W8 Homework graded

Attempt Score 18 / 20 - 90 % Overall Grade (Highest Attempt) 18 / 20 - 90 % 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. Enter the test statistic - round to 4 decimal places. Test statistic:___ ___ Answer: 7.3315 Hide question 1 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   Test Stat = (19−14)214+(28−42)242+(82−74)274+(71−70)270 Test Stat = 7.331532   n 2 1

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: 0.4798 Hide question 2 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

Test Stat = (188−165)2165+(153−137.5)2137.5+(83−99)299+(54−71.5)271.5+(61−66)266+(11−11)211 n 4 1 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 less 7-9 days 10-12 days 13 or more days Duration of Sore Throat 49 40 12 9 Expected Counts 46.2 34.1 17.6 12.1 Yes, the p-value = 0.287801 No, the p-value = 0.712199 Yes, the p-value = 0.712199 No, the p-value = 0.287801 Hide question 4 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) = 0.287801 0.287801 > .01, Do Not Reject Ho. No, this is not significant. n 5 1 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? Enter the test statistic - round to 2 decimal places. Test statistic=___ ___ Answer: 10.53 Hide question 5 feedback   Spring Summer Fall Winter Observed Counts 27 39 31 53 Expected Counts 150*.25 = 37.5 150*.25 = 37.5 150*.25 = 37.5 150*.25= 37.5   Test Stat = (27−37.5)237.5+(39−37.5)237.5+(31−37.5)237.5+(53−37.5)237.5 n 6 1 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

significance, can Mr. Goodman claim that his evaluations are significantly different than the History Department's? yes, the p-value = 0.5893 no, the p-value = 0.5893 no, the p-value = 0.3913 yes, the p-value = 0.3913 Hide question 7 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) p-value > .05, Do Not Reject Ho. No, this is not significant. n 8 1 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:

H 0 : Births in Pamplona ______ equally distributed throughout the year. H 1 : Births in Pamplona ______ 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 9 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 test statistic - round to 4 decimal places. Answer: 0.0060 (18.0789) 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.  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

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. 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. n 11 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

Enter the missing values in the expected matrix - round to 4 decimal places. Honors Math Regular Math General Math 6 th  Grade ___ ___ ___ 7 th  Grade 35 48 15 8 th  Grade ___ ___ ___ ___ Answer for blank # 1: 34.2857 (16.67 %) Answer for blank # 2: 47.0204 (16.67 %) Answer for blank # 3: 14.6939 (16.67 %) Answer for blank # 4: 35.7143 (16.67 %) Answer for blank # 5: 48.9796 (16.67 %) Answer for blank # 6: 15.3061 (16.67 %) 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.   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)  

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? Yes, it can be concluded that sales in the various departments are dependent on branch because the p-value = 0.6099 No, it cannot be concluded that sales in the various departments are dependent on branch because the p-value = 0.6099 No, it cannot be concluded that sales in the various departments are dependent on branch because the p-value = 0.3901 Yes, it can be concluded that sales in the various departments are dependent on branch because the p-value = 0.3901 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.   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 0.6099 > 0.05, Do Not Reject Ho.  No, it cannot be concluded that sales in the various departments are dependent on branch.   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 0.6099 > 0.05, Do Not Reject Ho. No, it cannot be concluded that sales in the various departments are dependent on branch. n 14 1 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? No, it cannot be concluded that participation in sports is dependent on grade level because the p-value = 0.0010. No, it cannot be concluded that participation in sports is dependent on grade level because the p-value = 0.0020. Yes, it can be concluded that participation in sports is dependent on grade level because the p- value = 0.0020. Yes, it can be concluded that participation in sports is dependent on grade level because the p- value = 0.0010. 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.  Next you need to sum the rows and columns.  Once

120 30 3 4 Hide question 15 feedback k is the number of groups n 16 1 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 SS within = 366.5 and SS total = 627.9 for this data. Identify the  SS between 261.4 3 30 994.4 255.4 Hide question 16 feedback SS total  =  SS within  +  SS between 627.9 = 366.5 +  SS between n 17 1 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? No Yes Hide question 17 feedback Use Excel to find the p-value df 1  = k - 1 = 4-1 = 3 df 2  = n - k = 100 - 4 = 96

=F.DIST.RT(4.72,3,96) = 0.004076343 0.004076343 < .01, Reject Ho, Yes, this is significant. n 18 0 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=___ ___ Answer: 7.0710 (7.0711) View question 18 feedback n 19 1 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: ___ 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