Week 1 Discussions

Week 1: Data Organization and Analysis - Discussion

Discussion of Week 1 Homework Problems (Graded)

Rather than doing a case, we are going to do the homework! My students grades are some of the best because of this in my opinion. Some students say, "It's redundant" talking about thehomework more. Trust me, the more you talk about it, the better you will be (I've seen the Final Exam).

Answer Only under the appropriate thread, DO NOT start a new thread. In other words, if you are answering Homework Question 1, Click My Post for Homework Question 1 and "Reply" to it.

There are 12 questions. We will cover all of them. I want to see three good posts with explanations of your answers or two questions answered and another post commenting on a fellow student's post (substantive).Please read about duplicating...

I am referring to the MyStatLab Homework Problems that can be found in the Assignments Link under Week 1 - See Attachment.

You have to cover three different days of the week and your posts must include explanations.

After ALL Exercises have been posted, you can start duplicating (BUT DO NOT POST THE SAME EXACT PROBLEM, YOU WILL HAVE TO PUSH THE "SIMILAR EXERCISE" BUTTON SO YOU WILL GETONE THAT IS A LITTLE DIFFERENT).

I will post a couple of examples by Monday evening so you will know what I am looking for in terms of posts.

FindingTheHomework.png

Responses

Response Author Date/Time

Homework Question 1 Professor Heard 4/30/2015 9:01:55 AM

All answers and reviews of other questions responses concerning this homework problem should take place under this thread. Please see instructions and do not duplicate exact questions; you can hit the push "SimilarProblem" button in the homework to get a problem that is not exactly the same. Also PLEASE make sure all of the homework problems have been attempted before you start duplicating.

RE: Homework Question 1 Ryan Frain 5/4/2015 2:37:30 PM

For the first problem, I completed the chart with the following information.

Grade on Business Statistics Exam Frequency Relative FrequencyA: 90-100 32 0.08B: 80-89 68 0.17C: 65-79 184 0.46D: 50-64 60 0.15F: Below 50 56 0.14Total 400 1

To get the Frequency of the students that go 90-100 I multiplied 400 by 0.08. To get the relative frequency I divided the frequency by the total so for the 80-89 range it would be 68/400.

RE: Homework Question 1 Professor Heard 5/5/2015 8:25:14 PM

Looking good here Ryan... A quality response

RE: Homework Question 1 Mohamed Ketat 5/10/2015 1:01:04 PM

Modified:5/10/2015 3:56 PM

Grade on Business Statistics Exam Frequency Relative FrequencyA: 90-100 24 0.06B: 80-89 88 .22C: 65-79 172 .43D: 50-64 60 .15F: Below 50 56 .14Total 400 1

To figure out the relative frequency you take the frequency divided by 400. To find the frequency you add up all of the known numbers and subtract from the total.

RE: Homework Question 1 Nicholas Payne 5/5/2015 9:03:12 PM

Very good posting. After looking at the work i must agree with the answer that was stated. I must applaud you for making the posting look very professional and taking the time to highlight the answers. Theexplanation was also quite well written. I did several of these exercises and seem to have come to the same conclusions that you have. It is interesting to look at this and see just how simple it really is.

RE: Homework Question 1 Kevin Gorman 5/6/2015 5:54:22 PM

Modified:5/6/2015 5:52 PM

Grade on Business Statistics Exam

Grade Freq. Relative Freq.

90-100 (A) 21 .07 80-87 (B) 66 .2265-79 (C) 114 .3850-64 (D) 57 .19Below 50 (F) 42 .14Total 300 1.0

The first part of the question asked me to find the frequency of the A grade at which point I added the rest of the frequency scores from the total of 300 and found the missing value. The second part of the questionasked me to find the missing relative frequencies from grades B and below, at which point I divided the frequency by 300 to find my answer.

RE: Homework Question 1 Paul Sandel 5/6/2015 7:40:45 PM

1 of 23 1 of 23

A: 90-100 28 0.07B: 80-89 96 0.24C: 65-79 156 0.39D: 50-64 64 0.16F: Below 50 56 0.14Total 400 1

These questions were simple for me. To figure out the relative frequency you take the frequency divided by 400. To find the frequency you add up all of the known numbers and subtract from the total.

RE: Homework Question 1 Jolene Whitmore 5/6/2015 8:17:04 PM

Grade on Business Statistics Exam Frequency Relative FrequencyA: 90-100 — 0.05B: 80-89 88 —C: 65-79 184 —D: 50-64 64 —F: Below 50 44 —Total 400 1

The first problem is asking to complete the table.

This is pretty simple to calculate in your head or calculator. To find the frequency for the letter grade A, I multiplied 400 by .05, or if you like to take the scenic route, you can add up all the number that are given foreach letter grade, then subtract that number from the total. Either way the answer is 20 for the frequency of the letter grade A.

To get the relative frequency of each letter grade, take the frequency data for each letter grade then divide by the total of 400. Completed relative frequency answers below. The relative frequency data will all add upto 1.

A: 20/400=.05B: 88/400 = .22C:184/400 = .46D: 64/400 = .16F: 44/400 = .11

RE: Homework Question 1 Nikki Sims 5/6/2015 10:59:35 PM

Here is my first question... To get the frequency of 15, I had to multiple 300 by 0.05. To get the rest of the relative frequency numbers, I had to divide the frequency number by 300.

