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Words 464

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In formulating the hypothesis statement using ANOVA, the team has decided on a null hypothesis that mean net income from the three different financial sectors are equal. After developing the null and alternate hypothesis, Team E preformed the five steps of hypothesis testing using MegaStat and Microsoft Excel (Doane & Seward, 2007).

Step 1: State the Hypotheses

The hypotheses to be tested are

Ho: μt =μf =μe

H1: Not all the means are equal (at least one mean is different).

Step 2: State the Decision Rule

There is c = 3 groups and n = 75 observations, so degrees of freedom for the F test are:

Numerator: d.f. 1 = c -1 = 3 -1 =2 (between treatments, factor)

Denominator: d.f.2 = n –c = 75 -3 =71 (within treatments, error)

Step 3: Select a Level of Significance

We will use a = .05 for the test. The 5 percent right-tail critical value from Appendix F is F(3,25) = 3.13

Step 4: Perform the Calculations/ Test Statistics

Using MegaStat for the calculations, we obtained the results shown in the Excel Spread sheet. Excels default is a = .05.

Step 5: Make the decision

Since the test statistic F = 7.60 exceeds the critical value F (.05) = 3.13, we can reject the hypotheses of equal means. The p-value (p = 0.001044) is less than the level of significance (a = .05) which confirms that we should reject the hypothesis of equal treatment means.

When using ANOVA. The null hypothesis is that all means are equal. The…...

...Variance and Nonparametric Tests Paper Andy Martinez Res/342 February 1, 2012 Victor Ornelas Most recent concepts are 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 away 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 to extrapolate the......

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...Team E has decided to use an ANOVA to see if there is a difference in the mean net income from the telecommunications sector, the financial sector, and the energy sector, we will be using the .05 significance level. The research question is “Does the financial sector, the telecommunication sector and the energy sector all share the same average net income.” ANOVA or analysis of variance allows team E to compare more than two means simultaneously. In formulating the hypothesis statement using ANOVA, the team has decided on a null hypothesis that mean net income from the three different financial sectors are equal. After developing the null and alternate hypothesis, Team E preformed the five steps of hypothesis testing using MegaStat and Microsoft Excel (Doane & Seward, 2007). Step 1: State the Hypotheses The hypotheses to be tested are Ho: μt =μf =μe H1: Not all the means are equal (at least one mean is different). Step 2: State the Decision Rule There is c = 3 groups and n = 75 observations, so degrees of freedom for the F test are: Numerator: d.f. 1 = c -1 = 3 -1 =2 (between treatments, factor) Denominator: d.f.2 = n –c = 75 -3 =71 (within treatments, error) Step 3: Select a Level of Significance We will use a = .05 for the test. The 5 percent right-tail critical value from Appendix F is F(3,25) = 3.13 Step 4: Perform the Calculations/ Test Statistics Using MegaStat for the calculations, we obtained the results shown in the Excel Spread sheet. Excels......

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...Topic: ANOVA Our topic is about the Analysis of Variance(ANOVA). This is use to compare means of 3 or more populations. You may encounter the word treatment. Treatment is a cause or specific source of variation in a set of data. Here are the assumptions: 1. the three or more population of interest are normally distributed. 2. These pop have equal standard deviation [pic]. 3. The sample we select from each of the population are random and independent. Test: F ratio of two variance F = Estimated pop variance based on variation bet the sample means Estimated pop variance bases on variation within samples ANOVA SV SS DF MS Treatment 1/2 (between columns) SST k-1 SST/k-1 Error (Between row) SSE N-k SSE/N-k Total SS F = SST/k-1 SSE/N-k SSTreat = [pic] SSTotal= [pic] SSError= SSTotal-SSTreat. Below is the example (You may also read the powerpoint on ANOVA for more examples. Example: Susan predicts that students will learn most effectively with a constant background sound, as opposed to an unpredictable sound or no sound at all. She randomly divides twenty-four students into three groups of eight. All students study a passage of text for 30 minutes. Those in group 1 study with background sound at a constant volume in the background. Those in group 2 study with noise that changes volume periodically. Those in group 3 study with no sound at...

