GPA Distribution Analysis for College Preparedness Assessment

Attempt Score 19 / 20 - 95 % Overall Grade (Highest Attempt) 19 / 20 - 95 % stion 1 1 / 1 p 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. Hypotheses: H 0 : There is ______ between the general population and the college prep students in GPA. H 1 : There is ______ between the general population and the college prep students in GPA. 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 2 1 / 1 point 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 p -value - round to 4 decimal places. Make sure you put a 0 in front of the decimal. P-value =___ Answer: 0.9960 Hide question 2 feedback Expected counts will be all the same. 300*(1/12) = 25. The 12 months are either equally distributed or they are not. 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) = 0.9960 n 3 1 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 3 feedback The observed counts are given to you, 398, 488, and 714 Calculate expected counts 1600*.27, 1600*.32 and 1600*.41 n 4 1 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 5 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

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

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 5 feedback Expected Counts = 110*.42 110*.31 110*.16 110*.11 n 6 0 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 6 feedback 1's 2's 3's 4's 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) n 7 1 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 8 1 / 1 point 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. After running a Goodness of Fit test, does the professor have evidence to conclude that location in the classroom has an impact on final grade and what is the p-value? Rows 1-2 Rows 3-4 Rows 5-6 Rows 7-8 Rows 9-10 # in Top 25% 14 8 13 10 15 Expected Counts 12 12 12 12 12 no, the p-value = 0.413907 yes, the p-value = 0.586093 no, the p-value = 0.58609 yes, the p-value = 0.413907 Hide question 8 feedback Use Excel to find the p-value you have the Observed and Expected Counts you can use =CHISQ.TEST( Highlight Observed Counts, Highlight

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Expected Counts) = 0.586093 0.586093 > .05, Do Not Reject Ho. No, this is not significant. n 9 1 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 Yes, you can reject the claim that the number of breakdowns is independent of the shift because the p-value = 0.0423 Yes, you can reject the claim that the number of breakdowns is independent of the shift because the p-value = 0.9577 No, you cannot reject the claim that the number of breakdowns is independent of the shift because the p-value = 0.0423 No, you cannot reject the claim that the number of breakdowns is independent of the shift because the p-value = 0.9577 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. Machine Shift A B C Sum 1 46 11 13 70 2 37 10 11 58 3 20 15 16 51

Sum 103 36 40 179 Shift A B C 1 =103*(70/179 ) =36*(70/179) =40*(70/179) 2 =103*(58/179 ) =36*(58/179) =40*(58/179) 3 =103*(51/179 ) =36*(51/179) =40*(51/179) 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) = 0.0423 0.0423 < .05, Reject Ho. You can reject the claim that the number of breakdowns is independent of the shift. n 10 1 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 No, the proportion of seeds that germinate are not dependent on the seed type because the p-value = 0.0132. No, the proportion of seeds that germinate are not dependent on the seed type because the p-value = 0.0265.

Yes, the proportion of seeds that germinate dependent on the seed type because the p-value = 0.0265. Yes, the proportion of seeds that germinate dependent on the seed type because the p-value = 0.0132. Hide question 10 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) 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) = 0.0265 0.0265 < .05, Reject Ho. Yes, the proportion of seeds that germinate dependent on the seed type. n 11 1 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

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Shift A B C 1 46 11 13 2 37 10 11 3 20 15 16 What is the Expected Count for Shift = 3 and Machine = C? 15.642 11.397 14.078 10.257 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. Machine Shift A B C Sum 1 46 11 13 70 2 37 10 11 58 3 20 15 16 51 Sum 103 36 40 179 Shift A B C 1 =103*(70/179 ) =36*(70/179) =40*(70/179) 2 =103*(58/179 ) =36*(58/179) =40*(58/179) 3 =103*(51/179 ) =36*(51/179) =40*(51/179) =40*(51/179) = 11.397 n 12 1

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 = 0.0017 No, it cannot 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 = 0.00085 Yes, it can be concluded that preference in learning management system is dependent on department because the p-value = 0.0017 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. 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 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. n 13 1 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? 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 = 0.24587 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 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. 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 =105*(96/294) =144*(96/294) =45*(96/294)

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Grade 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. n 14 1 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 Hypotheses: H 0 : Pass/fail rates are _______ school. H 1 : Pass/fail rates are _______ school. Which of the following best fits the blank spaces above? dependent of, independent on dependent of, dependent on independent of; independent on independent of, dependent on Question 15 1 / 1 point 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 Can it be concluded that medal won is dependent on country? No, it cannot be concluded that medal won is dependent on country because p = 0.1139. No, it cannot be concluded that medal won is dependent on country because p = 0.8861. Yes, it can be concluded that medal won is dependent on country because p = 0.1139. Yes, it can be concluded that medal won is dependent on country because p = 0.8861. Hide question 15 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 17 14 11 42 Germany 16 9 6 31 Canada 13 9 11 33 United States 12 10 7 29 Sum 58 42 35 135 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*(29/135) =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 0.8861 > 0.05, Do Not Reject Ho. No, it cannot be concluded that medal won is dependent on country. n 16 1 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 Enter the P -Value - round to 4 decimal places. Make sure you put a 0 in front of the decimal. ___ Answer: 0.6099 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. 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 1 56 28 63 24 Branch 2 44 22 55 27 Branch 3 53 17 49 33 Branch 4 51 31 66 29 Sum 204 98 233 113 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 n 17 1 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

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

X3 = score on exam #3 X4 = score on final exam Your professor wants to know if all tests are created equal. Looking at the Summary output, which two Exams have the biggest difference in average? See Attached Excel for Data. Exam data.xlsx Exam 1 and the Final Exam 2 and the Final Exam 3 and Exam 1 Exam 3 and the Final 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 4 columns including the top row with the Labels. Check the box Labels in the First Row and click OK If done correctly the ANOVA output is SUMMARY Groups Count Sum Average EXAM1 25 1976.00 79.04 EXAM2 25 1987.00 79.48 EXAM3 25 2037.00 81.48 FINAL 25 1755.43 70.22 The Final the lowest average of 70.22% Exam 3 has the highest average of 81.48% n 18 1 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. Looking at the Summary output, which two teams have the biggest difference in weight? See Attached Excel for Data. Reference: The Sports Encyclopedia Pro Football Football Weight data.xlsx Denver Broncos and Miami Dolphins San Francisco 49ers and Miami Dolphins San Francisco 49ers and Green Bay Packers Denver Broncos and San Francisco 49ers Hide question 18 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 you should get, SUMMARY Groups Count Sum Average Variance Dallas Cowboys Wt. 17 4202 247.18 235.28 Green Bay Packers Wt. 17 4267 251.00 291.63 Denver Broncos Wt. 17 4323 254.29 263.60 Miami Dolphins Wt. 17 4234 249.06 281.06

San Fran. 49ers Wt. 17 4094 240.82 288.53 Denver Broncos have the highest weight sum and average. San Francisco 49ers have the lowest weight sum and average. n 19 1 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 19 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. n 20 1 The F Statistic from an experiment with k = 3 and n = 75 is 2.99. At α = 0.05, will you reject the null hypothesis? Yes, because the p-value = 0.0566 Yes, because the p-value = 0283. No, because the p-value = 0.0283 No, because the p-value = 0.0566 Hide question 20 feedback Use Excel to find the p-value df 1 = k - 1 = 3 -1 = 2 df 2 = n - k = 75 - 3 = 72 =F.DIST.RT(2.99, 2, 72) = 0.0566 0.0566 > 0.05, Do Not Reject Ho, No, this is not significant.

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

week 8 Test

Related Documents