Grade on Business Statistics Exam Frequency Relative FrequencyA: 90-100 15 0.05B: 80-89 58 0.17C: 65-79 141 0.47D: 50-64 54 0.18F: Below 50 39 0.13Total 300 1

RE: Homework Question 1 Erick Johnson 5/7/2015 10:27:37 AM

A: 90-100 - Frequency = 24 (take the total # of measurements and multiply it by the relative frequency 400*0.06)B: 80-89 - Relative Frequency = 0.16 (take the frequency and divide it by the total number of measurements 64/400)C: 65-79 - Relative Frequency = 0.48 (take the frequency and divide it by the total number of measurements 192/400)D: 50-64 - Relative Frequency = 0.17 (take the frequency and divide it by the total number of measurements 68/400)E: Below 50 - Relative Frequency = 0.13 (take the frequency and divide it by the total number of measurements 52/400)

RE: Homework Question 1 Dennis Mellem 5/7/2015 6:29:53 PM

Many people have responded with the answer - I just have a technical question. This assignment did have a fairly small set of data so doing it manually was not too difficult. I also copied the data into excel tocalculate the relative frequency via a formula and to cross check my answers.

I am wondering if there was a way to move the base data provided into Minitab and have it generate the relative frequency? I played around with it for a while but could not find any way to do that. Did anyone finda way to make that work and if so, what were the steps?

RE: Homework Question 1 Tara Crase 5/8/2015 7:27:40 PM

Dennis I cannot get any data tables to copy out of the homework either. This would be helpful.



RE: Homework Question 2 Nicholas Payne 5/4/2015 5:17:48 PM


Level1- 38/273=.139Level2- 70/273=.256Level3- 52/273=.190Level4- 89/273=.326Level5- 17/273=.062Level6- 7/273 =.026

Level 4 has the highest frequency.

for this you take the amount of times that the item or occurance happened. In the case of level 1 it happened 38 times. you then divide this by the total number to get the percentage of the whole that was used inthat level. The level with the highest percentage has the highest frequency.


This needs more explanation... a sentence or two more

2 of 23 2 of 23

RE: Homework Question 2 Eric Deckwa 5/6/2015 1:25:28 AM

Modified:5/6/2015 10:49 AM

Q2 Using new data table:

1"I$copied$the$data$table$to$excel

2"$to$find$the$rela5ve$frequency,$I$added$the$number$of$observa5ons$to$calculate$the$freq

3"I$used$the$formula$to$divide$the$individual$observa5ons$by$the$whole$to$get$the$decimal$frequency

4"$I$used$the$round$func5on$to$round$the$freq$to$the$nearest$thousandth$(3$decimal$places)

5"I$constructed$a$frequency$bar$graph.$I$can$see$that$level$reading$4$is$the$highest$bar$indica5ng$most$frequency.$If$the$bars$were$closer$in$height,$it$would$be$difficultto$determine$which$bar$had$the$most$frequency.

6"$I$sorted$from$greates$to$least$the$frequency$data$to$construct$a$Pareto$graph

This$graph$arranges$categories$of$the$qualita5ve$variable$by$height$descending$from$leJ$to$right.

Reading$Level$4$can$easily$be$iden5fied$with$the$most$frequency$by$using$the$Pareto$graph.

Week 1 HW Question 2.xlsx


Eric, I am absolutely fine with Excel, Minitab, StatCrunch, Calculators, pencils, etc.

I just want you to learn!

Some instructors say "Minitab Only" ... I just think everyone should use what helps them the most. Due to time limitations, I have to show Minitab only in the lectures and there will betimes where Minitab is the best option.

Best,

Brent

RE: Homework Question 2 Regina Clayton 5/7/2015 3:33:19 AM

Professor Brent and Class,

a. Compute the relative frequencies in each category.Reading Level Relative FrequencyLevel 1 (Red) 0.148Level 2 (Blue) 0.277Level 3 (Yellow) 0.170Level 4 (Pink) 0.322Level 5 (Orange) 0.0 64Level 6 (Green) 0.019(Round to the nearest thousandth as needed.) Level 4 is the highest relative frequencies in this case. More occurrences is in number 4 for the books that were read at this reading level to improve for Japanese. Each student toread 40 books but all students were at different levels that show improvement.


red .134

blue .286

yell .192

pik .330

org .047

gre .011

total 276

these are the relative frequency for all of the reading levels. Relative frequency is calculated by determining the total number and then taking the amount that are red in this case, which is 37 anddivide that by the total which is 276. That would give you .134 rounding to the nearest thousandth. This problem also asked about bar graphs and Pareto charts. These were easily answered byfinding the corresponding correct graph if you followed the table correctly to find the reading level that occur the most often, which is level number 4

RE: Homework Question 2 Rhodora Frilles 5/6/2015 12:50:18 AM

Modified:5/6/2015 1:00 AM

Hi Class,

Since all 12 questions are already posted and answered by my classmates, I would like to start from this question #2 using a similar exercise. Here's similar exercise for Homework #2 below and mysolutions;

a. compute the relative frequencies in each category

step 1 -copy datausingdelimitedtab

step2 -

Divide eachnumber in"Col a" belowbox, forexampleLevel 1number 42divided bytotal numberof 271 toarrive at thedecimalfrequency of.155

col a Answer

3 of 23 3 of 23

ReadingLevel Number Fequencysolution:

Level 1(Red) 42 0.155

=+C15/C21

step 3 -(finalanswer) round off tothe nearestthreedecimalplaces

Level 2(Blue) 74 0.273=+C16/C21

Level 3(Yellow) 50 0.185=+C17/C21

Level 4(Pink) 89 0.328=+C18/C21

Level 5(Orange) 11 0.041=+C19/C21

Level 6(Green) 5 0.018=+C20/C21

Total 271 1.000

b. frequency bar graph. Letter c, is the correct answer as it shows exactly the percentage in purple col above, level 4 = .328(.3), as the highest frequency. So, letters a and b are wrong.

c. Pareto diagram , correct answer is letter a, because it shows level 4, as .328 or .3


Here is an example of what I am generally looking for..... Please do not duplicate this exact question unless you did before I posted.... You can always click the Similar Problem button (orwhatever it's called) and do a similar problem.