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...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: ...

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...change in mobility scores over 12 weeks? Provide a rationale for your answer.” “3. The researchers stated that the participants in the intervention group reported a reduction in mobility difficulty at week 12. Was this result statistically significant, and if so at what probability?” “4. If the researchers had set the level of significance or α= 0.01, would the results of p = 0.001 still be statistically significant? Provide a rationale for your answer.” “5. If F(3, 60) = 4.13, p = 0.04, and α = 0.01, is the result statistically significant? Provide a rationale for your answer. Would the null hypothesis be accepted or rejected?” “6. Can ANOVA be used to test proposed relationships or predicted correlations between variables in a single group? Provide a rationale for your answer.” “7. If a study had a result of F(2, 147) = 4.56, p = 0.003, how many groups were in the study, and what was the sample size?” “8. The researchers state that the sample for their study was 28 women with a diagnosis of OA, and that 18 were randomly assigned to the intervention group and 10 were randomly assigned to the control group. Discuss the study strengths and/or weaknesses in this statement.” “9. In your opinion, have the researchers established that guided imagery (GI) with progressive muscle relaxation (PMR)......

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...ANOVA Analysis of variance (ANOVA) is a collection of statistical models used to analyze the differences between group means and their associated procedures (such as "variation" among and between groups), developed by R.A. Fisher. In the ANOVA setting, the observed variance in a particular variable is partitioned into components attributable to different sources of variation. In its simplest form, ANOVA provides a statistical test of whether or not the means of several groups are equal, and therefore generalizes the t-test to more than two groups. As doing multiple two-sample t-tests would result in an increased chance of committing a statistical type I error, ANOVAs are useful in comparing (testing) three or more means (groups or variables) for statistical significance. The analysis of variance can be used as an exploratory tool to explain observations. A dog show provides an example. A dog show is not a random sampling of the breed: it is typically limited to dogs that are male, adult, pure-bred, and exemplary. A histogram of dog weights from a show might plausibly be rather complex, like the yellow-orange distribution shown in the illustrations. Suppose we wanted to predict the weight of a dog based on a certain set of characteristics of each dog. Before we could do that, we would need to explain the distribution of weights by dividing the dog population into groups based on those characteristics. A successful grouping will split dogs such that a) each group has a low......

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...ONE WAY ANOVA One-way analysis of variance (abbreviated one-way ANOVA) is a technique used to compare means of two or more samples (using the F distribution). This technique can be used only for numerical data. The ANOVA tests the null hypothesis that samples in two or more groups are drawn from populations with the same mean values. To do this, two estimates are made of the population variance. These estimates rely on various assumptions. The ANOVA produces an F-statistic, the ratio of the variance calculated among the means to the variance within the samples. If the group means are drawn from populations with the same mean values, the variance between the group means should be lower than the variance of the samples, following the central limit theorem. A higher ratio therefore implies that the samples were drawn from populations with different mean values. Descriptives | | N | Mean | Std. Deviation | Std. Error | 95% Confidence Interval for Mean | Minimum | Maximum | | | | | | Lower Bound | Upper Bound | | | QUALITY | 1 | 19 | 3.89 | .809 | .186 | 3.50 | 4.28 | 2 | 5 | | 2 | 12 | 3.83 | .937 | .271 | 3.24 | 4.43 | 1 | 5 | | Total | 31 | 3.87 | .846 | .152 | 3.56 | 4.18 | 1 | 5 | PRICE | 1 | 19 | 2.95 | .911 | .209 | 2.51 | 3.39 | 1 | 5 | | 2 | 12 | 2.75 | 1.055 | .305 | 2.08 | 3.42 | 1 | 5 | | Total | 31 | 2.87 | .957 | .172 | 2.52 | 3.22 | 1 | 5 | BRAND | 1 | 19 | 4.11 | .809 | .186 | 3.72 | 4.50 | 3 | 5 | | 2 | 12 | 4.17 | .577 |......

