Impact of Sore Throat Medicine on Healing Duration: A Study

Question 1 1 / 1 point 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? Enter the expected count for each category in the table below. Round to 1 decimal place. 6 days or less 7-9 days 10-12 days 13 or more days Duration of Sore Throat 49 40 12 9 Expected Counts ___ ___ ___ ___ Answer for blank # 1: 46.2 (25 %) Answer for blank # 2: 34.1 (25 %) Answer for blank # 3: 17.6 (25 %) Answer for blank # 4: 12.1 (25 %) Hide question 1 feedback

Expected Counts = 110*.42 110*.31 110*.16 110*.11 Question 2 1 / 1 point A college professor is curious if the location of 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. Hypotheses: H 0 : Location in the classroom __________ impact final grade. H 1 : Location in the classroom _________ impact final grade. Select the best fit choices that fit in the two blank spaces above. does not, does does, does not does, does does not, does not

Question 3 1 / 1 point The manager of a coffee shop wants to know if his customers' drink preferences have changed in the past year. He knows that last year the preferences followed the following proportions – 34% Americano, 21% Cappuccino, 14% Espresso, 11% Latte, 10% Macchiato, 10% Other. In a random sample of 450 customers, he finds that 115 ordered Americanos, 88 ordered Cappuccinos, 69 ordered Espressos, 59 ordered Lattes, 44 ordered Macchiatos, and the rest ordered something in the Other category. Run a Goodness of Fit test to determine whether or not drink preferences have changed at his coffee shop. Use a 0.05 level of significance. Hypotheses: H 0 : There is _______ in drink preference this year. H 1 : There is _______ in drink preference this year. Select the best fit choices that fit in the two blank spaces above. no difference, a difference a difference, no difference no difference, no difference a difference, a difference Question 4 1 / 1 point 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? Hypotheses: H 0 : Car accidents ________ equally distributed throughout the year.

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H 1 : Car accidents ________ equally distributed throughout the year. Select the best fit choices that fit in the two blank spaces above. are, are not are not, are are, are are not, are not Question 5 1 / 1 point The manager of a coffee shop wants to know if his customers' drink preferences have changed in the past year. He knows that last year the preferences followed the following proportions – 34% Americano, 21% Cappuccino, 14% Espresso, 11% Latte, 10% Macchiato, 10% Other. In a random sample of 450 customers, he finds that 115 ordered Americanos, 88 ordered Cappuccinos, 69 ordered Espressos, 59 ordered Lattes, 44 ordered Macchiatos, and the rest ordered something in the Other category. Run a Goodness of Fit test to determine whether or not drink preferences have changed at his coffee shop. Use a 0.05 level of significance. Americanos Capp. Espresso Lattes Macchiatos Other Observed Counts 115 88 69 59 44 75 Expected Counts 153 94.5 63 49.5 45 45 Enter the p-value - round to 5 decimal places. Make sure you put a 0 in front of the decimal.

P-value =___ Answer: 0.00001 Hide question 5 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) = 5.17696E-06 Or you can use the Test Statistic =CHISE.DIST.RT(32.3019,5) Where DF = row -1 = 6 - 1 = 5 Question 6 1 / 1 point 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 p -value - round to 4 decimal places. Make sure you put a 0 in front of the decimal. p-value=___ Answer: 0.0620 Hide question 6 feedback

0-0.99 1-1.99 2-2.99 3-4.00 Observed Counts 19 28 82 71 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) Question 7 0 / 1 point 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? Enter the p -value - round to 4 decimal places. Make sure you put a 0 in front of the decimal. p-value=___ ___ Answer: 0.2005 ( 0.3913) Hide question 7 feedback 1's 2's 3's 4's

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Observed Counts 10 13 48 52 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) Question 8 1 / 1 point 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? Test with a Goodness of Fit test at α=0.05 . Enter the observed and expected counts for each category in the table below. Round to whole numbers. Alone Roommates Parents/Guardians Observed Counts ___ ___ ___ Expected Counts ___ ___ ___ ___ Answer for blank # 1: 398 (16.67 %) Answer for blank # 2: 488 (16.67 %) Answer for blank # 3: 714 (16.67 %) Answer for blank # 4: 432 (16.67 %)

Answer for blank # 5: 512 (16.67 %) Answer for blank # 6: 656 (16.67 %) Hide question 8 feedback The observed counts are given to you, 398, 488, and 714 Calculate expected counts 1600*.27, 1600*.32 and 1600*.41 Question 9 1 / 1 point The data presented in the table below resulted from an experiment in which seeds of 4 different types were planted and the number of seeds that germinated within 4 weeks after planting was recorded for each seed type. At the .05 level of significance, is the proportion of seeds that germinate dependent on the seed type? Seed Type Observed Frequencies Germinated Failed to Germinate 1 39 9 2 54 34 3 88 63 4 57 42 What is the Expected Count for Seed Type = 3 and a Germinated Seed?

