Random samples were selected from two independent populations - 160 observations from population 1 (dual d students) and 140 from population 2 (management only degree). It is known from past studies that the population variance for dual degree students is 2 = 4.6 and the population variance for management only students is 2 = 5.0. The current study found that the mean score for the dual degree students is 81 and the mean score for the mana only students is 83. Research Question: At level of significance = 0.05, is there evidence that the dual degree students are receiving grades than regular management students? 1. Mark or highlight the correct hypotheses (Choose carefully as several answers depend on which HA is selected H0: 1 - 2 0, HA: 1 - 2 < 0 (dual degree less than single degree) H0: 1 - 2 0, HA: 1 - 2 > 0 (dual degree greater than single degree) H0: 1 - 2 = 0, HA: 1 - 2 0 (dual degree not equal to single degree) 2. The selected Hypotheses in question 1 Have a two-tailed test with lower and upper reject regions Have a one-tailed test with lower reject region Have a one-tailed test with upper reject region 3. Alpha is the probability of a type 1 error, the risk we are willing to take of rejecting H0 when it's true. The critical value(s) bound the reject region(s) with probability alpha. Pvalue is the probability of an equal or more extreme test statistic computed assuming H0 is true. Mark the two Decision Rules for rejecting H0 that apply to this problem. Test statistic > positive critical value of upper tail test Test statistic < negative critical value of lower tail test Test statistic falls outside interval (negative critical value of lower tail, positive critical value of upper tail) of two-tai pvalue < 4. In the copy of technology Statistics 250 below, update the Inputs to the information given in the problem abo Use the information in the problem as a checklist to verify that all the Inputs are updated (common error is to not 5. In Problem 1, the sample statistic is the difference of sample means. It is computed in H76 The standard deviation of the distribution of sample statistics is computed in M76, which is a function of H77. The distance in standard deviations of the difference of sample means from the numeric value in H0 is called the te It is computed in Q76. The test statistic has the probability distribution in I76. Each HA has different values of the Reject Region, critical value(s), and pvalue. These values are in one and the sam Critical values are computed in columns T and U using Excel function NORM.S.INV Pvalues are computed in column V using Excel function NORM.S.DIST for each of the HA's in column R. The Decision in column W about H0 turns red when H0 is rejected What are the following values as displayed below for the HA matching this problem? Test statistic = Reject Region = Critical Value(s) = pvalue = 6. The rules selected in question 3 indicate that The evidence against H0 is significant. H0 is rejected The evidence against H0 is not significant. H0 is not rejected 7. In every day non-statistical language, You conclude dual degree students are receiving lower grades than regular management students. You conclude dual degree students are not receiving lower grades than regular management students. Inputs Outputs Sample1 mean population variance v1 size n1 51.0000 Summary statistics 6.0000 _1 _2 pooled variance 200 Sample2 mean population variance v2 65.0000 6.2500 size n2 300 alpha 0.05 -14.000 0.05 tions from population 1 (dual degree s is 2 = 4.6 and the population d the mean score for the management degree students are receiving lower epend on which HA is selected) H0 when it's true. ng H0 is true. al value of upper tail) of two-tailed test. tion given in the problem above. ated (common error is to not update all of the Inputs). which is a function of H77. eric value in H0 is called the test statistic. values are in one and the same row as HA. HA's in column R. ement students. nagement students. Difference of 2 Means Confidence Limits Distribution std error (se) Margin Lower Upper test stat Z 0.2255 0.4419 -14.4419 -13.5581 -62.095 Reject Critical Values HA 1 - 2 < 0 Region Lower Upper p-value Decision about H0 Lower -1.6449 NA 0.0000 Reject 1 - 2 > 0 Upper NA 1.6449 1.0000 Accept 1 - 2 0 Both -1.9600 1.9600 0.0000 Reject 2015 Fall A well-known ice cream store tested two different methods for scooping ice cream so they can use the method t weight variability in their training of new and current employees. Ice cream cones scooped sample weights follo Research Question: At level of significance = 0.05, do the two methods have equal variability? Method 1 Method 2 4.7 3.6 3.7 4.1 3.2 3.9 3.8 5.5 3.9 4.1 4.8 4.7 3.5 4.9 5.1 3.5 4.8 3.9 5.3 3.8 3.3 4.8 4.2 5.0 