MATH 302 week 8 tests questions & answers | AMU Latest update

MATH 302 week 8 tests questions & answers | AMU Latest update Question 1 of 20 1.0/ 1.0 Points An urban economist is curious if the distribution in where Oregon residents live is different today than it was in 1990. She observes that today there are approximately 3,109 thousand residents in NW Oregon, 902 thousand residents in SW Oregon, 244 thousand in Central Oregon, and 102 thousand in Eastern Oregon. She knows that in 1990 the breakdown was as follows: 72.7% NW Oregon, 20.7% SW Oregon, 4.8% Central Oregon, and 2.8% Eastern Oregon. Can she conclude that the distribution in residence is different today at a 0.05 level of significance?  A. yes because the p-value = .0009  B. no, because the p-value = .0009  C. yes because the p-value = .0172  D. no because the p-value = .0172 Answer Key:C Feedback: NW Oregon SW Oregon Central Oregon Eastern Oregon Observed Counts Expected Counts 4357*.727 = 3167.539 4357*.207= 901.899 4357*.048= 209.136 4357*.028= 121.996 Use Excel to find the p-value =CHISQ.TEST(Highlight Observed, Highlight Expected) p-value is < .05, Reject Ho. Yes, this is significant. Question 2 of 20 1.0/ 1.0 Points Pamplona, Spain is the home of the festival of San Fermin – The Running of the Bulls. The town is in festival mode for a week and a half every year at the beginning of July. There is a running joke in the city, that Pamplona has a baby boom every April – 9 months after San Fermin. To test this claim, a resident takes a random sample of 300 birthdays from native residents and finds the following observed counts: January 25 February 25 March 27 April 26 May 21 June 26 July 22 August 27 September 21 October 26 November 28 December 26 At the 0.05 level of significance, can it be concluded that births in Pamplona are not equally distributed throughout the 12 months of the year? Hypotheses: H0: Births in Pamplona ______ equally distributed throughout the year. H1: Births in Pamplona ______ equally distributed throughout the year. Select the best fit choices that fit in the two blank spaces above.  A. are, are not  B. are not, are  C. are, are  D. are not, are not Answer Key:A Question 3 of 20 0.0/ 1.0 Points Students at a high school are asked to evaluate their experience in the class at the end of each school year. The courses are evaluated on a 1-4 scale – with 4 being the best experience possible. In the History Department, the courses typically are evaluated at 10% 1’s, 15% 2’s, 34% 3’s, and 41% 4’s. Mr. Goodman sets a goal to outscore these numbers. At the end of the year he takes a random sample of his evaluations and finds 10 1’s, 13 2’s, 48 3’s, and 52 4’s. At the 0.05 level of significance, can Mr. Goodman claim that his evaluations are significantly different than the History Department’s?  A. yes, the p-value = 0.5893  B. no, the p-value = 0.5893  C. no, the p-value = 0.3913  D. yes, the p-value = 0.3913 Answer Key:C Feedback: 1's 2's 3's 4's Observed Counts Expected Counts 123 *.10 = 12.3 123*.15 = 18.45 123*.34 = 41.82 123*.41 = 50.43 Use Excel to find the p-value =CHISQ.TEST(Highlight Observed, Highlight Expected) p-value > .05, Do Not Reject Ho. No, this is not significant. Question 4 of 20 0.0/ 1.0 Points Click to see additional instructions A college professor is curious if the location of a seat in class affects grades in the class. They are teaching in a lecture hall with 240 students. The lecture hall has 10 rows, so they split the rows into 5 sections – Rows 1-2, Rows 3-4, Rows 5-6, Rows 7-8, and Rows 9-10. At the end of the course, they determine the top 25% of grades in the class, and if the location of the seat makes no difference, they would expect that these top 25% of students would be equally dispersed throughout the classroom. Their observations are recorded below. Run a Goodness of Fit test to determine whether or not location has an impact on the grade. Let α=0.05. Enter the expected count for each section in the table below. Enter whole numbers without any decimals. Rows 1-2 Rows 3-4 Rows 5-6 Rows 7-8 Rows 9-10 # in Top 25% Expected Counts Answer Key:12, 12, 12, 12, 12 Feedback: Top 25% of 240 is 240*.25 = 60, then divide this equally among the 5 groups. 60/5 Question 5 of 20 1.0/ 1.0 Points The permanent residence of adults aged 18-25 in the U.S. was examined in a survey from the year 2000. The survey revealed that 27% of these adults lived alone, 32% lived with a roommate(s), and 41% lived with their parents/guardians. In 2008, during an economic recession in the country, another such survey of 1600 people revealed that 398 lived alone, 488 lived with a roommate(s), and 714 lived with their parents. Is there a significant difference in where young adults lived in 2000 versus 2008 and state the pvalue? Test with a Goodness of Fit test at α=0.05. Alon e Roommat es Parents/Guardi ans Observ ed Counts Expecte d Counts  A. yes, the p-value = 0.  B. No, the p-value = 0.  C. Yes, the p-value = 0.  D. Yes, the p-value = 0. Answer Key:C Feedback: Use Excel to find the p-value you have the Observed and Expected Counts you can use =CHISQ.TEST( Highlight Observed Counts, Highlight Expected Counts) = 0. 0. < .05, Reject Ho. Yes, this is significant. Question 6 of 20 0.0/ 1.0 Points A large department store is curious about what sections of the store make the most sales. The manager has data from ten years prior that show 30% of sales come from Clothing, 25% Home Appliances, 18% Housewares, 13% Cosmetics, 12% Jewelry, and 2% Other. In a random sample of 550 current sales, 188 came from Clothing, 153 Home Appliances, 83 Housewares, 54 Cosmetics, 61 Jewelry, and 11 Other. At α=0.10, can the manager conclude that the distribution of sales among the departments has changed?  A. yes because the p-value is .0006  B. no, because the p-value is .0006  C. yes because the p-value = .0321  D. no, because the p-value = .0321 Answer Key:C Feedback: Clothing Home App. Housewares Cosmetics Jewelry Other Observed Counts 11 Expected Counts 550*.30 = 165 550*.25 = 137.5 550*.18 = 99 550*.13 = 71.5 550*.12 = 66 550*.02= 11 Use Excel to find the p-value =CHISQ.TEST(Highlight Observed, Highlight Expected) The p-value is < .10, Reject Ho. Yes, this is significant. Question 7 of 20 0.0/ 1.0 Points Click to see additional instructions A college prep school advertises that their students are more prepared to succeed in college than other schools. To verify this, they categorize GPA’s into 4 groups and look up the proportion of students at a state college in each category. They find that 7% have a 0-0.99, 21% have a 1-1.99, 37% have a 2-2.99, and 35% have a 3-4.00 in GPA. They then take a random sample of 200 of their graduates at the state college and find that 19 has a 0-0.99, 28 have a 1-1.99, 82 have a 2-2.99, and 71 have a 3-4.00. Can they conclude that the grades of their graduates are distributed differently than the general population at the school? Test at the 0.05 level of significance. Enter the test statistic - round to 4 decimal places. Test statistic: Answer Key:7.3315 Feedback: 0-0.99 1-1.99 2-2.99 3-4.00 Observed Counts Expected Counts =200*0.07 =14 =200*.21= 42 =200*.37 = 74 =200*.35 = 70 Test Stat = Test Stat = 7. Question 8 of 20 0.0/ 1.0 Points A company that develops over-the-counter medicines is working on a new product that is meant to shorten the length of sore throats. To test their product for effectiveness, they take a random sample of 110 people and record how long it took for their symptoms to completely disappear. The results are in the table below. The company knows that on average (without medication) it takes a sore throat 6 days or less to heal 42% of the time, 7-9 days 31% of the time, 10-12 days 16% of the time, and 13 days or more 11% of the time. Can it be concluded at the 0.01 level of significance that the patients who took the medicine healed at a different rate than these percentages? Hypotheses: H0: There is __________ in duration of a sore throat for those that took the medicine. H1: There is __________in duration of a sore throat for those that took the medicine. Select the best fit choices that fit in the two blank spaces above.  