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MGMT 650Statistics for Managerial Decision Making UMGCStudy Guide for Final Exam Excel ExamplesUse the accompanying Excel file to review the examples in this presentationMGMT650 Review2 Student’s t Distributiont0t (df = 5) t (df = 13)t-distributions are bell-shaped and symmetric, but have ‘fatter’ tails than the normalStandard Normal(t with df = ∞)Note: t Z as n increasesMGMT650 Review3 Confidence IntervalsThe general formula for all confidence intervals is:Point Estimate ± (Critical Value)(Standard Error)Where:Point Estimate is the sample statistic estimating the population parameter of interestCritical Value is a table value based on the sampling distribution of the point estimate and the desired confidence levelStandard Error is the standard deviation of the point estimateMGMT650 Review4 Confidence LevelConfidence the interval will contain the unknown population parameterA percentage (less than 100%)MGMT650 Review5 Confidence Level, (1-)Confidence the interval will contain the unknown population parameterA percentage (less than 100%)Also written (1 - ) = 0.95, (so  = 0.05)A relative frequency interpretation95% of all the confidence intervals that can be constructed will contain the unknown true parameterA specific interval either will contain or will not contain the true parameterNo probability involved in a specific intervalMGMT650 Review6 Upper and Lower LimitsKnown population standard deviationUnknown population standard deviationMGMT650 Review7 Example – ResistanceA sample of 11 circuits from a large normal population has a mean resistance of 2.20 ohms. We know from past testing that the population standard deviation is 0.35 ohms. Determine a 95% confidence interval for the true mean resistance of the population.MGMT650 Review8 Solution - Resistance ExampleMGMT650 Review9 InterpretationWe are 95% confident that the true mean resistance is between 1.9932 and 2.4068 ohms Although the true mean may or may not be in this interval, 95% of intervals formed in this manner will contain the true meanMGMT650 Review10 Example of t distribution confidence interval A random sample of n = 25 has X = 50 and S = 8. Form a 95% confidence interval for μd.f. = n – 1 = 24, soThe confidence interval is 46.698 ≤ μ ≤ 53.302MGMT650 Review11 AMBA 600MGMT650 Review12Another ExampleSet up a 95% confidence interval estimate for the population mean, based on the following set of data, assuming that the population is normally distributed: 1, 1, 1, 1, 8, 8, 8, 8, 9, 9, 9, 9, 10, 10, 11 AMBA 600MGMT 650 Review13Answer, part 1The standard deviation is 3.758. (Calculated by Excel)The number of values is 15; the square root of 15 is 3.873.5. There are 14 degrees of freedom.5. The t value for (1 - ) / 2 with 14 degrees of freedom is 2.1448. 2.1448 * (3.758/3.873) = 2.081This number must be added and subtracted from thesample mean to find the range of values with a 95% confidence level of having the actual population mean. AMBA 600MGMT 650 Review14Answer, Part 2Since you are looking for a range of values,you must find the lower value by subtracting thenumber found on the last slide from the sample mean. Then find the upper value by adding the number to theSample mean. 4.785 ≤ 6.867 ≤ 8.948 There is a 95% chance that the true population mean is between 4.785 and 8.948. What is a Hypothesis?A hypothesis is a claim (assumption) about a population parameter:population meanpopulation proportionExample: The mean monthly cell phone bill of this city is μ = $52Example: The proportion of adults in this city with cell phones is p = .68MGMT650 Review15 Formal Steps in Conducting a Hypothesis TestState the Null HypothesisChoose the correct testing methodZ Test – one sample, σ knownT Test – one sample, σ unknownSelect the desired confidence level (significance)Obtain the critical test value or P-valueDecision: Reject or fail to rejectConclusion: Interpret the testMGMT650 Review16 Hypothesis Test – One SampleThis is called a one sample test because you have a single sample of data being used to test the value of a population parameter, like mean.MGMT650 Review17 Hypothesis Test – Two SamplesThis is called a two sample test because you have two samples of data, from two populations, which are being tested to see the values of their population parameters, like mean, are equal.MGMT650 Review18 Two Population Means, Independent SamplesPopulation variance unknownLower-tail test:H0: μ1 – μ2  0H1: μ1 – μ2 < 0Upper-tail test:H0: μ1 – μ2 ≤ 0H1: μ1 – μ2 > 0Two-tail test:H0: μ1 – μ2 = 0H1: μ1 – μ2 ≠ 0aa/2a/2a-ta-ta/2tata/2Reject H0 if tSTAT < -taReject H0 if tSTAT > taReject H0 if tSTAT < -ta/2 or tSTAT > ta/2 Hypothesis tests for μ1 – μ2 MGMT650 Review19 P-ValuesThe P-value is the probability of seeing the observed data (or something even less likely) given the null hypothesis.A low P-value says that the data we have observed would be very unlikely if our null hypothesis were true. When the P-value is high (or just not low