On this one, I decided to use Powerpoint, because it is easier to show some things. I don't expect you to go to this detail. (But it would be impressive if you did.)

Attached

Week_1_Number_2_Homework.pptx

RE: Homework Question 2 Shu-Harn Yang 5/5/2015 1:34:43 PM

Hi Professor, I did problem #3, is that enough information or what would you suggest?Thank you

RE: Homework Question 2 Venice Kane 5/7/2015 2:01:52 AM

Similar Problem:It took me a few attempts to get it right. I tried to capture it on video so I could remember how I did it...And I watched the lecture again, which helped.Loaded the data set into excel:Question had a scenario with students being asked to read out of a total of 271 booksRelative Freq... Divided the number of books that were read in each category by the total number of books.

Reading Level Number

Level 1 (Red) 34 0.125

Level 2 (Blue) 76 0.280

Level 3(Yellow) 49 0.181

Level 4 (Pink) 87 0.321

Level 5(Orange) 17 0.063

Level 6(Green) 8 0.030

Total 271 1.000

The screen capture video attached is from a Shockwave Flash file. Hope it will open, feedback is welcomed.

Homework_Wk1_Q2_2015-05-07_0045_VK.swf , Excel Pic HW Wk1Q2Statistics_VK.docx

RE: Homework Question 2 Charles Croy 5/6/2015 10:18:43 PM

Similar Problem.....

Professors collected 259 books written for Japanese children and required that the students read at least 40 as part of their grade. The books were categorized according to length and complexity into reading levelsand color coded for easy selection. The data set is.....Reading Level NumberLevel 1 (Red) 37Level 2 (Blue) 77Level 3 (Yellow) 49Level 4 (Pink) 84Level 5 (Orange) 10Level 6 (Green) 2Total 259

A. find the relative frequency for each reading level.....Take the Number of books read for each color and divide by the total number of books read to find the relative frequencyLevel 1(Red) 37 0.143

Level 2(Blue) 77 0.297

Level 3(Yellow) 49 0.189

Level 4(Pink) 84 0.324

4 of 23 4 of 23

Level 5(Orange) 10 0.039

Level 6(Green) 2 0.008

B. Construct a relative frequency graph... for this I used excel. I copied the results from part a into excel and created a graph. I then compared the excel graph to the selections available. I also attempted to copy asnip of the chart into the discussion but was having difficulties so included it as an attachment called capture

C. Convert the relative frequency into a Pareto Diagram. To do this you simply re-arrange the graph to display the frequencies from greatest to smallest. I attached a snip of this as well called capture2. Using thePareto Diagram we can determine that reading level 4 (pink) occurs most often.

Capture.PNG , Capture2.PNG

RE: Homework Question 2 Erick Johnson 5/7/2015 1:34:14 PM

Reading Level NumberLevel 1 (Red) 38Level 2 (Blue) 75Level 3 (Yellow) 45Level 4 (Pink) 83Level 5 (Orange) 19Level 6 (Green) 6Total 266

Part A: dividing the number of readers in each level by the total number of readers (266) gives us the relative frequency of each level (see answers in attachment). Part D: using the graphs in the attachment, the reading level that occurs most often is level 4.

Week1_Question2.docx

RE: Homework Question 2 Nimmi Prathap 5/7/2015 4:16:45 PM

Similar Problem

The total number of books (n) in this case is 255. The table with reading levels is given below.

Reading Level NumberLevel 1 (Red) 39Level 2 (Blue) 74Level 3 (Yellow) 45Level 4 (Pink) 80Level 5 (Orange) 14Level 6 (Green) 3Total 255

a. Compute the relative frequencies:

Relative Frequency shows the proportion of observations that falls under a particular category, in this case , reading level. It can be obtained by dividing the number of observation in a category by the total number ofobservations. The table given below shows the relative frequencies in each category.

Reading(Level Number(of(books((Frequency) Relative(frequency1 39 0.1532 74 0.2903 45 0.1764 80 0.3145 14 0.0556 3 0.012Total 255 1.00

b. Construct a relative frequency bar graph.

I used excel to construct the rel. frequency bar graph, where the heights of bars correspond to the relative frequencies. As evident from the diagram (Please see attachment) level 4 (pink) is the mostfrequently occurring reading level.

c. Convert the rel. frequency bar graph in to a Pareto diagram.

A Pareto diagram is a bar graph with the categories of variables ( the bars) are arranged by height in descending order from left to right. Pareto diagram facilitates easy interpretation of the data set. Again, the pinkbar which corresponds to level 4 is the highest, implying that it is the most frequently occurring reading level.

Reading Level_ HW.Qn.2.docx


This was excellent Nimmi


Thank you, Professor....