5.3 4.4 4.6 1. Mark or highlight the correct hypotheses (Choose carefully as several answers depend on HA) H0: 12 22, HA: 12 < 22 H0: 12 22, HA: 12 > 22 H0: 12 = 22, HA: 12 22 2. The selected Hypotheses in question 1 Have a two-tailed test with lower and upper reject regions Have a one-tailed test with lower reject region Have a one-tailed test with upper reject region 3. Alpha is the probability of a type 1 error, the risk we are willing to take of rejecting H0 when it's true. The critical value(s) bound the reject region(s) with probability alpha. Pvalue is the probability of an equal or more extreme test statistic computed assuming H0 is true. Mark the two Decision Rules for rejecting H0 that apply to this problem. Test statistic > positive critical value of upper tail test Test statistic < negative critical value of lower tail test Test statistic falls outside interval (negative critical value of lower tail, positive critical value of upper tail) of two-tai pvalue < 4. In the copy of technology Statistics 250 below, update the Inputs to the information given in the problem abo Use the information in the problem as a checklist to verify that all the Inputs are updated (common error is to not 5. In Problem 2, the sample statistic is the ratio of variances. It is computed in Q86, which is a function of the samp It has the probability distribution in L86. Each HA has different values of the Reject Region, critical value(s), and pvalue. These values are in one and the sam Critical values are computed in columns T and U using Excel function F.INV Pvalues are computed in column V using Excel function F.DIST for each of the HA's in column R. The Decision in column W about H0 turns red when H0 is rejected What are the following values as displayed below for the HA matching this problem? Test statistic = Reject Region = Critical Value(s) = pvalue = 6. The rules selected in question 3 indicate that The evidence against H0 is significant. H0 is rejected The evidence against H0 is not significant. H0 is not rejected 7. In every day non-statistical language, You conclude the two scooping methods have equal variability. You conclude the two scooping methods have unequal variability. Inputs Outputs alpha 0.05 Data If > 250 observations, update data ranges in orange cells in Outputs Do not write below the data column(s) Sample1 4.7000 3.7000 3.2000 3.1000 3.9000 4.8000 3.1000 5.1000 Sample2 3.7000 4.1000 3.5000 5.5000 4.1000 4.7000 4.9000 3.5000 Summary statistics Sample1 standard deviation s sample variance s2 Sample2 standard deviation s sample variance s2 Do not write below the data column(s) 4.9000 5.3000 2.8000 4.2000 15.0000 3.9000 3.7000 4.8000 5.0000 5.3000 4.4000 4.6000 so they can use the method that has the least scooped sample weights follow. al variability? pend on HA) H0 when it's true. g H0 is true. value of upper tail) of two-tailed test. tion given in the problem above. ated (common error is to not update all of the Inputs). hich is a function of the sample variances in K88 and K94. values are in one and the same row as HA. Summary statistics Sample1 standard deviation s sample variance s2 size n DF Distribution F 3.149 9.914 13 12 Sample2 standard deviation s sample variance s2 size n DF 0.652 0.425 15 14 test stat F 23.3508 HA 1 < 22 2 12 > 22 12 22 Reject Region Lower Upper Both Critical Values Lower 0.379201 Upper p-value Decision about H0 0.9999996 Accept 2.5342433 3.54E-007 Reject 0.3118946 3.0501548 7.08E-007 Reject A random sample of IQs in a population resulted in the following data: 118, 98, 112, 114, 106, and 111. Research Question: At level of significance = 0.05, is the mean IQ different from 100? 1. Mark or highlight the correct hypotheses (Choose carefully as several answers depend on HA) H0: 100, HA: < 100 H0: 100, HA: > 100 H0: = 100, HA: 100 2. The selected Hypotheses in question 1 Have a two-tailed test with lower and upper reject regions Have a one-tailed test with lower reject region Have a one-tailed test with upper reject region 3. Alpha is the probability of a type 1 error, the risk we are willing to take of rejecting H0 when it's true. The critical value(s) bound the reject region(s) with probability alpha. Pvalue is the probability of an equal or more extreme test statistic computed assuming H0 is true. Mark the two Decision Rules for rejecting H0 that apply to this problem. Test statistic > positive critical value of upper tail test Test statistic < negative critical value of lower tail test Test statistic falls outside interval (negative critical value of lower tail, positive critical value of upper tail) of two-tai pvalue < 4. In the copy of technology Statistics 250 below, update the Inputs to the information given in the problem abo Use the information