A. no difference, a difference  B. a difference, no difference  C. no difference, no difference  D. a difference, a difference Answer Key:A Part 2 of 4 - Chi Square Test for Independence 0.0/ 6.0 Points Question 9 of 20 0.0/ 1.0 Points Click to see additional instructions A public opinion poll surveyed a simple random sample of 550 voters in Oregon. The respondents were asked which political party they identified with most and were categorized by residence. Results are shown below. Decide if voting preference is independent from location of residence. Let α=0.05. Republican Democrat Independent NW Oregon 85 103 22 SW Oregon 45 66 10 Central Oregon 46 53 9 Eastern Oregon 67 33 11 Enter the test statistic - round to 4 decimal places. Answer Key:18.0789 Feedback: We are running a Chi-Square Test for Independence. Copy and Paste the table into Excel. You are given the Observed Counts in the table. Next you need to sum the rows and columns. Once you have those you need to calculate the Expected Counts. You need to find the probability of the row and then multiple it by the column total. Republican Democrat Independent Sum NW Oregon SW Oregon Central Oregon Eastern Oregon Sum Republican Democrat Independent NW Oregon =243*(210/550) =255*(210/550) =52*(210/550) SW Oregon =243*(121/550) =255*(121/550) =52*(121/550) Central Oregon =243*(108/550) =255*(108/550) =52*(108/550) Eastern Oregon =243*(111/550) =255*(111/550) =52*(111/550) Now that we calculated the Expected Counts we need to find the Test Statistic. Test Stat = You will need to use all 12 Count values but I am only showing you 3 because there isn't room to write out the entire equation. Question 10 of 20 0.0/ 1.0 Points Click to see additional instructions A local gym is looking in to purchasing more exercise equipment and runs a survey to find out the preference in exercise equipment amongst their members. They categorize the members based on how frequently they use the gym each month – the results are below. Run an independence test at the 0.01 level of significance. Free Weights Weight Machines Endurance Machines Aerobics Equipment 0-10 Uses 11-30 Uses 20 18 9 9 31+ Uses 26 12 11 9 Enter the P-Value - round to 4 decimal places. Make sure you put a 0 in front of the decimal. Answer Key:0.0144 Feedback: We are running a Chi-Square Test for Independence. Copy and Paste the table into Excel. You are given the Observed Counts in the table. Next you need to sum the rows and columns. Once you have those you need to calculate the Expected Counts. You need to find the probability of the row and then multiple it by the column total. Free Weights Weight Machines Endurance Machines Aerobics Equipment 0-10 Uses 11-30 Uses 20 18 9 9 31+ Uses 26 12 11 9 Sum Free Weights Weight Machines Endurance Machines Aerobics Equipment 0-10 Uses =58*(67/181) =47*(67/181) =45*(67/181) =31*(67/181) 11-30 Uses =58*(56/181) =47*(56/181) =45*(56/181) =31*(56/181) 31+ Uses =58*(58/181) =47*(58/181) =45*(58/181) =31*(58/181) Now that we calculated the Expected Count we can use Excel to find the p-value. Use =CHISQ.TEST(highlight actual counts, highlight expected counts) = 0.0144 Question 11 of 20 0.0/ 1.0 Points Click to see additional instructions A restaurant chain that has 3 locations in Portland is trying to determine which of their 3 locations they should keep open on New Year’s Eve. They survey a random sample of customers at each location and ask each whether or not they plan on going out to eat on New Year’s Eve. The results are below. Run a test for independence to decide if the proportion of customers that will go out to eat on New Year’s Eve is dependent on location. Use α=0.05. Enter the P-Value - round to 4 decimal places. Make sure you put a 0 in front of the decimal. 0.4387 Answer Key:0.1294 Feedback: We are running a Chi-Square Test for Independence. Copy and Paste the table into Excel. You are given the Observed Counts in the table. Next you need to sum the rows and columns. Once you have those you need to calculate the Expected Counts. You need to find the probability of the row and then multiple it by the column total. NW Location NE Location SE Location Sum Will Go Out Won’t Go Out Sum NW Location NE Location SE Location Will Go Out =86*(151/216) =65*(151/216) =65*(151/216) Won’t Go Out =86*(65/216) =65*(65/216) =65*(65/216) Now that we calculated the Expected Count we can use Excel to find the p-value. Use =CHISQ.TEST(highlight actual counts, highlight expected counts) = 0.129 Question 12 of 20 0.0/ 1.0 Points Click to see additional instructions NW Location NE Location SE Location Will Go Out 66 40 45 Won’t Go Out 20 25 20 An electronics store has 4 branches in a large city. They are curious if sales in any particular department are different depending on location. They take a random sample of 4 purchases throughout the 4 branches – the results are recorded below. Run an independence test for the data below at the 0.05 level of significance. Appliances TV Computers Cell Phones Branch Branch Branch Branch Enter the test statistic - round to 4 decimal places. .0568 Answer Key:7.2612 Feedback: We are running a Chi-Square Test for Independence. Copy and Paste the table into Excel. You are given the Observed Counts in the table. Next you need to sum the rows and columns. Once you have those you need to calculate the Expected Counts. You need to find the probability of the row and then multiple it by the column total. Appliances TV Computers Cell Phones Branch Branch Branch Branch Sum Appliances TV Computers Cell Phones Branch 1 =204*(171/648) =98*(171/648) =233*(171/648) =113*(171/648) Branch 2 =204*(148/648) =98*(148/648) =233*(148/648) =113*(148/648) Branch 3 =204*(152/648) =98*(152/648) =233*(152/648) =113*(152/648) Branch 4 =204*(177/648) =98*(177/648) =233*(177/648) =113*(177/648) Now that we calculated the Expected Counts we need to find the Test Statistic. Test Stat = You will need to use all 16 Count values but I am only showing you the first and last value because there isn't room to write out the entire equation. Question 13 of 20 0.0/ 1.0 Points Click to see additional instructions A university changed to a new learning management system during the past school year. The school wants to find out how it’s working for the different departments – the results in preference found from a survey are below. Run a test for independence at α=0.05. Prefers Old LMS Prefers New LMS No Preference School of Business 18 29 8 School of Science 41 11 4 School of Liberal Arts 25 20 7 Enter the 4 missing values in the expected matrix. Round to four decimal places. Prefers Old LMS Prefers New LMS No Preference School of Business 20.2454 School of Science 28.8589 6.5276 School of Liberal Arts 19.1411 6.0613 Answer Key:28.3436, 6.4110, 20.6135, 26.7975 Feedback: We are running a Chi-Square Test for Independence. Copy and Paste the table into Excel. You are given the Observed Counts in the table. We need to calculate the Expected Counts. Then sum up the rows and column. You need to find the probability of the row and then multiple it by the column total. Prefers Old LMS Prefers New LMS No Preference Sum School of Business 18 29 8 55 School of Science 41 11 4 