RE: Homework Question 2 Kevin Gorman 5/7/2015 7:21:04 PM

Compute the Relative Frequency in each Catagory

Reading Level Number of Books Relative Frequency(1)Red 39 .146(2)Blue 77 .287(3)Yellow 48 .179(4)Pink 81 .302(5)Orange 16 .060(6)Green 7 .26

Total Number of Books= 268

To find the relative frequency in this exercise we had to divide the number of books in each colored classification by the total of number of books in the study (268) to find the relative frequency, which was to berounded to the nearest thousandth. The exercise also asked us to select a correct graph in the exercise demonstrating the relationship between the relative frequency and the correct colored classification. Thereading level in this study that occurred the most often in this study was #4(Pink) at .302, or 30% with 81 books.

5 of 23 5 of 23

RE: Homework Question 2 Paul Sandel 5/9/2015 7:21:18 PM

red .132 36blue .275 75yell .20155pik .32288org .059 16gre .0113 total 273 these are the relative frequency for all of the reading levels. Relative frequency is calculated by determining the total number and then taking the amount that are red in this case, which is 36 and divide that by thetotal which is 273. That would give you .132 rounding to the nearest thousandth. This problem also asked about bar graphs and Pareto charts. These were easily answered by finding the corresponding correct graph ifyou followed the table correctly.




There were 14 observations in the original data set. Counting the numbers in the leaf column. The bottom row of the original data represents stem is 0, leaf is 0, 3, and 6. The numbers in the original data set are 0,3,6The dot plot is option A, the number break down is 31, 31, 36, 20,20,20, 22, 11,11,13,17, 0, 3 and 6


Looks pretty good, but you can easily do a simple stem and leaf in the posting window

RE: Homework Question 3 Arnas Eidukaitis 5/6/2015 8:40:16 PM


Consider stem-and -leaf display to the bottom.a. How many observations were in the original data set?b. In the bottom of the row of the stem-and-leaf display, identify the stem, the leaves, and the number in the original data set represented by this stem and its leaves.c. Re-create all the numbers in the data set and construct a dot plot.

Stem Leaf3 1442 17771 1550 78

a.Leaf for each observation is recorded in the row of the display corresponding to the observation’s stem, therefore there are 14 observations (number of leafs).b.The stem is 0 and the leaves are 0,7,8. The numbers in original data set are 0, 7, 8 .

For example, suppose we had defined the stem as the tens digit for the R&D percentage data, rather than the ones and tens digits. With this definition, the stems and leaves corresponding to the measurements 13.5and 8.4 would be as follows:c . See attachment

ex.png


This was a good job!

I can't respond to every post, but I try to find some good examples and some that need more. Just to let you know what we're looking for....

Best,

Brent


STEM LEAF3 1192 07771 05660 025

The number of observations is equal to the number of entries (leaves). In this case, its is 14. I the bottom row of the display is stem 0 with leaves 0, 2, and 5. The original data set is 0, 2, and 5. The dot plot is attached.

Week1_Question3Graph.docx


3| 223

2|1118

1|1344

0|049

Use the stem and leaf diagram to a) determine the total number of observations b) find the original dataset where the stem equals zero c) recreate the entire dataset and create a dot plot diagram.

A) Answer is 14 Each stem represents a group of observations so you do not count the stems, only count the leaves.

B) Answer is 0,4,9 Combine the stem with each leaf to recreate the data. 00,04 and 09

C) Answer is 32,32,33,21,21,21,28,11,13,14,14,00,04,09 Do the same thing as we did in part B but for the entire data set then construct the dot plot. The dot plot is attached.

6 of 23 6 of 23

I'm still trying to figure out how to paste pics into a post. If anyone can help I'd appreciate it.

Capture2.GIF

RE: Homework Question 3 Nicole Reed 5/7/2015 8:16:34 PM

Stem | Leaf3 5582 14441 01190 045

There were 14 observations. 35,35,38,21,24,24,24,10,11,11,19,0,4,5.

The original data set is 0,4,5 and the stem is 0.

If you were to plot it on a dot plot there would be 2 dots at 35,1 at 38,1 at 21,3 at 24,1 at 10,2 at 11, 1 at 19,1 at 0,1 at 4 and 1 at 5.

RE: Homework Question 3 Nimmi Prathap 5/10/2015 3:26:55 AM

Similar Question.

The stem and leaf display is given below.

Stem Leaf3 0$7$72 0$0$0$61 0$2$2$90 0$4$9

a. How many observations were in the original data set?

There are 14 observations in the original data set. This can be easily obtained by counting the leaves.

b. In the bottom row, identify the stem , the leaves and the numbers in the original data set.

The stem is 0 and the leaves are 0,4 and 9. The numbers in the original data set represented by this stem and leaves are 0, 4 and 9..

c. Re-create all the numbers in the data set and construct a dot plot.I recreated all numbers (30,37,37,20,20,20,26,10,12,12,19,0,4,9) and constructed a dot plot in Minitab using the Graph>Dot Plot option. Somehow my snapshot is not getting attached.