in the problem as a checklist to verify that all the Inputs are updated (common error is to not 5. In Problem 3, the sample statistic is the sample mean. It is computed in I72. The standard deviation of the distribution of sample means is computed in L72, which is a function of I73 and I74. The distance in standard deviations of the sample mean from the numeric value in H0 is called the test statistic. It is computed in Q72. The test statistic has the probability distribution in J72 and K72. Each HA has different values of the Reject Region, critical value(s), and pvalue. These values are in one and the sam Critical values are computed in columns T and U using Excel function T.INV Pvalues are computed in column V using Excel function T.DIST for each of the HA's in column R. The Decision in column W about H0 turns red when H0 is rejected What are the following values as displayed below for the HA matching this problem? Test statistic = Reject Region = Critical Value(s)= pvalue = 6. The rules selected in question 3 indicate that The evidence against H0 is significant. H0 is rejected The evidence against H0 is not significant. H0 is not rejected 7. In every day non-statistical language, You conclude mean IQ is equal to 100 You conclude mean IQ is not equal to 100 d value of mean in H0 alpha 40.0000 0.05 Data 1.0000 2.0000 3.0000 4.0000 5.0000 6.0000 7.0000 8.0000 9.0000 10.0000 11.0000 12.0000 Data only in the green cells. Don't type anything under the data in column C or it will be read as data. Summary Statistics sample mean sample standard deviation s sample size n 6.5000 3.6056 12 2, 114, 106, and 111. 100? pend on HA) H0 when it's true. g H0 is true. value of upper tail) of two-tailed test. tion given in the problem above. ated (common error is to not update all of the Inputs). h is a function of I73 and I74. 0 is called the test statistic. values are in one and the same row as HA. Mean Confidence Limits Distribution T df 11 std error (se) 1.0408 Margin 2.2909 Lower 4.2091 Upper 8.7909 test stat -32.1858 HA < 40 > 40 40 Reject Critical Values Region Lower Lower -1.7959 Upper NA p-value 0.0000 Decision about H0 Upper NA 1.7959 1.0000 Accept Both -2.2010 2.2010 0.0000 Reject Reject 2015 Fall The administration of UMUC is concerned that a significant number of their students are not graduating in 4 yea may contribute to the low graduation rate could be the number of students that change their majors after the fir A nationwide study found that about 49% of first year students at any university change their major after the firs UMUC sampled 100 students and found that 47 had changed their major after their first year. Research Question: At level of significance = 0.05, is the proportion of students at UMUC that change their ma 49%? 1. Mark or highlight the correct hypotheses (Choose carefully as several answers depend on HA) H0: 0.49, HA: < 0.49 H0: 0.49, HA: > 0.49 H0: = 0.49, HA: 0.49 2. The selected Hypotheses in question 1 Have a two-tailed test with lower and upper reject regions Have a one-tailed test with lower reject region Have a one-tailed test with upper reject region 3. Alpha is the probability of a type 1 error, the risk we are willing to take of rejecting H0 when it's true. The critical value(s) bound the reject region(s) with probability alpha. Pvalue is the probability of an equal or more extreme test statistic computed assuming H0 is true. Mark the two Decision Rules for rejecting H0 that apply to this problem. Test statistic > positive critical value of upper tail test Test statistic < negative critical value of lower tail test Test statistic falls outside interval (negative critical value of lower tail, positive critical value of upper tail) of two-tai pvalue < 4. In the copy of technology Statistics 250 below, update the Inputs to the information given in the problem abo Confidence level = 100% minus alpha (5%) Use the information in the problem as a checklist to verify that all the Inputs are updated (common error is to not 5. In Problem 4, the sample statistic is the sample proportion in D77. The standard deviation of the distribution of sample proportions is computed in L77, which is a function of D76 an The distance in standard deviations of the sample proportion from the numeric value in H0 is called the test statisti It is computed in Q77. The test statistic has the probability distribution in J77. Each HA has different values of the Reject Region, critical value(s), and pvalue. These values are in one and the sam Critical values are computed in columns T and U using Excel function NORM.S.INV Pvalues are computed in column V using Excel function NORM.S.DIST for each of the HA's in column R. The Decision in column W about H0 turns red when H0 is rejected What are the following values as displayed below for the HA matching this problem? 