56 School of Liberal Arts 25 20 7 52 Sum Prefers Old LMS Prefers New LMS No Preference School of Business =84*(55/163) =60*(55/163) =19*(55/163) School of Science =84*(56/163) =60*(56/163) =19*(56/163) School of Liberal =84*(52/163) =60*(52/163) =19*(52/163) Arts Question 14 of 20 0.0/ 1.0 Points Click to see additional instructions The medal count for the 2018 winter Olympics is recorded below. Run an independence test to find out if the medal won is dependent on country. Use α=0.10. Gold Silver Bronze Norway 17 14 11 Germany 16 9 6 Canada 13 9 11 United States 12 10 7 Enter the P-Value - round to 4 decimal places. Make sure you put a 0 in front of the decimal. Answer Key:0.8861 Feedback: We are running a Chi-Square Test for Independence. Copy and Paste the table into Excel. You are given the Observed Counts in the table. Next you need to sum the rows and columns. Once you have those you need to calculate the Expected Counts. You need to find the probability of the row and then multiple it by the column total. Gold Silver Bronze Sum Norway Germany 16 9 6 31 Canada 13 9 11 33 United States 12 10 7 29 Sum Gold Silver Bronze Norway =58*(42/135) =42*(42/135) =35*(42/135) Germany =58*(31/135) =42*(31/135) =35*(31/135) Canada =58*(33/135) =42*(33/135) =35*(33/135) United States =58*(33/135) =42*(29/135) =35*(29/135) Now that we calculated the Expected Count we can use Excel to find the p-value. Use =CHISQ.TEST(highlight actual counts, highlight expected counts) = 0.8861 Part 3 of 4 - F-Distribution and Three or More Means 2.0/ 3.0 Points Question 15 of 20 0.0/ 1.0 Points You are conducting a study of three types of feed supplements for cattle to test their effectiveness in producing weight gain among calves whose feed includes one of the supplements. You have four groups of 30 calves (one is a control group receiving the usual feed, but no supplement). You will conduct a one-way ANOVA after one year to see if there are difference in the mean weight for the four groups. If SSwithin = 366.5 and SStotal = 627.9 for this data. Identify the SSbetween  A. 261.4  B. 3  C. 30  D. 994.4  E. 255.4 Answer Key:A Feedback: SStotal = SSwithin + SSbetween 627.9 = 366.5 + SSbetween Question 16 of 20 1.0/ 1.0 Points The χ 2 Independence Test is always:  A. Right-Tailed  B. Left-Tailed  C. A Confidence Interval  D. Two-Tailed Answer Key:A Question 17 of 20 1.0/ 1.0 Points The F Statistic from an experiment with k = 4 and n = 100 is 4.72. At α = 0.01, will you reject the null hypothesis?  A. No  B. Yes Answer Key:B Feedback: Use Excel to find the p-value df1 = k - 1 = 4-1 = 3 df2 = n - k = 100 - 4 = 96 =F.DIST.RT(4.72,3,96) = 0. 0. < .01, Reject Ho, Yes, this is significant. Part 4 of 4 - Chi-Square Distribution for Independence 0.33/ 3.0 Points Question 18 of 20 0.0/ 1.0 Points Click to see additional instructions If the number of degrees of freedom for a chi-square distribution is 25, what is the standard deviation? Round to four decimal places. Standard Deviation= 0.3167 Answer Key:7.0711 Feedback: SQRT(2*25) Question 19 of 20 0.33/ 1.0 Points Click to see additional instructions Staples, a chain of large office supply stores, sells a line of desktop and laptop computers. Company executives want to know whether the demands for these two types of computers are dependent on one another. Each day's demand for each type of computers is categorized as Low, Medium-Low, MediumHigh, or High. The data shown in the table below is based on 205 days of operation. Based on these data, can ? Test at the 5% level of significance. desktops low med-low med-high high low laptops med-low med-high high What is the test value for this hypothesis test? Answer: Round your answer to two decimal places. What is the critical value for this hypothesis test? Answer: Round your answer to two decimal places. What is the conclusion for this hypothesis test? Choose one. 1. At the .05 level of significance, Staples can conclude that demands for these two types of computers are independent. 2. At the .05 level of significance, Staples can conclude that demands for these two types of computers are dependent. Answer: 2 Enter only a 1 or 2 for your answer. Answer Key:17.05, 16.92, 2 Feedback: We are running a Chi-Square Test for Independence. Copy and Paste the table into Excel. You are given the Observed Counts in the table and the rows and columns are already summed for you. We need to calculate the Expected Counts. You need to find the probability of the row and then multiple it by the column total. low med-low med-high high low =30*(36/205) =65*(36/205) =57*(36/205) =53*(36/205) med-low =30*(65/205) =65*(65/205) =57*(65/205) =53*(65/205) med-high =30*(57/205) =65*(57/205) =57*(57/205) =53*(57/205) high =30*(47/205) =65*(47/205) =57*(47/205) =53*(47/205) Now that we calculated the Expected Counts we need to find the Test Statistic. Test Stat = You will use this equation to find the Test Stat. You will need to all 16 Count values but I am only showing you 3 because there isn't room to write out the entire equation. To find the Chi-Square Critical Value use =CHISQ.INV.RT( ) function in Excel. The probability is .05 and df = (4-1)*(4- 1) = 9. =CHISQ.INV.RT(.05,9) = 16.92 17.05 > 16.92. The Test Stat is greater than the Critical Value. Reject Ho. This is significant and enough evidence that Staples conclude that demands for these two types of computers are dependent. Question 20 of 20 0.0/ 1.0 Points In the graph, the df is most likely:  A. 90  B. 45  C. 20  D. 2 Answer Key:D Feedback:When the curve is nonsymmetrical and skewed to the right the df is low. Question 1 of 20 0.0/ 1.0 Points A college prep school advertises that their students are more prepared to succeed in college than other schools. To verify this, they categorize GPA’s into 4 groups and look up the proportion of students at a state college in each category. They find that 7% have a 0-0.99, 21% have a 1-1.99, 37% have a 2-2.99, and 35% have a 3-4.00 in GPA. They then take a random sample of 200 of their graduates at the state college and find that 19 has a 0-0.99, 28 have a 1-1.99, 82 have a 2-2.99, and 71 have a 3-4.00. Can they conclude that the grades of their graduates are distributed differently than the general population at the school? Test at the 0.05 level of significance.  A. yes because the p-value is .0620  B. no because the p-value is .0620  C. yes because the p-value is .9379  D. no because the p-value is .9379 Answer Key:B Feedback: 0-0.99 1-1.99 2-2.99 3-4.00 Observed Counts Expected Counts =200*0.07 =14 =200*.21= 42 =200*.37 = 74 =200*.35 = 70 You can use Excel to find the p-value =CHISQ.TEST(Highlight Observed, Highlight Expected) p-value = .0620 > .05, Do Not Reject Ho. No, this is not significant. Question 2 of 20 0.0/ 1.0 Points A color code personality test categorizes people into four colors – Red (Power), Blue (Intimacy), Green (Peace), and Yellow (Fun). In general, 25% of people are Red, 35% Blue, 20% Green, and 20% Yellow. An art class of 45 students is tested at a university and 7 are found to be Red, 18 Blue, 9 Green, and 11 Yellow. Can it be concluded that personality type has an impact on students’ areas of interest and talents, such as artistic students and state the p-value? Test at a 0.05 level of significance. Red Blue GreenYellow Observed Counts 7 18 9 11 Expected Counts 11.2515.75 9 9  A. Yes, the p-value = 0.  B. No, the p-value = 0.  C. No, the p-value = 0.  D. Yes, the p-value = 0. Answer Key:C Feedback: Use Excel to find the p-value you have the Observed and Expected Counts you can use =CHISQ.TEST( Highlight Observed Counts, Highlight Expected Counts) = 0. 