54.259 93.104 61.041 29.596 Hide question 9 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. Observed Frequencies Germinated Failed to Germinate Sum 1 39 9 48 2 54 34 88 3 88 63 151 4 57 42 99 Sum 238 148 386 Germinated Failed to Germinate 1 =238*(48/386) =148*(48/386) 2 =238*(88/386) =148*(88/386) 3 =238*(151/386) =148*(151/386) 4 =238*(99/386) =148*(99/386)

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=238*(151/386) = 93.104 Question 10 1 / 1 point A company operates three machines during three shifts each day. From production records, the data in the table below were collected. At the .05 level of significance test to determine if the number of breakdowns is independent of the shift. Machine Shift A B C 1 46 11 13 2 37 10 11 3 20 15 16 Hypotheses: H 0 : The Machine ________ the Shift. H 1 : The Machine ________ the Shift. Which of the following best fits the blank spaces above? independent of, independent on

dependent of, independent on independent of, dependent on dependent of, dependent on Question 11 1 / 1 point The data presented in the table below resulted from an experiment in which seeds of 4 different types were planted and the number of seeds that germinated within 4 weeks after planting was recorded for each seed type. At the .05 level of significance, is the proportion of seeds that germinate dependent on the seed type? Seed Type Observed Frequencies Germinated Failed to Germinate 1 39 9 2 54 34 3 88 63 4 57 42 What is the Expected Count for Seed Type = 1 and a Failed to Germinated Seed?

18.404 33.741 57.896 37.959 Hide question 11 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. Observed Frequencies Germinated Failed to Germinate Sum 1 39 9 48 2 54 34 88 3 88 63 151 4 57 42 99 Sum 238 148 386 Germinated Failed to Germinate 1 =238*(48/386) =148*(48/386) 2 =238*(88/386) =148*(88/386) 3 =238*(151/386) =148*(151/386) 4 =238*(99/386) =148*(99/386)

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=148*(48/386) = 18.404 Question 12 0 / 1 point 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 P -Value - round to 4 decimal places. Make sure you put a 0 in front of the decimal. ___ Answer: 0.0114 ( 0.0060) Hide question 12 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 85 103 22 210 SW Oregon 45 66 10 121 Central Oregon 46 53 9 108

Eastern Oregon 67 33 11 111 Sum 243 255 52 550 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 Question 13 1 / 1 point 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 P -Value - round to 4 decimal places. Make sure you put a 0 in front of the decimal. ___

Answer: 0.5736 Hide question 13 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 Question 14 1 / 1 point 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 After running an independence test, can it be concluded that level of math is dependent on grade level?

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No, it cannot be concluded that level of math is dependent on grade level, because the p-value = 0.24587 Yes, it can be concluded that level of math is dependent on grade level, because the p-value = 0.75413 No, it cannot be concluded that level of math is dependent on grade level, because the p-value = 0.75413 Yes, it can be concluded that level of math is dependent on grade level, because the p-value = 0.24587 Hide question 14 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 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 ) 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) = 0.75413 0.75413 > .05, Do Not Reject Ho. No, it cannot be concluded that level of math is dependent on grade level. Question 15 1 / 1 point 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 Hypotheses: H 0 : Level of math is ________ grade level. H 1 : Level of math is ________ grade level. Which of the following best fits the blank spaces above? dependent of, dependent on independent of, dependent on dependent of, independent on independent of; independent on Question 16 1 / 1 point

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 After running an independence test, can it be concluded that preference in learning management system is dependent on department? No, it cannot be concluded that preference in learning management system is dependent on department because the p-value = 0.00085 No, it cannot be concluded that preference in learning management system is dependent on department because the p-value = 0.0017 Yes, it can be concluded that preference in learning management system is dependent on department because the p-value = 0.00085 Yes, it can be concluded that preference in learning management system is dependent on department because the p-value = 0.0017 Hide question 16 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 25 20 7 52