RE: Homework Question 4 Kristian Martin 5/4/2015 7:10:16 PM


Based on the histogram generated by Minitab it is a frequency histogram with 12 measurement classes and a total of 43 measurements in the data set.

a. The y-axis title (frequency) dictates the type of histogram.

b. The measurement classes were determined by counting the number of bars in the histogram. A total of 12 blue bars.

c. The measurements in the data set were calculated by performing a summation of the height of each bar (i.e. the frequency per measurement class). ∑ (2+2+4+4+4+4+6+5+5+3+3+1) = 43


Very good

RE: Homework Question 4 Venice Kane 5/10/2015 9:09:15 AM

Similar Exercise:In the example I was given, the frequency axis and measurements were also shown in a histogram, (a)identified as the frequency histogram (not relative frequency histogram)(b)The measurement was to count the number of bars on the histogram which was 15(c)Then adding up where each bar landed on the frequency axis the total was 47


Everyone:

Having trouble copying or pasting? Don't worry about it..... Here is a way to do a good job on this one.....

a. It is a frequency histogram because it shows the class frequency of the observations. It would have been a relative frequency histogram if it showed the relative frequency of eachclass.

b. There are 9 measurement classes in this histogram (24, 26, 28, 30, 32, 34, 36, 38 and 40). Count the bars!

c. There are 40 measurements in the data set. By summing the height of each bar (looking at where it goes up to on the y axis) you can find the total number of measurements in theset.


Thank you professor, I tried to paste the picture of the histogram but for some reason when I posted my response the picture didn't show up.

7 of 23 7 of 23

RE: Homework Question 4 Roderick Harris 5/4/2015 10:03:45 PM

Modified:5/4/2015 10:14 PM

I could not paste the graph either, so I added it as an attachment. Professor, I got 38 instead of 40 data sets for part c.

Histogram1.docx


Never thought about posting it as an attachment Roderick, good idea, I'll have to do that next time!For the problem I had mine is attached

Screen Shot 2015-05-04 at 8.00.28 PM.png

RE: Homework Question 4 Professor Heard 5/6/2015 8:24:29 AM

Explain in a little more in your main post... How did you get 38? You simply ____ the __________ etc.

RE: Homework Question 4 Roderick Harris 5/8/2015 6:15:14 AM

In question 4 I added, from left to right, 3+4+4+4+5+4+6+3+5 = 38I simply added the values of the heights of each to get 38. I just double checked my calculations to be sure.


You're good here Roderick. Simple and answers how and what you did.


Best,

Brent


Screenshots don't provide answers... You need to explain what you did to come up with your answers.


Ok, fine to use attachments but always explain what you did in the posting window (to make us want to look)


Similar exercise for no. 4 is attached please.

a. answer is frequency histogramb. there are 11 measurements classes in this histogram. How did I arrive at 11? I count the number of bars, there 11 bars.c. how many measurements are in the data set described by this histogram? there 44. How did I arrive at number 44? =2+2+3+3+3+4+4+5+6+6+6

What it means, there are 2 bars at frequency 2, so 2+2. There are three bars at frequency 3, so 3+3+3, there are two bars in frequency 4 so, 4+4 and 1 at 5 so +5, then 3 bars atfrequency 6, so 6+6+6= all in all equals, 44.

screen shot of question no. 4.docx


Rhodora, you did a great job (no need for the screenshot). I am more interested in "how you solved the problem."


Based on the histogram generated by Minitab it is a frequency histogram with 8 measurement classes and a total of 36 measurements in the data set.

a. The (frequency) is the type of histogram, Because it showing the class frequency of the observation

b. The measurement classes were determined by counting the number of bars in the histogram. A total of 8 bars.

c. The measurements in the data set were calculated by performing a summation of the height of each bar(5+3+4+7+4+5+4+4) = 36

RE: Homework Question 4 Arnas Eidukaitis 5/8/2015 8:29:56 AM

Similar exercise for # 4 (please see the attachment)

a. Answer is frequency histogram, because Y-axis is labeled as Frequency

b. There are 13 measurements classes in this histogram. Each bar in histogram represents one measurement class, therefore I counted 13 bars.

c. How many measurements are in the data set described by this histogram? There are 45.Here is the calculation: each bar represents measurement class at different frequencies, that being said first and second bars taken at frequency 1 would translate to 1 + 1 =2 measurements (each ofthese measurements were taken once only (frequency =1).Third and fourth bars were taken at frequency rate 3, meaning 3 times,and this translates to 3 + 3=6 and so on.Here is recap:

8 of 23 8 of 23

1st bar =12nd bar=23rd bar= 34th bar= 35th bar=46th bar=67th bar= 58th bar= 59th bar= 610th bar =411th bar =312th bar =313th bar= 2

Total = 45

Histogram.jpg

RE: Homework Question 4 Abigail Hernandez 5/10/2015 11:37:06 PM

Professor Heard,

This is great advice! Counting bars to determine the measurement classes and counting the data, height of each bar, for the measurement of the data set. I found that very helpful for the homework andquiz. The part that I had most trouble with was determining what kind of graph was displayed, frequency vs. relative frequency histogram. But I found page 44 on the e-book very helpful. I realized that for afrequency histogram it shows the class frequency on the y axis while, on the x-axis measures the class interval. I kept thinking the x-axis was measuring frequency and the y-axis was relative frequency. Ithought I would share in case anyone else found it difficult to determine what time of graph is given.

RE: Homework Question 4 Nicholas Payne 5/6/2015 6:57:01 AM

Based on the histogram that was shown it is a frequency histogram with 14 measurements and 48 measurements in the data set. The frequency that is shown on the y-axis shows how many total measurements arein the data set. By adding up all of the measurements in this axis you arrive at a total of 48 measurements. On the X-axis it is much the same. Each bar on the histogram is a measurement and by adding them upyou get a total of unique measurements.