0. > .05, Do Not Reject Ho. No, this is not significant. Question 3 of 20 1.0/ 1.0 Points Click to see additional instructions A college professor is curious if the location of a seat in class affects grades in the class. They are teaching in a lecture hall with 240 students. The lecture hall has 10 rows, so they split the rows into 5 sections – Rows 1-2, Rows 3-4, Rows 5-6, Rows 7-8, and Rows 9-10. At the end of the course, they determine the top 25% of grades in the class, and if the location of the seat makes no difference, they would expect that these top 25% of students would be equally dispersed throughout the classroom. Their observations are recorded below. Run a Goodness of Fit test to determine whether or not location has an impact on the grade. Let α=0.05. Enter the expected count for each section in the table below. Enter whole numbers without any decimals. Rows 1-2 Rows 3-4 Rows 5-6 Rows 7-8 Rows 9-10 # in Top 25% Expected Counts Answer Key:12, 12, 12, 12, 12 Feedback: Top 25% of 240 is 240*.25 = 60, then divide this equally among the 5 groups. 60/5 Question 4 of 20 1.0/ 1.0 Points Pamplona, Spain is the home of the festival of San Fermin – The Running of the Bulls. The town is in festival mode for a week and a half every year at the beginning of July. There is a running joke in the city, that Pamplona has a baby boom every April – 9 months after San Fermin. To test this claim, a resident takes a random sample of 300 birthdays from native residents and finds the following observed counts: January 25 February 25 March 27 April 26 May 21 June 26 July 22 August 27 September 21 October 26 November 28 December 26 At the 0.05 level of significance, can it be concluded that births in Pamplona are not equally distributed throughout the 12 months of the year? Hypotheses: H0: Births in Pamplona ______ equally distributed throughout the year. H1: Births in Pamplona ______ equally distributed throughout the year. Select the best fit choices that fit in the two blank spaces above.  A. are, are not  B. are not, are  C. are, are  D. are not, are not Answer Key:A Question 5 of 20 0.0/ 1.0 Points Students at a high school are asked to evaluate their experience in the class at the end of each school year. The courses are evaluated on a 1-4 scale – with 4 being the best experience possible. In the History Department, the courses typically are evaluated at 10% 1’s, 15% 2’s, 34% 3’s, and 41% 4’s. Mr. Goodman sets a goal to outscore these numbers. At the end of the year he takes a random sample of his evaluations and finds 10 1’s, 13 2’s, 48 3’s, and 52 4’s. At the 0.05 level of significance, can Mr. Goodman claim that his evaluations are significantly different than the History Department’s?  A. yes, the p-value = 0.5893  B. no, the p-value = 0.5893  C. no, the p-value = 0.3913  D. yes, the p-value = 0.3913 Answer Key:C Feedback: 1's 2's 3's 4's Observed Counts Expected Counts 123 *.10 = 12.3 123*.15 = 18.45 123*.34 = 41.82 123*.41 = 50.43 Use Excel to find the p-value =CHISQ.TEST(Highlight Observed, Highlight Expected) p-value > .05, Do Not Reject Ho. No, this is not significant. Question 6 of 20 1.0/ 1.0 Points Click to see additional instructions A college prep school advertises that their students are more prepared to succeed in college than other schools. To verify this, they categorize GPA’s into 4 groups and look up the proportion of students at a state college in each category. They find that 7% have a 0-0.99, 21% have a 1-1.99, 37% have a 2-2.99, and 35% have a 3-4.00 in GPA. They then take a random sample of 200 of their graduates at the state college and find that 19 has a 0-0.99, 28 have a 1-1.99, 82 have a 2-2.99, and 71 have a 3-4.00. Can they conclude that the grades of their graduates are distributed differently than the general population at the school? Test at the 0.05 level of significance. Enter the test statistic - round to 4 decimal places. Test statistic: 7.3315 Answer Key:7.3315 Feedback: 0-0.99 1-1.99 2-2.99 3-4.00 Observed Counts Expected Counts =200*0.07 =14 =200*.21= 42 =200*.37 = 74 =200*.35 = 70 Test Stat = Test Stat = 7. Question 7 of 20 0.0/ 1.0 Points A company that develops over-the-counter medicines is working on a new product that is meant to shorten the length of sore throats. To test their product for effectiveness, they take a random sample of 110 people and record how long it took for their symptoms to completely disappear. The results are in the table below. The company knows that on average (without medication) it takes a sore throat 6 days or less to heal 42% of the time, 7-9 days 31% of the time, 10-12 days 16% of the time, and 13 days or more 11% of the time. Can it be concluded at the 0.01 level of significance that the patients who took the medicine healed at a different rate than these percentages? After running a Goodness of Fit test, can it be concluded that there is a statistically significant difference in duration of a sore throat for those that took the medicine and what is the p-value? 6 days or less7-9 days10-12 days13 or more days Duration of Sore Throat 49 40 12 9 Expected Counts 46.2 34.1 17.6 12.1  A. Yes, the p-value = 0.  B. No, the p-value = 0.  C. Yes, the p-value = 0.  D. No, the p-value = 0. Answer Key:D Feedback: Use Excel to find the p-value you have the Observed and Expected Counts you can use =CHISQ.TEST( Highlight Observed Counts, Highlight Expected Counts) = 0. 0. > .01, Do Not Reject Ho. No, this is not significant. Question 8 of 20 0.0/ 1.0 Points Click to see additional instructions A company manager believes that a person’s ability to be a leader is directly correlated to their zodiac sign. He never selects someone to chair a committee without first evaluating their zodiac sign. An irate employee sets out to prove her manager wrong. She claims that if zodiac sign truly makes a difference in leadership, then a random sample of 210 CEO’s in our country would reveal a difference in zodiac sign distribution. She finds the following zodiac signs for her random sample of 210 CEO’s: Births Signs 25 Aries 13 Taurus 17 Gemini 21 Cancer 16 Leo 18 Virgo 15 Libra 16 Scorpio 20 Sagittarius 11 Capricorn 23 Aquarius 15 Pisces Can she conclude that zodiac sign makes a difference in whether or not a person makes a good leader? Enter the p-value - round to 4 decimal places. Make sure you put the 0 in front of the decimal. p-value = Answer Key:0.4798 Feedback: The Expected Count is all the same value. 210*(1/12) = 17.5 Births Signs Expected Count 25 Aries 17.5 13 Taurus 17.5 17 Gemini 17.5 21 Cancer 17.5 16 Leo 17.5 18 Virgo 17.5 15 Libra 17.5 16 Scorpio 17.5 20 Sagittarius 17.5 11 Capricorn 17.5 23 Aquarius 17.5 15 Pisces 17.5 Use Excel to find the p-value =CHISQ.TEST(Highlight Observed, Highlight Expected) Part 2 of 4 - Chi Square Test for Independence 1.0/ 6.0 Points Question 9 of 20 0.0/ 1.0 Points A high school runs a survey asking students if they participate in sports. The results are found below. Run an independence test for the data at α=0.01. Freshmen Sophomores Juniors Seniors Yes No Can it be concluded that participation in sports is dependent on grade level?  A. No, it cannot be concluded that participation in sports is dependent on grade level because the p-value = 0.0020.  B. No, it cannot be concluded that participation in sports is dependent on grade level because the p-value = 0.0010.  C. Yes, it can be concluded that participation in sports is dependent on grade level because the p-value = 0.0010.  