enough), data are consistent with the model from the null hypothesis.We reject the null hypothesis when the p-value is “low”P-value of .10 is some evidence to reject the null hypothesis.P-value of .05 is strong evidence to reject the null hypothesisP-value of .01 is very strong evidence to reject the null hypothesisP-value of .001 is extremely strong evidence to reject the null hypothesisMGMT650 Review20 Two Types of ErrorsWe can make mistakes in two ways:I. (False Positive or False Hypothesis) The null hypothesis is true, but we mistakenly reject it.II. (False Negative) The null hypothesis is false, but we fail to reject it.MGMT650 Review21 Review of ANOVAAnalysis of VarianceMGMT650 Review22 Analysis of Variance (ANOVA)z-Tests and t-Tests are limited to analysis of one or two samples/groups.ANOVATests the null hypothesis that the means of three or more populations are equal. (It can also be used on two groups.)MGMT650 Review23 MGMT650 Review24The ANOVA test uses a 5-step process, similar to the two-sample testStep 1: State the null hypothesis and the alternate hypothesisStep 2: Select the level of significanceStep 3: Determine the test statisticStep 4: Formulate a decision ruleStep 5: Make the decisionAnalysis of Variance MGMT650 Review25Things to RememberYou will be using the F distribution with ANOVA.The F critical value will always be on the right side of the distribution in the right tail even though you cannot have a directional alternative hypothesis. There are no negative values on the F distribution. (It looks like a ski slope)For three population means: Ho: 1=2=3 H1: At least two population means are different. MGMT650 Review26Degrees of FreedomThere are two degrees of freedom in the F tableDF1 = k – 1 ; DF2 = n – kk – Number of groupsn – Total number of observations in all cells ANOVA by HandWe need to account for the variation The F value is – (Estimate of the population variance based on difference among sample means) divided by (Estimate of the population variance based on variation within the sample)F = MSA/MSWSee full explanation here:http://davidmlane.com/hyperstat/intro_ANOVA.htmlMGMT650 Review27 MGMT650 Review28Calculating ANOVA by ExcelIt is possible to calculate ANOVA by hand, but if you ever tried you would agree it is MUCH easier in ExcelUse the data analysis toolsANOVA: Single Factor MGMT650 Review29ExampleWe have three sets of observations:1 4 32 5 43 66 59 5 610 6 7Is there evidence that the population means are equal? MGMT650 Review30Results of ANOVAAnova: Single Factor SUMMARY Groups Count Sum Average Variance Column 1 5 25 5 17.5 Column 2 5 86 17.2 744.7 Column 3 5 25 5 2.5 ANOVA Source of Variation SS df MS F P-value F critBetween Groups 496.133 2 248.067 0.97319 0.4058 3.885Within Groups 3058.8 12 254.9 Total 3554.933333 14 MGMT650 Review31Studying the resultsThe actual value of the Fstat is 0.973The F critical value is 3.885So, the actual value is in the acceptance region and we do not reject the null hypothesis that the means are equalANOVA proved our initial feeling MGMT650 Review32Changing the exampleLet’s change the example data1 222 32 5 4443 66 59 5 66610 6 888 MGMT650 Review33Running ANOVA againAnova: Single Factor SUMMARY Groups Count Sum Average Variance Column 1 5 25 5 17.5 Column 2 5 304 60.8 8810.7 Column 3 5 2006 401.2 156115.7 ANOVA Source of Variation SS df MS F P-value F critBetween Groups 459933.7333 2 229966.8667 4.182637854 0.041854843 3.885293835Within Groups 659775.6 12 54981.3 Total 1119709.333 14 MGMT650 Review34Studying the resultsThe actual value of the Fstat is 4.182The F critical value is 3.885This time, the actual value is in the rejection region and we reject the null hypothesis that the means are equal MGMT650 Review35Simple Linear Regression“Regression analysis is a statistical tool that utilizes the relation between two or more quantitative variables so that one variable (dependent variable) can be predicted from the others (independent variables). For example, if one knows the relationship between advertising expenditures and sales, one can predict sales by regression analysis once the level of advertising expenditures has been set.”Lethen, J. (1996, November 13). The simple linear regression. Retrieved from http://stat.tamu.edu/stat30x/notes/node156.html#SECTION001210000000000000000Since you look for a relationship between two variables when doing simple linear regression, linear regression is shown as a graph, and the relationship is expressed as an algebraic expression. MGMT650 Review36Graphing“Regression” consists of graphing data numbers and then looking for a relationship between the numbers.