The histogram is a frequency histogram based on the label of the y axis. The histogram contained 10 measurement classes (total number of columns in the graph) and the number of measurements are equal to thesum of all measurement totals (3+5+4+5+6+6+6+5+5+2=43




Data set = 12 15 19 17 17 20 18 13 17 11 12, 11 sets of measurementsMean = sum of measurements divided by the number of measurements:12+15+19+17+17+20+18+13+17+11+12+11 = 171/11 = 15.54545454545 rounded to the nearest 1000th = 15.55Median = the middle number when the data is arranged in ascending or descending order:Ascending order: 11 12 12 13 15 17 17 17 18 19 20, the value in the 6th position is the median = 17Mode = most frequent measurement in the data set, 17 occurs 3 times, 12 occurs twice all other values occur once. The mode is 17


If you were to use Minitab on this one (I recommend it), you could just tell us the steps you used in Minitab and the results and what they mean (basics).

RE: Homework Question 5 Eric Deckwa 5/6/2015 11:01:29 AM

Similar Excersie:

Data set given: 15 18 13 13 17 20 19 10 11 13 19

First step was to copy the data set onto my clipboard.Next, I opened minitab and in the first column, I pasted the data set. labeled the column as "Data"Next, In minitab, on the toolbar, I selected the following path: Stat> Basics Statistics > Display Descriptive Statistics> clicked the Variables box to place cursor inside> double clicked Data next to C1> clickedon the statistics button next to graphs button> only selected Mean, Median, Mode, hit ok twice to get results:

Mean = 15.27

Median = 15.00

Mode = 13

N for Mode = 3


Nice job Eric, everyone please note that I don't respond to every single post (it would make for a big mess). I try to find examples of good posts to serve as examples.

RE: Homework Question 5 Regina Clayton 5/7/2015 11:07:58 PM

Professor Brent and Class,

Data Set: 13,20,14,19,17,13,18,13,10,16,10

Ascending order: 10,10,13,13,13,14,16,17,18,19,20

Mean= 14.82

9 of 23 9 of 23

(Round to two decimal places as needed.)Median= 14(Round to two decimal places as needed.)

The mode is 13 as it appears most often in this problem.


1815131519151910201712

The Mean (the sum of the measurements divided by the total number of measurements) is 15.73The Median (the number in the center of the sample when the samples are put in numeric order) is 15. The group does have a Mode of 15 (which is the measurement that occurs most often.


Erick,

This was a good job!


Best,

Brent

RE: Homework Question 5 Larry Parks 5/9/2015 9:07:19 AM

Modified:5/9/2015 9:07 AM

20,14,17,11,10,17,14,17,19,13,16

MEAN: ADD ALL NUMBERS AND DIVIDE BY TOTAL20,14,17,11,10,17,14,17,19,13,16 = 168 / 11 = 15.27

MEDIAN: ARRANGE IN ASCENDING ORDER. IF ODD USE MIDDLE. IF EVEN, USE MIDDLE 2.10,11,13,14,14, 16, 17,17,17,19,20

MODE: IS THE NUMBER THAT OCCURS MOST FREQUENTLY10,11,13,14,14,16, 17,17,17, 19,20

RE: Homework Question 5 Nimmi Prathap 5/10/2015 3:46:26 AM

The given data set is (15,20,15,18,12,14,17,12,10,16,15)

The Number of Observations is 11.

Mean is obtained by dividing the sum of measurements/observations by the number of observations. Mean for the above data set is 164/11=14.91 (Rounded to two decimal points)

If n is odd, Median is the observation in the middle when the data is arranged in an ascending or descending order. If n is even, then median is the mean of two middle numbers. In the given data set, which has oddnumber of observations, median is 15.(10, 12, 12,14,15,15,15,16,17,18,20)

Mode is the most frequently occurring observation in a data set. Here, it is 15.


MEAN: ADD ALL NUMBERS AND DIVIDE BY TOTAL 13+12+17+13+12+18+12+11+19+20+16/ 11 = 163/11=14.82

MEDIAN: ARRANGE IN ASCENDING ORDER. IF ODD USE MIDDLE. IF EVEN, USE MIDDLE 2.11,12,12,12,13,13,16,17,18,19,20=13

MODE: IS THE NUMBER THAT OCCURS MOST FREQUENTLY

11,12,12,12,13,13,16,17,18,19,20


Here is the set of data for my question number 5:

11 11 12 12 12 13 15 16 17 18 20

I was asked to identify the mean, median, and mode.

Mean is the average number of the data set given. I would add all the numbers up and divide by 11, since there are 11 number given of the data. 11+11+12+12+12+13+15+16+17+18+20= 157/11 = 14.2727

Median is the middle number of the data. The middle number of the data given is number 13.

Mode is the the number that appears the most. In this given data the number most repeated was the number 12. It was repeated three times.

This was probably one of the easier problems, but it was helpful because it refreshed my memory. I haven't taken stat or similar math course since three years ago.

10 of 23 10 of 23





The Mean Value for the data set ,148.16 ,shows the average number of semester hours of college credit for the candidates. It is obtained by dividing the sum of all candidates' credit hours by the total number ofcandidates.

The median value is 142 which signifies that half the candidates had more than 142 credit hours and half the candidates had less than that.

The Median value is less than the Mean value, so it can be concluded that the distribution is skewed to the right.


Hi Nimmi - you got a detailed explanation of your answers above. Good post!

My exercise is similar to yours, different numbers:

The Mean Value for the data set, 130.39 hours ,shows the sum of all the candidates credit hours divided by the the total humber of candidates is equal to 130.39 hours.