D. Yes, it can be concluded that participation in sports is dependent on grade level because the p-value = 0.0020. Answer Key:C Feedback: We are running a Chi-Square Test for Independence. Copy and Paste the table into Excel. You are given the Observed Counts in the table. Next you need to sum the rows and columns. Once you have those you need to calculate the Expected Counts. You need to find the probability of the row and then multiple it by the column total. Freshmen Sophomores Juniors Seniors Yes No Sum Freshmen Sophomores Juniors Seniors Yes =105*(260/396) =116*(260/396) =93*(260/396) =82*(260/396) No =105*(136/396) =116*(136/396) =93*(136/396) =82*(136/396) Now that we calculated the Expected Count we can use Excel to find the p-value. Use =CHISQ.TEST(highlight actual counts, highlight expected counts) = 0.0010 0.0010 < .01,Reject Ho. Yes, it can be concluded that participation in sports is dependent on grade level. Question 10 of 20 0.0/ 1.0 Points An electronics store has 4 branches in a large city. They are curious if sales in any particular department are different depending on location. They take a random sample of 4 purchases throughout the 4 branches – the results are recorded below. Run an independence test for the data below at the 0.05 level of significance. Appliances TV Computers Cell Phones Branch Branch Branch Branch Can it be concluded that sales in the various departments are dependent on branch?  A. No, it cannot be concluded that sales in the various departments are dependent on branch because the p-value = 0.6099  B. No, it cannot be concluded that sales in the various departments are dependent on branch because the p-value = 0.3901  C. Yes, it can be concluded that sales in the various departments are dependent on branch because the p-value = 0.3901  D. Yes, it can be concluded that sales in the various departments are dependent on branch because the p-value = 0.6099 Answer Key:A Feedback: We are running a Chi-Square Test for Independence. Copy and Paste the table into Excel. You are given the Observed Counts in the table. Next you need to sum the rows and columns. Once you have those you need to calculate the Expected Counts. You need to find the probability of the row and then multiple it by the column total. Appliances TV Computers Cell Phones Branch Branch Branch Branch Sum Appliances TV Computers Cell Phones Branch 1 =204*(171/648) =98*(171/648) =233*(171/648) =113*(171/648) Branch 2 =204*(148/648) =98*(148/648) =233*(148/648) =113*(148/648) Branch 3 =204*(152/648) =98*(152/648) =233*(152/648) =113*(152/648) Branch 4 =204*(177/648) =98*(177/648) =233*(177/648) =113*(177/648) Now that we calculated the Expected Count we can use Excel to find the p-value. Use =CHISQ.TEST(highlight actual counts, highlight expected counts) = 0.6099 0.6099 > 0.05, Do Not Reject Ho. No, it cannot be concluded that sales in the various departments are dependent on branch. Question 11 of 20 0.0/ 1.0 Points Click to see additional instructions A high school runs a survey asking students if they participate in sports. The results are found below. Run an independence test for the data at α=0.01. Freshmen Sophomores Juniors Seniors Yes No Enter the test statistic - round to 4 decimal places. 16.2407 Answer Key:16.2406 Feedback: We are running a Chi-Square Test for Independence. Copy and Paste the table into Excel. You are given the Observed Counts in the table. Next you need to sum the rows and columns. Once you have those you need to calculate the Expected Counts. You need to find the probability of the row and then multiple it by the column total. Freshmen Sophomores Juniors Seniors Yes No Sum Freshmen Sophomores Juniors Seniors Yes =105*(260/396) =116*(260/396) =93*(260/396) =82*(260/396) No =105*(136/396) =116*(136/396) =93*(136/396) =82*(136/396) Now that we calculated the Expected Counts we need to find the Test Statistic. Test Stat = You will need to all 8 Count values but I am only showing you 3 because there isn't room to write out the entire equation. Question 12 of 20 0.0/ 1.0 Points A public opinion poll surveyed a simple random sample of 550 voters in Oregon. The respondents were asked which political party they identified with most and were categorized by residence. Results are shown below. Decide if voting preference is independent from location of residence. Let α=0.05. Republican Democrat Independent NW Oregon 85 103 22 SW Oregon 45 66 10 Central Oregon 46 53 9 Eastern Oregon 67 33 11 Can it be concluded that voting preference is dependent on location of residence?  A. Yes, it can be concluded that voting preference is dependent on location of residence because the p-value = 0.0030  B. Yes, it can be concluded that voting preference is dependent on location of residence because the p-value = 0.0060  C. No, it cannot be concluded that voting preference is dependent on location of residence because the p-value = 0.0060  D. No, it cannot be concluded that voting preference is dependent on location of residence because the p-value = 0.0030 Answer Key:B Feedback: We are running a Chi-Square Test for Independence. Copy and Paste the table into Excel. You are given the Observed Counts in the table. Next you need to sum the rows and columns. Once you have those you need to calculate the Expected Counts. You need to find the probability of the row and then multiple it by the column total. Republican Democrat Independent Sum NW Oregon SW Oregon Central Oregon Eastern Oregon Sum Republican Democrat Independent NW Oregon =243*(210/550) =255*(210/550) =52*(210/550) SW Oregon =243*(121/550) =255*(121/550) =52*(121/550) Central Oregon =243*(108/550) =255*(108/550) =52*(108/550) Eastern Oregon =243*(111/550) =255*(111/550) =52*(111/550) Now that we calculated the Expected Count we can use Excel to find the p-value. Use =CHISQ.TEST(highlight actual counts, highlight expected counts) = 0.0060 0.0060 < 0.05. Reject Ho. Yes, it can be concluded that voting preference is dependent on location of residence. Question 13 of 20 1.0/ 1.0 Points A high school offers math placement exams for incoming freshmen to place students into the appropriate math class during their freshman year. Three different middle schools were sampled and the following pass/fail results were found. Run a test for independence at the 0.10 level of significance. School A School B School C Pass 40 33 50 Fail 59 45 67 Hypotheses: H0: Pass/fail rates are _____ school. H1: Pass/fail rates are _____school. Which of the following best fits the blank spaces above?  A. independent of; independent on  B. independent of, dependent on  C. dependent of, dependent on  D. dependent of, independent on Answer Key:B Question 14 of 20 0.0/ 1.0 Points A high school offers math placement exams for incoming freshmen to place students into the appropriate math class during their freshman year. Three different middle schools were sampled and the following pass/fail results were found. Run a test for independence at the 0.10 level of significance. School A School B School C Pass 40 33 50 Fail 59 45 67 After running an independence test, can it be concluded that pass/fail rates are dependent on school?  A. No, it cannot be concluded that pass/fail rates are dependent on school because the pvalue = 0.9373.  B. Yes, it can be concluded that pass/fail rates are dependent on school because the p-value = 0.0627.  C. Yes, it can be concluded that pass/fail rates are dependent on school because the p-value = 0.9373.  