MATH 130 WEEK 8 PRACTICE 1| 2022 latest update | 100% correct

MATH 130 WEEK 8 PRACTICE 1 Question 1 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= Answer Key:10.1714 Feedback: NW Oregon SW Oregon Central Oregon Eastern Oregon Observed 3109 902 244 102 Counts Expected 4357*.727 4357*.207= 4357*.048=4357*.028= Counts Test Stat = = 3167.539 901.899 209.136 121.996 Question 2 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 3 of 20 1.0/ 1.0 Points Click to see additional instructions 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? Enter the test statistic - round to 2 decimal places. 2.48 Answer Key:2.48 Feedback: The Expected Counts are all the same. 300*(1/12) = 25. Test Stat = You need to use this equation to find the Test Statistic and use all the values from January to December. The "..." doesn't show all 12 value but you will need to include all of them. Question 4 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: Observed Counts Spring Summer Fall Winter 27 39 31 53 Expected 150*.25 = 150*.25 = 150*.25 = 150*.25= Counts 37.5 37.5 37.5 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 5 of 20 1.0/ 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 188 153 83 54 61 11 Counts 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 6 of 20 1.0/ 1.0 Points Click to see additional instructions 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? Enter the test statistic - round to 4 decimal places. Test statistic = 12.2012 Answer Key:12.2012 Feedback: Clothing Home App. Housewares Cosmetics Jewelry Other Observed 188 153 83 54 61 11 Counts Expected Counts 550*.30 = 165 550*.25 = 137.5 550*.18 = 99 550*.13 = 71.5 550*.12 = 66 550*.02= 11 Test Stat = 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: Observed Counts 0-0.99 1-1.99 2-2.99 3-4.00 19 28 82 71 Expected =200*0.07 =200*.21==200*.37 =200*.35 Counts =14 42 = 74 = 70 Test Stat = Test Stat = 7. Question 8 of 20 0.0/ 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 p- value? Test with a Goodness of Fit test at α=0.05. Alon Roommat Parents/Guardi e es Observ ans ed Counts Expecte d Counts • 398 488 714 432 512 656 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. Part 2 of 4 - Chi Square Test for Independence 4.0/ 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 75 88 55 42 No 30 28 38 40 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 = 0.0020. • D. Yes, it can be concluded that participation in sports is dependent on grade level because the p-value = 0.0010. 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. 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 75 88 55 42 No 30 28 38 40 Sum 105 116 93 82 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 1.0/ 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 1 56 28 63 24 Branch 2 44 22 55 27 Branch 3 53 17 49 33 Branch 4 51 31 66 29 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 = 0.3901 • B. Yes, it can 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 = 0.6099 • D. No, it cannot be concluded that sales in the various departments are dependent on branch because the p-value = 0.6099 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. 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. Branch 1 Appliances 56 TV 28 Computers 63 Cell Phones 24 Branch 2 44 22 55 27 Branch 3 53 17 49 33 Branch 4 51 31 66 29 Sum 204 98 233 113 Branch 1 Appliances =204*(171/648) TV =98*(171/648) Computers =233*(171/648) Cell Phones =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 1.0/ 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 12 17 25 13 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. 0.0144 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 Aerobics 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 12 of 20 1.0/ 1.0 Points A manufacturing company knows that their machines produce parts that are defective on occasion. They have 4 machines producing parts, and want to test if defective parts are dependent on the machine that produced it. They take a random sample of 321 parts and find the following results. Test at the 0.05 level of significance. Machine 1 Machine 2 Machine 3 Machine 4 Defective 10 15 16 9 Non-Defective 72 75 66 58 Can it be concluded that number of defective parts is dependent on machine? • A. Yes, it can be concluded concluded that number of defective parts is dependent on machine because the p-value = 0.4264. • B. No, it cannot be concluded concluded that number of defective parts is dependent on machine because the p-value = 0.5736. • C. Yes, it can be concluded concluded that number of defective parts is dependent on machine because the p-value = 0.5736. • D. No, it cannot be concluded concluded that number of defective parts is dependent on machine because the p-value = 0.4264. 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. Machine 1 Machine 2 Machine 3 Machine 4 Sum Defective 10 15 16 9 50 Non-Defective 72 75 66 58 271 Sum 82 90 82 67 321 Machine 1 Machine 2 Machine 3 Machine 4 Defective =82*(50/321) =90*(50/321) =82*(50/321) =67*(50/321) Non-Defective =82*(271/321) =90*(271/321) =82*(271/321) =67*(271/321) 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) = 0.5736 0.5736 > 0.05, Do Not Reject Ho. No, it cannot be concluded concluded that number of defective parts is dependent on machine. Question 13 of 20 1.0/ 1.0 Points Click to see additional instructions A manufacturing company knows that their machines produce parts that are defective on occasion. They have 4 machines producing parts, and want to test if defective parts are dependent on the machine that produced it. They take a random sample of 321 parts and find the following results. Test at the 0.05 level of significance. Machine 1 Machine 2 Machine 3 Machine 4 Defective 10 15 16 9 Non-Defective 72 75 66 58 Enter the test statistic - round to 4 decimal places. 1.9943 Answer Key:1.9943 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. Machine 1 Machine 2 Machine 3 Machine 4 Sum Defective 10 15 16 9 50 Non-Defective 72 75 66 58 271 Sum 82 90 82 67 321 Machine 1 Machine 2 Machine 3 Machine 4 Defective =82*(50/321) =90*(50/321) =82*(50/321) =67*(50/321) Non-Defective =82*(271/321) =90*(271/321) =82*(271/321) =67*(271/321) 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 0.0/ 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 6th Grade 35 47 14 7th Grade 37 49 12 8th Grade 33 48 19 Enter the missing values in the expected matrix - round to 4 decimal places. Honors Math Regular Math General Math 6th Grade 7th Grade 35 48 15 8th Grade 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 35 47 14 96 7th Grade 37 49 12 98 8th Grade 33 48 19 100 Sum 105 144 45 294 6th Grade 7th Grade 8th Grade Honors Math Regular Math General Math =105*(96/294) =144*(96/294) =45*(96/294) =105*(98/294) =144*(98/294) =45*(98/294) =105*(100/294)=144*(100/294)=45*(100/294) Part 3 of 4 - F-Distribution and Three or More Means 1.67/ 3.0 Points Question 15 of 20 1.0/ 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 χ2 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 χ2 test statistic of 5.491. What is the p-value and conclusion? • A. 0., Reject Ho. • B. 0., Do Not Reject Ho. • C. 0., Reject Ho • D. 0., Do Not Reject Ho Answer Key:D 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 0.67/ 1.0 Points Which of the following numbers are possible F Statistics? Select all that apply. A. 2 B. -7.28 C. 0.0834 D. 3.61 Answer Key:A, C, D Feedback:The F-stat must be positive. Part 4 of 4 - Chi-Square Distribution for Independence 1.33/ 3.0 Points Question 18 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, Medium- High, 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 4 15 14 3 36 laptops med-low 6 18 18 23 65 med-high 13 17 10 17 57 high 7 15 15 10 47 30 65 57 53 205 What is the test value for this hypothesis test? Answer: 15.31 Round your answer to two decimal places. What is the critical value for this hypothesis test? Answer: 16.92 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: 1 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. 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 19 of 20 1.0/ 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= 7.0711 Answer Key:7.0711 Feedback: SQRT(2*25) Question 20 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 p- value = 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 p- value = 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 Sum 38 90 147 96 29 163 400 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.