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Arts Sum 84 60 19 163 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) 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) = 0.00085 0.00085 < 0.05, Reject Ho. Yes, it can be concluded that preference in learning management system is dependent on department. Question 17 0 / 1 point The following data represent weights (pounds) of a random sample of professional football players on the following teams. X1 = weights of players for the Dallas Cowboys X2 = weights of players for the Green Bay Packers X3 = weights of players for the Denver Broncos X4 = weights of players for the Miami Dolphins X5 = weights of players for the San Francisco Forty Niners You join a Fantasy Football league and you are wondering if weight is a factor in winning Football games. At a 5% level of Significant, determine if the weights of football players on each of these teams is same or different? See Attached Excel for Data. Reference: The Sports Encyclopedia Pro Football

Football Weight data.xlsx Do Not Reject Ho. The football player's weights for each team are the same because the p-value = 0.1890 Do Not Reject Ho. The football player's weights for each team are the same because the p-value = 0.0472 Reject Ho. The football player's weights for each team are differnt because the p-value = 0.1890 Reject Ho. The football player's weights for each team are differnt because the p-value = 0.0472 Hide question 17 feedback Run a F-Distribution ANOVA analysis using Excel. Data -> Data Analysis -> the first option is Anova: Single Factor -> Click OK In the Input Range: Highlight all 5 columns including the top row with the Labels. Check the box Labels in the First Row and click OK If done correctly the p-value from the ANOVA output is P-value 0.1890 .1890 > .05, Do Not Reject Ho. This is not significant. The weight of the football players on each team is the same. Question 18 1 / 1 point The F Statistic from an experiment with k = 7 and n = 42 is 2.55. At α = 0.05, will you reject the null hypothesis? Yes, because the p-value = 0.0373 No, because the p-value = 0.0373 Yes, because the p-value = 0.0187 Yes, because the p-value = 0.0187

Hide question 18 feedback Use Excel to find the p-value df 1 = k - 1 = 7-1 = 6 df 2 = n - k = 42-7 = 35 =F.DIST.RT(2.55,6,35) = 0.0373 0.0373 < .05, Reject Ho, Yes, this is significant. Question 19 1 / 1 point The Test Scores for a Statistics course are given in the Excel below. The data (X1, X2, X3, X4) are for each student. X1 = score on exam #1 X2 = score on exam #2 X3 = score on exam #3 X4 = score on final exam Your professor wants to know if all tests are created equal. What is the F-Stat ? See Attached Excel for Data. Exam data.xlsx 4.521

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5.532 3.331 4.872 Hide question 19 feedback Run a F-Distribution ANOVA analysis using Excel. Data -> Data Analysis -> the first option is Anova: Single Factor -> Click OK In the Input Range: Highlight all 4 columns including the top row with the Labels. Check the box Labels in the First Row and click OK If done correctly the F-Stat from the ANOVA output is ANOVA Source of Variation SS df MS F Between Groups 1878.945633 3 626.3152111 4.521 Within Groups 13298.84916 96 138.5296787 Total 15177.79479 99 Question 20 0 / 1 point The Test Scores for a Statistics course are given in the Excel below. The data (X1, X2, X3, X4) are for each student. X1 = score on exam #1 X2 = score on exam #2 X3 = score on exam #3 X4 = score on final exam

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Your professor wants to know if all tests are created equal. At a 5% level of Significant, run a test to determine if the mean difference between the test scores is the same. See Attached Excel for Data. Exam data.xlsx The mean difference between the test scores is the same because the p-value = 0.9948 The mean difference between the test scores is the not same because the p-value = 0.0052 The mean difference between the test scores is the not same because the p-value = 0.9948 The mean difference between the test scores is the same because the p-value = 0.0052 Hide question 20 feedback Run a F-Distribution ANOVA analysis using Excel. Data -> Data Analysis -> the first option is Anova: Single Factor -> Click OK In the Input Range: Highlight all 4 columns including the top row with the Labels. Check the box Labels in the First Row and click OK If done correctly the p-value from the ANOVA output is P-value 0.0052 0.0052 < .05, Reject Ho. This is significant. The mean difference between the test scores is the not same.

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