D. No, it cannot be concluded that pass/fail rates are dependent on school because the pvalue = 0.0627. Answer Key:A Feedback: We are running a Chi-Square Test for Independence. Copy and Paste the table into Excel. You are given the Observed Counts in the table. We need to calculate the Expected Counts. Then sum up the rows and column. You need to find the probability of the row and then multiple it by the column total. School A School B School C Sum Pass Fail Sum School A School B School C Pass =99*(123/294)=78*(123/294) =117*(123/294) Fail =99*(171/294)=78*(171/294) =117*(171/294) Now that we calculated the Expected Count we can use Excel to find the p-value. Use =CHISQ.TEST(highlight actual counts, highlight expected counts) = 0.9373 0.9373 > .10, Do Not Reject Ho. No, it cannot be concluded that pass/fail rates are dependent on school. Part 3 of 4 - F-Distribution and Three or More Means 0.0/ 3.0 Points Question 15 of 20 0.0/ 1.0 Points You are conducting a study of three types of feed supplements for cattle to test their effectiveness in producing weight gain among calves whose feed includes one of the supplements. You have four groups of 30 calves (one is a control group receiving the usual feed, but no supplement). You will conduct a one-way ANOVA after one year to see if there are difference in the mean weight for the four groups. What is k for this experiment?  A. 120  B. 30  C. 3  D. 4 Answer Key:D Feedback:k is the number of groups Question 16 of 20 0.0/ 1.0 Points You’re running a χ 2 Independence Test to see if there is an association between age (Under 50/50+) and type of car owned (Sedan/SUV/Truck/Other). You find a χ 2 test statistic of 5.491. What is the p-value and conclusion?  A. 0., Reject Ho  B. 0., Reject Ho.  C. 0., Do Not Reject Ho  D. 0., Do Not Reject Ho. Answer Key:C Feedback: row = 2, for Under 50 and 50+ column = 4, for 4 types of trucks, Sedan/SUV/Truck/Other. df = (r - 1) (c- 1) df = (2-1)(4-1) df = 1*3 df = 3 Use Excel to find the p-value =CHISQ.DIST.RT(5.491,3) = 0. 0. > .05, Do Not Reject Ho. This is Not Significant. Question 17 of 20 0.0/ 1.0 Points The null hypothesis for the χ 2 Independence Test always states that:  A. The expected values and observed values are the same.  B. One variable is dependent of another variable.  C. One variable is independent of another variable.  D. The two values are equal. Answer Key:C Part 4 of 4 - Chi-Square Distribution for Independence 1.0/ 3.0 Points Question 18 of 20 0.0/ 1.0 Points The number of degrees of freedom for a test of independence is equal to:  A. The sample size minus one  B. The sample size plus one  C. (number of columns - 1)(number of rows - 1) Answer Key:C Question 19 of 20 0.0/ 1.0 Points The data presented in the table below resulted from an experiment in which seeds of 5 different types were planted and the number of seeds that germinated within 5 weeks after planting was recorded for each seed type. At the .01 level of significance, is the proportion of seeds that germinate dependent on the seed type? Seed Type Observed Frequencies Germinated Failed to Germinate 1 31 7 2 57 33 3 87 60 4 52 44 5 10 19  A. No, the proportion of seeds that germinate are not dependent on the seed type because pvalue = 0.00205.  B. No, the proportion of seeds that germinate are not dependent on the seed type because the test value 17.99 is greater than the critical value of 13.28.  C. Yes, the proportion of seeds that germinate are dependent on the seed type because the test value 17.99 is greater than the critical value of 13.28.  D. Yes, the proportion of seeds that germinate are dependent on the seed type because pvalue = 0.00205. Answer Key:D Feedback: We are running a Chi-Square Test for Independence. Copy and Paste the table into Excel. You are given the Observed Counts in the table. We need to calculate the Expected Counts . First sum up the rows and column. Then you need to find the probability of the row and then multiple it by the column total. Germinated Failed to Germinate Sum 1 31 7 38 2 57 33 90 4 52 44 96 5 10 19 29 Sum Germinated Failed to Germinate 1 =237*(38/400) =163*(38/400) 2 =237*(90/400) =163*(90/400) 3 =237*(147/400)=163*(147/400) 4 =237*(96/400) =163*(96/400) 5 =237*(29/400) =163*(29/400) Now that we calculated the Expected Count we can use Excel to find the p-value. Use =CHISQ.TEST(highlight actual counts, highlight expected counts) = 0.00205 p-value = 0.00205 < .01, Reject Ho. Yes, the proportion of seeds that germinate dependent on the seed type. Question 20 of 20 1.0/ 1.0 Points When is μ located on a chi-square curve in the middle?  A. when the number of degrees of freedom is less than 90  B. when the number of degrees of freedom is less than 80  C. when the number of degrees of freedom is greater than 90  D. when the number of degrees of freedom is greater than 80 Answer Key:C Feedback: When df > 90, the chi-square curve approximates the normal. Question 1 of 20 1.0/ 1.0 Points An urban economist is curious if the distribution in where Oregon residents live is different today than it was in 1990. She observes that today there are approximately 3,109 thousand residents in NW Oregon, 902 thousand residents in SW Oregon, 244 thousand in Central Oregon, and 102 thousand in Eastern Oregon. She knows that in 1990 the breakdown was as follows: 72.7% NW Oregon, 20.7% SW Oregon, 4.8% Central Oregon, and 2.8% Eastern Oregon. Can she conclude that the distribution in residence is different today at a 0.05 level of significance?  A. yes because the p-value = .0009  B. no, because the p-value = .0009  C. yes because the p-value = .0172  D. no because the p-value = .0172 Answer Key:C Feedback: NW Oregon SW Oregon Central Oregon Eastern Oregon Observed Counts Expected Counts 4357*.727 = 3167.539 4357*.207= 901.899 4357*.048= 209.136 4357*.028= 121.996 Use Excel to find the p-value =CHISQ.TEST(Highlight Observed, Highlight Expected) p-value is < .05, Reject Ho. Yes, this is significant. Question 2 of 20 1.0/ 1.0 Points A Driver’s Ed program is curious if the time of year has an impact on number of car accidents in the U.S. They assume that weather may have a significant impact on the ability of drivers to control their vehicles. They take a random sample of 150 car accidents and record the season each occurred in. They found that 27 occurred in the Spring, 39 in the Summer, 31 in the Fall, and 53 in the Winter. Can it be concluded at the 0.05 level of significance that car accidents are not equally distributed throughout the year?  A. yes because the p-value = 0.0145  B. no, because the p-value = 0.0145  C. yes, because the p-value = 0.0291  D. no, because the p-value = 0.0291 Answer Key:A Feedback: Spring Summer Fall Winter Observed Counts Expected Counts 150*.25 = 37.5 150*.25 = 37.5 150*.25 = 37.5 150*.25= 37.5 You can use Excel to find the p-value =CHISQ.TEST(Highlight Observed, Highlight Expected) p-value = 0.0145 < .05, Reject Ho, Yes, this is significant Question 3 of 20 1.0/ 1.0 Points A company that develops over-the-counter medicines is working on a new product that is meant to shorten the length of sore throats. To test their product for effectiveness, they take a random sample of 110 people and record how long it took for their symptoms to completely disappear. The results are in the table below. The company knows that on average (without medication) it takes a sore throat 6 days or less to heal 42% of the time, 7-9 days 31% of the time, 10-12 days 16% of the time, and 13 days or more 11% of the time. Can it be concluded at the 0.01 level of significance that the patients who took the medicine healed at a different rate than these percentages? After running a Goodness of Fit test, can it be concluded that there is a statistically significant difference in duration of a sore throat for those that took the medicine and what is the p-value? 6 days or less7-9 days10-12 days13 or more days Duration of Sore Throat 49 40 12 9 Expected Counts 46.2 34.1 17.6 12.1  A. Yes, the p-value = 0.  B. No, the p-value = 0.  C. Yes, the p-value = 0.  D. No, the p-value = 0. Answer Key:D Feedback: Use Excel to find the p-value you have the Observed and Expected Counts you can use =CHISQ.TEST( Highlight Observed Counts, Highlight Expected Counts) = 0. 