“Regression analysis is used primarily for the purpose of prediction. The goal in regression analysis is to develop a statistical model that can be used to predict values of a dependent or response variable based on the values of at least one explanatory or independent variable.”The y variable is the dependent variable on the graph, while the x is independent. Be careful to not mix up the variables or you will get incorrect results.Levine, D.M., Krehbiel, T.C., & Berenson, M.L. (2003). Business statistics: A first course (3rd ed.). Upper Saddle River: Prentice Hall. MGMT650 Review37Regression - No relationship MGMT650 Review38Regression - Positive Relationship MGMT650 Review39ExampleManagement of a soft-drink bottling company wished to develop a method for allocating delivery costs to customers. Although one cost clearly relates to travel time within a particular route, another variable cost reflects the time required to unload the cases of soft drink at the delivery point. A sample of 20 customers was selectedfrom routes within a territory and the delivery time and number of cases delivered were measured, with the following results: MGMT650 Review40Data MGMT650 Review41Scatter DiagramAssume that you wanted to develop a model to predict delivery time based on the number of cases delivered.Set up a scatter diagram.Using Excel…Insert  Chart  XY (Scatter)  Next MGMT650 Review42Scatter Diagram MGMT650 Review43Regression CoefficientsTools  Data Analysis… Regression(In this problem) The Y range (dependent variable) consists of the values for the delivery time in minutes.(In this problem) The X range (independent variable) consists of the values for the number of cases delivered. MGMT650 Review44Regression Output MGMT650 Review45Regression Equationb0 = 24.83b1 = 0.14Therefore, the regression equation isy = 0.14*X + 24.83 MGMT650 Review46Delivery time for 500 casesWould it be appropriate to use the model to predict the delivery time for a customer who is receiving 500 cases of soft drink? Why?No, 500 cases is outside the relevant range of the data used to fit the regression equation.Our range goes up to 300.For 500, there could be a need for a second tripWe don’t know the size of the truck Chi Squared DistributionExplains how well a theoretical distribution explains an observed distributionThree types of Chi Square tests:Goodness of Fit (one group, one variable)Homogeneity (several groups, one variable)Independence (one group, two variables)𝑥2=obs−ⅇ𝑥𝑝2ⅇ𝑥𝑝 MGMT650 Review47 Chi-Square Test of HomogeneityA bank wants to know if the distributions of applications is the same from three marketing campaignsA sample of 200 applications from each campaign has these results:What are the expected values?Find the test statistic and p-value. (Use .1 significance level)State conclusions.MGMT650 Review48{5C22544A-7EE6-4342-B048-85BDC9FD1C3A}TypeSilverGoldPlatinumTotalCampaign 11205030200Campaign 21155035200Campaign 31055540200Total340155105600 Homogeneity Test – Expected Values Observed ExpectedMGMT650 Review49{5C22544A-7EE6-4342-B048-85BDC9FD1C3A}TypeSilverGoldPlatinumTotalCampaign 1120 113.33 50 51.6730 35200Campaign 2115 113.3350 51.6735 35200Campaign 3105 113.3355 51.6740 35200Total340155105600 Find the Test StatisticCritical value found in chi square table for 4 degrees of freedom and .1 significance is 7.77944P Valuefound with Excel functionCHISQ.DIST.RT(2.7806,4)0.595MGMT650 Review50 ConclusionDo not reject the NULL hypothesisThere is insufficient evidence to suggest that the distributions are different for the three campaignsMGMT650 Review51 Goodness of Fit TestMarket Up DaysSample of 1000 “up” daysDo stocks behave different on different days?MGMT650 Review52{5C22544A-7EE6-4342-B048-85BDC9FD1C3A}Day of Week# of “up” days Observed% of trading daysExpected up days out of 1000Monday19219.37%193.369Tuesday18920.26%202.582Wednesday20220.37%203.695Thursday19920.06%200.607Friday21819.97%199.747 DecisionChi Square value (See Excel): 2.615Critical value (4 Degrees of freedom, significance 0.05): 9.488P-Value: 0.624Do not reject Null HypothesisThere is no evidence that stocks behave different on different daysMGMT650 Review53 Chi Square Independence TestAre women more likely to graduate high school than men?Graduation rates:Men – 84.9%Women – 88.1%Overall – 86.51%MGMT650 Review54{5C22544A-7EE6-4342-B048-85BDC9FD1C3A}MenWomenTotalGraduated HS105791116921748Did not Graduate188115093390Total124601267825138 Expected ValuesObserved ExpectedMGMT650 Review55{5C22544A-7EE6-4342-B048-85BDC9FD1C3A}MenWomenTotalGraduated HS10579 10779.711169 10968.321748Did not Graduate1881 1680.31509 1709.73390Total124601267825138 DecisionChi Square critical value (from table) – 3.84146(1 Degree of freedom, significance 0.05)Chi Square actual value (from Excel) - 54.94092P-value – 1.24208E-13 (Extremely small)Both Chi Square value and p-value are extremely strong evidence to reject H0Women are graduating at a higher rate than menMGMT650 Review56 Understanding the ExampleWe had a small difference (84.9 vs. 88.1) so just looking at the graduation rates does not explain the overwhelming resultWhy was the result so strong?Hint: Take the Excel and change the numbers to a smaller sample size, keeping the same ratios. What happened to the p-value? To the chi-square actual value?MGMT650 Review57
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