : What is the alternate hypothesis in a problem where sales group two is predicted to be “. . . significantly less productive than sales group one?” @Answer found in sections 4.3 The One-sample t-Test and 4.4 Hypothesis Testing, in Statistics for Managers Student Answer: HA: μ1 ≠ μ 2 HA: μ 1= μ 2 HA: μ 1> μ2 HA: μ 1< μ 2 Points Received: 0 of 1 Comments: Question 4. Question : Which of the following defines statistical significance? Student Answer: The outcome is......

Words: 606 - Pages: 3

Premium Essay

Bus 599 Week 3 Bus599 Week 3

...BUS 599 Complete Course BUS599 Complete Course Click Link for the Answer: http://workbank247.com/q/bus-599-complete-course-bus599-complete-course/21454 http://workbank247.com/q/bus-599-complete-course-bus599-complete-course/21454 BUS 599 Week 1 Discussion "Company Description" Throughout this course, you will develop a series of written papers / projects that you will later combine into a complete business plan for a Non-Alcoholic Beverage company. For this discussion, you must first review the “NAB Company Portfolio”.  The mentioned portfolio contains the company parameters and details you must follow when developing your company. Provide the following information to set the foundation for your non-alcoholic beverage (NAB) business plan. Please respond to the following: * Create your NAB company name and explain its significance. * Develop your company’s Mission Statement and provide a rationale for its components. BUS 599 Week 2 Discussion "Growing Honest Tea"  Please respond to the following: * Review the following documents: * Honest Tea’s business plan for 1999 (PDF). * A strengths, weaknesses, opportunities, and threats (SWOT) analysis based on Honest Teas’ business plan (PDF). * Suppose Honest Tea has hired you as a consultant to evaluate the completeness of their strategy for future growth. Base your evaluation on the provided SWOT analysis. Provide a rationale for your response. BUS 599 Week 3 Discussion "Don't Miss the Mark"  Please respond...

Words: 5722 - Pages: 23

Premium Essay

Bus 519 Week 3 Bus519 Week 3

...BUS 519 Complete Course BUS519 Complete Course Click Link for the Answer: http://workbank247.com/q/bus-519-complete-course-bus519-complete-course/22300 http://workbank247.com/q/bus-519-complete-course-bus519-complete-course/22300 BUS 519 Week 1 Discussion "What is Risk?"  Please respond to the following: * There are three (3) schools of thought regarding risk. The first considers the positive and negative aspects of risk, but sees them as separate. The second group believes that there are benefits from treating threats and opportunities together, while the third school does not label uncertainties, but addresses uncertainty as part of “doing the job.” Argue the value of having a risk strategy despite the cost associated with it. Include an example to support your response. Provide a rationale for your selection and determine how this approach helps a project to be successful. BUS 519 Week 2 Discussion "Need for Risk Management" Please respond to the following: * There are four (4) critical success factors that are important for effective risk management: supportive organization; competent people; appropriate methods, tools and techniques; and simple, scalable process. Determine three (3) obstacles for an organization to manage risk effectively. Suggest strategies from the perspective of a project manager to avoid the obstacles. BUS 519 Week 3 Discussion "Project Initiation" Please respond to the following: * Using the “Stakeholder Analysis Template” (Appendix B2 in...

Words: 2422 - Pages: 10

Premium Essay

Bus 308 Week 3 Anova and Paired T-Test

...BUS 308 Week 3 ANOVA and Paired T-test Click Link Below To Buy: http://hwcampus.com/shop/bus-308-week-3-anova-paired-t-test/ 1Week 3 ANOVA and Paired T-test At this point we know the following about male and female salaries. a. Male and female overall average salaries are not equal in the population. b. Male and female overall average compas are equal in the population, but males are a bit more spread out. c. The male and female salary range are almost the same, as is their age and service. d. Average performance ratings per gender are equal. Let's look at some other factors that might influence pay - education(degree) and performance ratings. 1 Last week, we found that average performance ratings do not differ between males and females in the population. Now we need to see if they differ among the grades. Is the average performace rating the same for all grades? (Assume variances are equal across the grades for this ANOVA.) You can use these columns to place grade Perf Ratings if desired. A B C D E F Null Hypothesis: Alt. Hypothesis: Place B17 in Outcome range box. Interpretation: What is the p-value: Is P-value < 0.05? Do we REJ or Not reject the null? If the null hypothesis was rejected, what is the effect size value (eta squared): Meaning of effect size measure: What does that decision mean in terms of our equal pay question: 2 While it appears that average salaries per each grade differ, we need to test this...

Words: 1045 - Pages: 5

Free Essay

Bus 415 Week 3

...pain and suffering. She also can sue them for distress she is old and might never recuperate from the injuries. Anna vs. Hospital Non-Intentional Anna will win her malpractice law suit because the doctor preformed the wrong surgery. Anna was under the impression that she was having surgery to stop the bleeding and allowing the doctors to save her tooth. She didn’t give them permission to amputate her leg. She will be affected for the rest of her life and not to mention without her leg. Anna will be able to live as she was prior to this surgery. These are the many reasons she will be able to sue for a large lump sum of money. The doctor was hired by the hospital so that’s why the hospital and the doctor are responsible. Scenario 3 Scenario 4 The possible potential plaintiffs in this case are the boy, husband, and his wife. The potential defendants in the case are Randy, Lee, and the security guard. Lee's defense could possibly be that he didn't commit the murder. It is Lee’s word against Randy, the security guard, the wife, the husband, the boy, and whoever witnessed the incident. Randy's possible defense could be that he didn't know that the man was going to shoot the plaintiffs. He has a case, because he has several witnesses that saw what happened. The security guard's possible defense could be that he was just doing his job, the way he was trained to do it. The security guard has several witnesses in this case to protect him and his job. In this case, I......

Words: 1228 - Pages: 5

Isidora Goreshter | Noble (13) | DISH Anywhere