0. > .01, Do Not Reject Ho. No, this is not significant. Question 4 of 20 0.0/ 1.0 Points Click to see additional instructions An urban economist is curious if the distribution in where Oregon residents live is different today than it was in 1990. She observes that today there are approximately 3,109 thousand residents in NW Oregon, 902 thousand residents in SW Oregon, 244 thousand in Central Oregon, and 102 thousand in Eastern Oregon. She knows that in 1990 the breakdown was as follows: 72.7% NW Oregon, 20.7% SW Oregon, 4.8% Central Oregon, and 2.8% Eastern Oregon. Can she conclude that the distribution in residence is different today at a 0.05 level of significance? Enter the test statistic - round to 4 decimal places. Test statistic= 4.5262 Answer Key:10.1714 Feedback: NW Oregon SW Oregon Central Oregon Eastern Oregon Observed Counts Expected Counts 4357*.727 = 3167.539 4357*.207= 901.899 4357*.048= 209.136 4357*.028= 121.996 Test Stat = Question 5 of 20 1.0/ 1.0 Points Click to see additional instructions A company manager believes that a person’s ability to be a leader is directly correlated to their zodiac sign. He never selects someone to chair a committee without first evaluating their zodiac sign. An irate employee sets out to prove her manager wrong. She claims that if zodiac sign truly makes a difference in leadership, then a random sample of 210 CEO’s in our country would reveal a difference in zodiac sign distribution. She finds the following zodiac signs for her random sample of 210 CEO’s: Births Signs 25 Aries 13 Taurus 17 Gemini 21 Cancer 16 Leo 18 Virgo 15 Libra 16 Scorpio 20 Sagittarius 11 Capricorn 23 Aquarius 15 Pisces Can she conclude that zodiac sign makes a difference in whether or not a person makes a good leader? Enter the p-value - round to 4 decimal places. Make sure you put the 0 in front of the decimal. p-value = 0.4798 Answer Key:0.4798 Feedback: The Expected Count is all the same value. 210*(1/12) = 17.5 Births Signs Expected Count 25 Aries 17.5 13 Taurus 17.5 17 Gemini 17.5 21 Cancer 17.5 16 Leo 17.5 18 Virgo 17.5 15 Libra 17.5 16 Scorpio 17.5 20 Sagittarius 17.5 11 Capricorn 17.5 23 Aquarius 17.5 15 Pisces 17.5 Use Excel to find the p-value =CHISQ.TEST(Highlight Observed, Highlight Expected) Question 6 of 20 1.0/ 1.0 Points A college prep school advertises that their students are more prepared to succeed in college than other schools. To verify this, they categorize GPA’s into 4 groups and look up the proportion of students at a state college in each category. They find that 7% have a 0-0.99, 21% have a 1-1.99, 37% have a 2-2.99, and 35% have a 3-4.00 in GPA. They then take a random sample of 200 of their graduates at the state college and find that 19 has a 0-0.99, 28 have a 1-1.99, 82 have a 2-2.99, and 71 have a 3-4.00. Can they conclude that the grades of their graduates are distributed differently than the general population at the school? Test at the 0.05 level of significance.  A. yes because the p-value is .0620  B. no because the p-value is .0620  C. yes because the p-value is .9379  D. no because the p-value is .9379 Answer Key:B Feedback: 0-0.99 1-1.99 2-2.99 3-4.00 Observed Counts Expected Counts =200*0.07 =14 =200*.21= 42 =200*.37 = 74 =200*.35 = 70 You can use Excel to find the p-value =CHISQ.TEST(Highlight Observed, Highlight Expected) p-value = .0620 > .05, Do Not Reject Ho. No, this is not significant. Question 7 of 20 1.0/ 1.0 Points A company that develops over-the-counter medicines is working on a new product that is meant to shorten the length of sore throats. To test their product for effectiveness, they take a random sample of 110 people and record how long it took for their symptoms to completely disappear. The results are in the table below. The company knows that on average (without medication) it takes a sore throat 6 days or less to heal 42% of the time, 7-9 days 31% of the time, 10-12 days 16% of the time, and 13 days or more 11% of the time. Can it be concluded at the 0.01 level of significance that the patients who took the medicine healed at a different rate than these percentages? Hypotheses: H0: There is __________ in duration of a sore throat for those that took the medicine. H1: There is __________in duration of a sore throat for those that took the medicine. Select the best fit choices that fit in the two blank spaces above.  A. no difference, a difference  B. a difference, no difference  C. no difference, no difference  D. a difference, a difference Answer Key:A Question 8 of 20 1.0/ 1.0 Points Students at a high school are asked to evaluate their experience in the class at the end of each school year. The courses are evaluated on a 1-4 scale – with 4 being the best experience possible. In the History Department, the courses typically are evaluated at 10% 1’s, 15% 2’s, 34% 3’s, and 41% 4’s. Mr. Goodman sets a goal to outscore these numbers. At the end of the year he takes a random sample of his evaluations and finds 10 1’s, 13 2’s, 48 3’s, and 52 4’s. At the 0.05 level of significance, can Mr. Goodman claim that his evaluations are significantly different than the History Department’s?  A. yes, the p-value = 0.5893  B. no, the p-value = 0.5893  C. no, the p-value = 0.3913  D. yes, the p-value = 0.3913 Answer Key:C Feedback: 1's 2's 3's 4's Observed Counts Expected Counts 123 *.10 = 12.3 123*.15 = 18.45 123*.34 = 41.82 123*.41 = 50.43 Use Excel to find the p-value =CHISQ.TEST(Highlight Observed, Highlight Expected) p-value > .05, Do Not Reject Ho. No, this is not significant. Part 2 of 4 - Chi Square Test for Independence 4.75/ 6.0 Points Question 9 of 20 1.0/ 1.0 Points Click to see additional instructions The medal count for the 2018 winter Olympics is recorded below. Run an independence test to find out if the medal won is dependent on country. Use α=0.10. Gold Silver Bronze Norway 17 14 11 Germany 16 9 6 Canada 13 9 11 United States 12 10 7 Enter the P-Value - round to 4 decimal places. Make sure you put a 0 in front of the decimal. 0.8861 Answer Key:0.8861 Feedback: We are running a Chi-Square Test for Independence. Copy and Paste the table into Excel. You are given the Observed Counts in the table. Next you need to sum the rows and columns. Once you have those you need to calculate the Expected Counts. You need to find the probability of the row and then multiple it by the column total. Gold Silver Bronze Sum Norway Germany 16 9 6 31 Canada 13 9 11 33 United States 12 10 7 29 Sum Gold Silver Bronze Norway =58*(42/135) =42*(42/135) =35*(42/135) Germany =58*(31/135) =42*(31/135) =35*(31/135) Canada =58*(33/135) =42*(33/135) =35*(33/135) United States =58*(33/135) =42*(29/135) =35*(29/135) Now that we calculated the Expected Count we can use Excel to find the p-value. Use =CHISQ.TEST(highlight actual counts, highlight expected counts) = 0.8861 Question 10 of 20 1.0/ 1.0 Points Click to see additional instructions The following sample was collected during registration at a large middle school. At the 0.05 level of significance, can it be concluded that level of math is dependent on grade level? Honors Math Regular Math General Math 6 th Grade 35 47 14 7 th Grade 37 49 12 8 th Grade 33 48 19 Enter the missing values in the expected matrix - round to 4 decimal places. Honors Math Regular Math General Math 6 th Grade 34.2857 47.0204 14.6939 7 th Grade 35 48 15 8 th Grade 35.7143 48.9796 15.3061 Answer Key:34.2857, 47.0204, 14.6939, 35.7143, 48.9796, 15.3061 Feedback: We are running a Chi-Square Test for Independence. Copy and Paste the table into Excel. You are given the Observed Counts in the