The median value is 125 hours which means half the candidates had more than 125 credit hours and half of candidates had less than 125 credit hours.

What type of skewness, if any, exists in the distribution of total semester hours? The mean(130.39) exceeds the median (125) value, so distribution is skewed to the right or positive.

RE: Homework Question 6 Jolene Whitmore 5/7/2015 8:14:29 AM

Hi Nimmi & Rhodora. I did this problem and kept getting the skewness backwards. I had to make myself a "cheat sheet". Silly I know, but Professor Heard had a great power point slide from hislecture that helped me understand the visual of the "left tailed", & "right tailed" skewness. Left tailed is when the mean is smaller than the median, and right tailed is when the mean is greater thanthe median. Because I am a visual person, It better suits me to have the picture and explanation in front of me. You both did great and explained very well. Thank you


Remember the cat.....


Rhonda,

I agree with your post. My question of number 6 had "The mean and median for the data set were

131.1 and 125 hours. The total number of 131.1 was the sum of all the candidates' credit hours divided by the total number of candidates is equal 131.1 and the 125 hours was the candidates thathad the amount of hours of 125.

The skewed is to the right it means that the median is less than the mean.

mathxl.com


Very nice Nimmi!


Interpret the mean value of 149.36 hours: The sum of all the candidates' credit hours divided by the total number of candidates is equal to 149.36 hours.

Interpret the median value of 155 hours: Half the candidates had more than 155 credit hours and half the candidates had less than 155 hours.

What type of skewness, if any, exists in the distribution of total semester hours? because the median is more than the mean, the distribution is skewed to the left.

RE: Homework Question 6 Eric Deckwa 5/8/2015 8:39:15 PM

A sampling produced a mean of 143.55 hrs. which is the sum of all candidates hrs divided by total candidates.

The median of 138 hrs. of the data sampling is the halfway point, from an ascending order, of the candidates sampled.

When the mean < median, this interprets that the skew is to the left.

When the mean> median, this interprets that the skew is to the right.

Since the mean of 143.55 is greater than the median of 138, the skew of the candidates sampled is to the left.

RE: Homework Question 6 Tariq Sabir 5/10/2015 8:58:51 PM

Nice example Eric, sometimes it is how something is presented that will make it easier to understand. The greater than and less than signs help me better understand. When the mean < median the skew isto the left. This means that there is more data points to the left of the median than to the right of the median. I look at the skew as a build up (if the shew is to the left) to the median. That means there ismore that is needed to build up to the median. When the mean > median then the skew is to the right. So there is a longer descent from the median, so there is more data to the right of the median. Maybe a better way of saying it is where is the data spread most, so when the mean < median the data is more spread to the left and when the mean > median the data is more spread to the right.

RE: Homework Question 6 Nikki Sims 5/8/2015 10:44:32 PM

11 of 23 11 of 23

Question 6

144.74- is interpreted as the sum of the candidates credit hours added up together and divided by the numbers of credit hours. The mean is the number that is in the middle of the set of numbers and that number is150. The distribution is skewed to the left because the mean is less than the median.





https://www.dropbox.com/s/zt68b8et7chl00z/Screenshot%202015-05-05%2012.36.08.png?dl=0

4, 6, 3, 3, 2, 7, 2 -> re-arrange to 2, 2, 3, 3, 4, 6, 7Range is the difference between the largest measurement and lowest measurement, therefore it is 7-2 =5

is the formula to find sample variance, S^2 while Xi equals individual measurement and X is the mean of the data, and N is the total number of measurements. mean = (2+2+3+3+4+6+7)/7 = 3.857 = 3.86[(2-3.86)^2+(2-3.86)^2+(3-3.86)^2+(3-3.86)^2+(4-3.86)^2+(6-3.86)^2+(7-3.86)^2]/7-1 = 3.81(S^2)

Sample standard deviation is S, it is square root of variance, therefore √3,81 = 1.95 (rounded to the nearest hundredth)





***Please disregard previous entry***

Calculate$the$range,$variance,$and$standard$devia5on$for$the$following$sample.0.4,$0,$0,$−3,$3,$−1,$3,$0,$−3,$3,$−3,$0,$−4,$−1,$−3,$0,$3 This$was$my$problem.$$First$off$I$had$to$go$though$the$IT$Helpdesk$for$two$days$to$try$to$get$Minitab$installed$in$working.$$So$to$this$point$I$have$not$used$Minitab$yet.$$The$problems$which$I$have$completed$I$have$used$strictly$Excel.

To$calculate$the$range$you$want$to$first$sort$the$numbers$in$ascending$order,$the$purpose$is$so$it$is$easier$to$be$able$to$see$the$upper$and$lower$limits

"4 What$you$want$to$do$next$is$subtract$the$lower$limit$from$the$upper$limit$so$you$will$have$the$numerical$answer$for$the$range

"3

"3 3"("4)=7 The$range$for$this$data$set$is$equal$to$7

"3

"3 The$purpose$of$the$range$is$to$tell$you$how$spread$out$the$data$is. This$is$very$simple,$but$with$simplicity$comes$sensi5vity

"1 The$range$can$be$easily$affected$by$one$number$that$is$so$far$outside$of$the$range$of$the$other$numbers$that$it$throws$it$all$off

"1 Imagine$if$we$added$one$number,$say$100$to$this$data$set,$now$the$range$would$be$104. This$greatly$affects$the$spread$of$the$data.