1.8257 Test statistic = Upper Reject Region = Critical Value(s)= pvalue = 1.645 0.0339 6. The rules selected in question 3 indicate that The evidence against H0 is significant. H0 is rejected The evidence against H0 is not significant. H0 is not rejected 7. In every day non-statistical language, You conclude the proportion of students at UMUC that change their major is greater than 49% You conclude the proportion of students at UMUC that change their major is not greater than 49% Inputs Outputs tested value of proportion in H0 sample proportion p sample size n 0.1000 0.2000 30 confidence level 95% margin of error to est. n 0.03 ts are not graduating in 4 years. One factor that ange their majors after the first year at UMUC. ange their major after the first year. r first year. UMUC that change their major greater than pend on HA) H0 when it's true. g H0 is true. value of upper tail) of two-tailed test. tion given in the problem above. ated (common error is to not update all of the Inputs). which is a function of D76 and D78. in H0 is called the test statistic. values are in one and the same row as HA. HA's in column R. eater than 49% Distribution Z alpha 5% std error (se) 0.0548 Margin 0.1431 est. n 683 Proportion Confidence Limits Lower Upper 0.0569 0.3431 test stat 1.8257 HA < 0.1 Testing > 0.1 0.0730 For CI 0.1 Reject Region Lower Critical Values Lower Upper -1.6449 NA p-value 0.9661 Decision about H0 Accept Upper NA 1.6449 0.0339 Reject Both -1.9600 1.9600 0.0679 Accept 2015 Fall Graph the following data using the XY chart feature in Excel. Add the chart name, Axes labels, R2, and the linear equation to your chart. X Year Y Highest Grossing Movies 1997 2015 2015 2012 2015 2015 2011 2013 2013 2015 2016 2011 2003 2012 2014 2012 2006 2010 2011 1993 1999 2010 2016 2012 2008 2001 2013 1994 2007 2010 2013 2014 2007 2016 2003 2009 2002 $2,386.80 $2,068.20 $1,670.40 $1,519.60 $1,516.00 $1,405.40 $1,341.50 $1,276.50 $1,255.40 $1,159.40 $1,151.10 $1,123.80 $1,119.90 $1,118.60 $1,104.10 $1,084.90 $1,076.20 $1,063.20 $1,045.70 $1,029.20 $1,027.00 $1,024.50 $1,022.60 $1,021.10 $1,004.60 $974.80 $970.80 $968.50 $963.40 $960.30 $958.40 $956.00 $949.90 $937.90 $936.70 $934.40 $926.00 Scatterplot of Hi $3,000.00 $2,500.00 $2,000.00 Grossing Movies by Price $1,500.00 $1,000.00 f(x) = 2.0616714905x - 32 R = 0.0026106934 $500.00 $0.00 1975 1980 1985 1990 Y 2004 2005 2007 2009 2015 2002 2012 2016 2001 2013 2015 2005 2009 2012 2010 2002 1996 2007 2004 1982 2013 2008 2004 2016 1977 2014 2009 2014 2006 2012 2014 2010 2014 2012 2005 2013 2003 2009 2016 2013 2014 2011 2014 2009 2007 $919.80 $896.90 $890.90 $886.70 $880.70 $879.00 $877.20 $872.70 $871.50 $865.00 $857.40 $848.80 $836.30 $829.70 $825.50 $821.70 $817.40 $799.00 $796.70 $792.90 $788.70 $786.60 $783.80 $782.40 $775.40 $773.30 $769.70 $758.50 $758.20 $757.90 $755.40 $752.60 $747.90 $746.90 $745.00 $743.60 $742.10 $731.30 $727.60 $723.20 $714.40 $712.20 $710.60 $709.70 $709.70 2014 2010 2011 2012 2015 1994 2014 1999 2013 2011 2006 2014 2003 2015 2002 2013 2008 $709.00 $698.50 $694.70 $694.40 $682.30 $677.90 $675.10 $672.80 $668.00 $665.70 $660.90 $657.80 $654.30 $653.40 $649.40 $644.60 $631.70 Scatterplot of Highest Grossing Movies by Year 3,000.00 2,500.00 2,000.00 Highest Grossing Movies Linear (Highest Grossing Movies) Linear (Highest Grossing Movies) 1,500.00 1,000.00 f(x) = 2.0616714905x - 3229.9817434465 R = 0.0026106934 $500.00 $0.00 1975 1980 1985 1990 1995 Years 2000 2005 2010 2015 2020 Graph the following data using the XY chart feature in Excel. Add the chart name, Axes descriptions, R2, and the linear equation to your chart. X Hours Studied Y Test Grade 3 5 2 6 7 1 2 7 1 7 2 6 7 1 5 5 5 3 3 9 5 6 2 5 2 3 8 4 5 6 75 90 75 80 90 50 65 85 40 100 70 85 85 70 80 85 80 50 60 95 75 80 50 65 40 70 80 75 60 95 Grades depending on am 120 100 f(x) = 5.5147901857x + 48.8 R = 0.600138932 80 Test Grades 60 40 20 0 0 1 2 3 4 H ades depending on amount of time studying f(x) = 5.5147901857x + 48.8844301766 R = 0.600138932 1 2 3 4 5 Hours Studied 6 7 8 9 10 Graph the following data using the XY chart feature in Excel. Add the chart name, Axes descriptions, R2, and the linear equation to your chart. X Humidity Y Time Spent Outside 50% 40% 70% 20% 90% 30% 70% 30% 20% 30% 40% 10% 90% 70% 80% 50% 70% 40% 30% 20% 20% 70% 70% 80% 50% 6 7 0.5 4.5 0.25 4 5.5 2 2 2.75 2.75 8.5 1 2.2 1.5 3 1.25 4 3.75 6 5 2.5 2 1 4 Spending Time Outside based on H 9 8 7 6 5 Amount of time spendt outside f(x) = - 5.6244481236x + 6.10772 R = 0.4210007165 4 3 2 1 0 0% 10% 20% 30% 40% 50% Humidi Outside based on Humidity - 5.6244481236x + 6.1077262693 0.4210007165 20% 30% 40% 50% 60% Humidity Level 70% 80% 90% 100%