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...Mean | | 3.16 | Attractive | Acceptable | Table1.1 shows the ratings of the respondents on the different treatments of potato fries on sensory evaluation in terms of color. Treatment 2 got the highest mean of 3.53 followed by Treatment 3 with a mean of 3.20. Treatment 4 obtained a mean of 3.00 treatment 1 got the lowest mean of 2.90. The result implies that the color of the treatment of potato fries are “attractive” and interpreted as “acceptable” as reflected by the section mean of 3.16. Table 1.2 Analysis of variance (ANOVA) on the different treatments of Potato fries in terms of Color. Sources of Variance | Sum of Squares | DF | Means of Squares | F- computed | F- tab0.05 0.01 | Treatments | 14.05 | 3 | 4.68 | 6.88 | 3.13 4.58 | Error | 159.93 | 236 | 0.68 | | | Total | 173.98 | 239 | | | | * *Significant Difference ( At 0.05 and 0.01 level of significance) | Table 1.2 shows the analysis of variance (ANOVA) on the different treatments of potato fries in terms of color. Statistical analysis shows the f- computed of 6.88. The tabular value of f is 3.13 at 0.05 level of significance and 4.58 at 0.01 level of significance. Since the computed value is greater than the tabular value, this means that there is “significant” difference among the treatments of potato fries in terms of color Table 1.3 Ratings of the respondents to the different treatments of Potato fries in terms of Texture. Aroma | Rating Scale4 ......

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...Applying ANOVA and Nonparametric Tests Simulation Many organizations use various tools to ensure quality assurance and management for their business. The challenge for them is to ensure that they provide the best quality of service to their clients in a time effective manner. As such, having a diversity of tool options in place helps the organization identify daily challenges and increase overall effectiveness practices in their decision making processes. Implicitly, identifying the problems is the first key component towards making a sound decision. Once the problems are identified organizations can use tests such as ANOVA, nonparametric test and Kruskal-Wallis test for operational research methods and total quality management. These methods will allow researchers to analyze significant data that will subsequently result in implementation of the found solutions. Accordingly, the Praxidike Systems Corporation has identified a problem with their turn-around time in delivering products to their clients. With that, using the ANOVA, nonparametric and Kruskal- Wallis test has taught me that utilizing these tools can assist in analyzing information in order to make the best possible decision for the organization. Furthermore, using these tools help make the process easier to control by breaking down the data into various groups in order to manage the information. When using these tools, it is easy to apply what was reviewed to everyday life......

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...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...

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...ANOVA Excel Worksheet The following table contains a random sample of 40 women partitioned into three groups: Group 1: ages below 20 Group 2: ages 20 through 40 Group 3: ages over 40 the values in the table are systolic blood pressure levels The Hypothesis test: H0: μ1=μ2=μ3 H1: at least one of the treatment means is different use the Excel Analysis Toolpak to create an Anova- Single factor table. is there sufficient evidence to support the claim that women in different age categories have different mean blood pressure levels? Give for your decision. Group 1 Group 2 Group 3 [Place your Anova table here] 104 97 123 106 116 107 Anova: Single Factor 104 98 127 92 95 133 SUMMARY 112 108 114 Groups Count Sum Average Variance 107 106 93 Group 1 102 113 Group 2 108 114 Group 3 94 116 100 155 101 105 ANOVA 119 Source of Variation SS df MS F P-value F-crit 89 Between Groups 113 Within Groups 99 93 Total ......