table. We need to calculate the Expected Counts. Then sum up the rows and column. You need to find the probability of the row and then multiple it by the column total. Honors Math Regular Math General Math Sum 6th Grade 7th Grade 8th Grade Sum Honors Math Regular Math General Math 6th Grade =105*(96/294) =144*(96/294) =45*(96/294) 7th Grade =105*(98/294) =144*(98/294) =45*(98/294) 8th Grade =105*(100/294)=144*(100/294)=45*(100/294) Question 11 of 20 0.75/ 1.0 Points Click to see additional instructions A university changed to a new learning management system during the past school year. The school wants to find out how it’s working for the different departments – the results in preference found from a survey are below. Run a test for independence at α=0.05. Prefers Old LMS Prefers New LMS No Preference School of Business 18 29 8 School of Science 41 11 4 School of Liberal Arts 25 20 7 Enter the 4 missing values in the expected matrix. Round to four decimal places. Prefers Old LMS Prefers New LMS No Preference School of Business 28.3436 20.2454 6.4110 School of Science 28.8589 20.6135 6.5276 School of Liberal Arts 26.7976 19.1411 6.0613 Answer Key:28.3436, 6.4110, 20.6135, 26.7975 Feedback: We are running a Chi-Square Test for Independence. Copy and Paste the table into Excel. You are given the Observed Counts in the table. We need to calculate the Expected Counts. Then sum up the rows and column. You need to find the probability of the row and then multiple it by the column total. Prefers Old LMS Prefers New LMS No Preference Sum School of Business 18 29 8 55 School of Science 41 11 4 56 School of Liberal Arts 25 20 7 52 Sum Prefers Old LMS Prefers New LMS No Preference School of Business =84*(55/163) =60*(55/163) =19*(55/163) School of Science =84*(56/163) =60*(56/163) =19*(56/163) School of Liberal Arts =84*(52/163) =60*(52/163) =19*(52/163) Question 12 of 20 1.0/ 1.0 Points A public opinion poll surveyed a simple random sample of 550 voters in Oregon. The respondents were asked which political party they identified with most and were categorized by residence. Results are shown below. Decide if voting preference is independent from location of residence. Let α=0.05. Republican Democrat Independent NW Oregon 85 103 22 SW Oregon 45 66 10 Central Oregon 46 53 9 Eastern Oregon 67 33 11 Can it be concluded that voting preference is dependent on location of residence?  A. Yes, it can be concluded that voting preference is dependent on location of residence because the p-value = 0.0030  B. Yes, it can be concluded that voting preference is dependent on location of residence because the p-value = 0.0060  C. No, it cannot be concluded that voting preference is dependent on location of residence because the p-value = 0.0060  D. No, it cannot be concluded that voting preference is dependent on location of residence because the p-value = 0.0030 Answer Key:B Feedback: We are running a Chi-Square Test for Independence. Copy and Paste the table into Excel. You are given the Observed Counts in the table. Next you need to sum the rows and columns. Once you have those you need to calculate the Expected Counts. You need to find the probability of the row and then multiple it by the column total. Republican Democrat Independent Sum NW Oregon SW Oregon Central Oregon Eastern Oregon Sum Republican Democrat Independent NW Oregon =243*(210/550) =255*(210/550) =52*(210/550) SW Oregon =243*(121/550) =255*(121/550) =52*(121/550) Central Oregon =243*(108/550) =255*(108/550) =52*(108/550) Eastern Oregon =243*(111/550) =255*(111/550) =52*(111/550) Now that we calculated the Expected Count we can use Excel to find the p-value. Use =CHISQ.TEST(highlight actual counts, highlight expected counts) = 0.0060 0.0060 < 0.05. Reject Ho. Yes, it can be concluded that voting preference is dependent on location of residence. Question 13 of 20 0.0/ 1.0 Points Click to see additional instructions A high school runs a survey asking students if they participate in sports. The results are found below. Run an independence test for the data at α=0.01. Freshmen Sophomores Juniors Seniors Yes No Enter the test statistic - round to 4 decimal places. 18.2578 Answer Key:16.2406 Feedback: We are running a Chi-Square Test for Independence. Copy and Paste the table into Excel. You are given the Observed Counts in the table. Next you need to sum the rows and columns. Once you have those you need to calculate the Expected Counts. You need to find the probability of the row and then multiple it by the column total. Freshmen Sophomores Juniors Seniors Yes No Sum Freshmen Sophomores Juniors Seniors Yes =105*(260/396) =116*(260/396) =93*(260/396) =82*(260/396) No =105*(136/396) =116*(136/396) =93*(136/396) =82*(136/396) Now that we calculated the Expected Counts we need to find the Test Statistic. Test Stat = You will need to all 8 Count values but I am only showing you 3 because there isn't room to write out the entire equation. Question 14 of 20 1.0/ 1.0 Points Click to see additional instructions An electronics store has 4 branches in a large city. They are curious if sales in any particular department are different depending on location. They take a random sample of 4 purchases throughout the 4 branches – the results are recorded below. Run an independence test for the data below at the 0.05 level of significance. Appliances TV Computers Cell Phones Branch Branch Branch Branch Enter the test statistic - round to 4 decimal places. 7.2612 Answer Key:7.2612 Feedback: We are running a Chi-Square Test for Independence. Copy and Paste the table into Excel. You are given the Observed Counts in the table. Next you need to sum the rows and columns. Once you have those you need to calculate the Expected Counts. You need to find the probability of the row and then multiple it by the column total. Appliances TV Computers Cell Phones Branch Branch Branch Branch Sum Appliances TV Computers Cell Phones Branch 1 =204*(171/648) =98*(171/648) =233*(171/648) =113*(171/648) Branch 2 =204*(148/648) =98*(148/648) =233*(148/648) =113*(148/648) Branch 3 =204*(152/648) =98*(152/648) =233*(152/648) =113*(152/648) Branch 4 =204*(177/648) =98*(177/648) =233*(177/648) =113*(177/648) Now that we calculated the Expected Counts we need to find the Test Statistic. Test Stat = You will need to use all 16 Count values but I am only showing you the first and last value because there isn't room to write out the entire equation. Part 3 of 4 - F-Distribution and Three or More Means 2.0/ 3.0 Points Question 15 of 20 0.0/ 1.0 Points You are conducting a study of three types of feed supplements for cattle to test their effectiveness in producing weight gain among calves whose feed includes one of the supplements. You have four groups of 30 calves (one is a control group receiving the usual feed, but no supplement). You will conduct a one-way ANOVA after one year to see if there are difference in the mean weight for the four groups. What is k for this experiment?  A. 120  B. 30  C. 3  D. 4 Answer Key:D Feedback:k is the number of groups Question 16 of 20 1.0/ 1.0 Points The F Statistic from an experiment with k = 4 and n = 100 is 4.72. At α = 0.01, will you reject the null hypothesis?  A. No  B. Yes Answer Key:B Feedback: Use Excel to find the p-value df1 = k - 1 = 4-1 = 3 df2 = n - k = 100 - 4 = 96 =F.DIST.RT(4.72,3,96) = 0. 0. < .01, Reject Ho, Yes, this is significant. Question 17 of 20 1.0/ 1.0 Points The χ 2 Independence Test is always:  A. Right-Tailed  B. Left-Tailed  C. A Confidence Interval  D. Two-Tailed Answer Key:A Part 4 of 4 - Chi-Square Distribution for Independence 1.0/ 3.0 Points Question 18 of 20 0.0/ 1.0 Points Click to see additional instructions