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...of time spent for weekend evening. * Identify the appropriate statistical test to accept or reject the null hypothesis. * The test that I would use to test to accept or reject the null hypothesis is the one-way ANOVA test. * Calculate the appropriate statistical test values to accept or reject your null and alternative hypotheses. Analysis of Variance (One-Way) | | | | | | | | Summary | | | | | | | Groups | Sample size | Sum | Mean | Variance | | | Midweek, Day | 25 | 2,087.89898 | 83.51596 | 1,300.16709 | | | Midweek, Evening | 25 | 2,384. | 95.36 | 1,479.65667 | | | Weekend, Day | 25 | 2,528. | 101.12 | 2,434.61 | | | Weekend, Evening | 25 | 3,074. | 122.96 | 2,608.45667 | | | | | | | | | | ANOVA | | | | | | | Source of Variation | SS | df | MS | F | p-level | F crit | Between Groups | 20,487.11927 | 3 | 6,829.03976 | 3.49182 | 0.01862 | 3.43446 | Within Groups | 187,749.37009 | 96 | 1,955.72261 | | | | | | | | | | | Total | 208,236.48936 | 99 | | | | | * What should you tell the Mall of Elbonia's food court stores managers in terms of the mall's high-traffic times and customer tendencies? Based upon the results of the one-way ANOVA test we can reject the null hypothesis and accept the alternative. This means that there is not equal time spent at the mall for midweek day, midweek evening, weekend day, and weekend evening. I would advise that there is......

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...|Applying Analysis of Variance | |[Anova test simulation] | |Rochelle Kuebler | |[September 23, 2011] | Praxidike is a software company, which has concern defining why their assignments are not done on time. The set-up starts by handing out two nonparametric analysis methods that include ANOVA and Kruskal-Wallis. Nonparametric testing procedures need definite requests to make use of efficiently. The key norms of ANOVA testing consist of the following: the population consumes a standard distribution, mistakes are independent, and population consumes the same variance. The Kruskal-Wallis test, instead, does not involve the hypothesis of a common distribution and the facts need to be on an ordinal measure. This is characteristically a superior choice if the expectations of ANOVA will not be met. This setup runs three examples of how certain testing systems can be practical to real-world circumstances. The first part of the situation is to relate the Kruskal-Wallis test because the expectations of ANOVA may not be seen. After studying the facts, it was obvious that the level of capability dealing with the......

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...variable Experimental and observational design Factor/factor level Treatment/Treatment combination Experimental unit Balanced design Randomization Control group Source of variations Linear model Indicator variable ANOVA table decomposition Interpreting SAS output with respect to 1-way and 2-way ANOVA model Interpret data plots to identify visual differences between groups Understand the cell means model and the factor effects model: parameters and estimates Compute parameter estimates for cell means/factor effects model given a table of means Conduct the F-test for testing the equality of factor level means Compute CI’s for differences between means Multiple comparison methods: LSD, Tukey, Bonferroni, and Scheffe Diagnostics of model assumptions and remedial measures: three key assumptions on the error terms of ANOVA/ transformation methods Residual analysis Regression model corresponding to ANOVA model Review problems: These are the minimum knowledge to get ready for the exam. You should study thoroughly text book material and lecture notes to be well prepared for the exam. 1. Indicate how you would check that the three key assumptions on the error terms for ANOVA model are satisfied. 2. Conducting the one-way ANOVA for the following data, write the cell means and factor effects model and the corresponding parameter estimates. Factor level size mean 1 5 12 2 5 10 3 5 15 ......

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...ANOVA Simulation Working as a quality control manager is no easy task. There are a lot of variables that are involved in it. Knowing how to research specific problems truned out to be crucial when it came to solve a specific problem, which opened my eyes as to the complication involved in it. There are two specific test used in this particular simulation which are Anova and Non-parametric. “Anova is defined as a statistical method for making simultaneous comparisons between two or more means; a statistical method that yields values that can be tested to determine whether a significant relation exists between variables.” (dictionary.com) “A non parametric test is a branch of statistics that are applied when populations are not normal or they are severely skewed data.” There were three lessons learned during this simulation which were how to monitor a situation. The importance to measuring the accumulated data and how to provide a reliable solution to the problem. It is important that we learn how to apply things that we learn in everyday life as well as the workplace. Using an appropriate test can help us whenever it is time to make a decision in business, which is one other thing learn while going over the simulation. This helps at the time of determining how to correct a stiuation in the workplace. Training in customer service and technical support has shown to be crucial in the company shown in the simulation this information is shown during......

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