0

0

0

0

0

0.4

3

3

3

3

To$calculate$variance$you$want$to$take$the$summa5on$of$each($integer$minus$the$mean)$squared$divided$by$the$number$of$intervals$minus$one

Simply$the$variance$is$the$average$of$the$squared$differences$from$the$mean

What$does$variance$tell$us? It$explains$how$different$or$how$much$individuals$in$a$group$vary. The$larger$the$variance$the$larger$the$differences$are$from$the$mean

This$is$another$way$to$see$how$spread$out$the$data$you$are$using$is

5.519706 is$the$answer$and$luckily$there$is$a$func5on$in$Excel$that$will$calculate$that$number$for$you (var$func5on)

Standard$devia5on$is$another$measurement$of$varia5on$of$the$data$set$that$you$are$working$with,$or$in$other$words$how$spread$out$the$data$is

The$higher$the$standard$devia5on$the$more$spread$out$the$data$set$is. A$real$world$example$that$helped$me$be_er$understand$is$stocks.

The$higher$the$standard$devia5on$is$for$a$stock$that$means$it$is$more$volitale$or$the$prices$change$more$frequently$and$dras5cally

To$calculate$standard$devia5on$it$is$simply$the$square$root$of$the$variance$=$2.35

12 of 23 12 of 23

All$three$of$these$items:$range,$variance,$and$standard$devia5on$basically$tells$you$how$spread$out$your$data$is$as$you$go$down$the$line

each$one$is$more$accurate$than$the$next$and$tells$a$be_er$story.


This was a good job! Sorry about your Minitab issues.


Best,

Brent

RE: Homework Question 7 Larry Parks 5/8/2015 5:41:49 PM

3,2,1,0,1R=3-0R=3

3,2,1,0,19,4,1,0,1∑×=7∑×^2=15

S^2=15-(7)^2/5/5-1=15-9/5/5-1=15-9.8/5-1=5.2/4=1.3

SD=1.14

RE: Homework Question 7 Rhodora Frilles 5/8/2015 11:48:05 PM

Similar exercise for no. 7 is below. I got this after third try I think and viewing the sample many times;

calculate range, variance, and standard deviation for the following sample: 7,6,2,2,4,5,4

Range = 5

13 of 23 13 of 23

Sample variance = 3.57(rounded off to two decimal places)standard deviation = 1.89

Step 1 - Range? get the highest number which is 7 minus the lowest number from the sample which is 2, equals, 5Step 2 - a) get the sum of the sample like 7+6+2+2+4+5+4 then divided by the number of sample which is 7, equals 4.29 b) then calculate sample variance of it like "(7-4.29)^2 + (6-4.29)^2+(2-4.29)^2+(2-4.29)^2+(4-4.29)^2+(5-4.29)^2+(4-4.29)^2 divided by 6 (7-1)=3.57 (rounded off to two decimal places)Step 3 - calculate the STd deviation of 3.57, which I use my scientific calculator so that gives me 1.889 or 1.89(rounded off to two decimal places). Thank you, I hope I get accross what I am trying toexplain:)


This question was asking us to compute the range, variance and standard deviation of a data sample.

17 data points were provided as follows:

0.1, 0, 0, -2, 2, -1, 2, 0, -2, 2,-2,0,-4,-1,-2,0,2

I copied the data into Minitab and it computed the Range of 6, a Sample variance of 3.00 and a Standard Deviation of 1.73. These were the correct numbers.

The range of 6 is the difference between the extreme values, so -4 is the lowest number and 2 is the highest number, so the range is 6.

We do not know what the units represent in this example – are they distances from a point or how much money someone made or lost at a lemonade stand.. We don’t know really know. However if we did knowwhat we were measuring, the variance would be that measure squared (inches squared for example), while the standard deviation would retain the original unit measure (inches). The range also retains the originalunit of measure.

The calculations for the std deviation and the variance are complex.

The sample variance calc would be

(0.1 -(-.347))2 + (0 -(-.347))2 + (0 -(-.347))2 + (-2 -(-.347))2 + (2 -(-.347))2 + (-1 -(-.347))2 + (2 -(-.347))2 + (0 -(-.347))2 + (-2 -(-.347))2 + (2 -(-.347))2 + (-2 -(-.347))2 + (0 -(-.347))2 + (-4 -(-.347))2 + (-1 -(-.347))2 +(-2 -(-.347))2 + (0 -(-.347))2 + (2 -(-.347))2_____________________________________________________________________________________________

17 – 1

.1998 + .1204 + .1204 + 2.7324 + 5.5084 + .4264 + 5.5084 + .1204 + 2.7324 + 5.5084 + 2.7324 + .1204 + 13.3444 + .4264 + 2.7324 + .1204 + 5.5084_____________________________________________________________________________________________

16

47.9612 __________

16

Which equals 2.9975 – rounded to a variance of 3.

The standard deviation is then the square root of 3 or 1.73


7,-1,2,6,8,4,4,2,6,6,Range is the difference between the largest and lowest measurement, which is 8 – (-1) = 9

The Mean would be the sum of all measurements divided by the total number of measurements, which is 4.4

The sample variance would be as follows: [(7-4.4)^2+(-1-4.4)^2+(2-4.4)^2+(6-4.4)^2+(8-4.4)^2+(4-4.4)^2+(4-4.4)^2+(2-4.4)^2+(6-4.4)^2+(6-4.4)^2]/10-1 = 7.60

Sample standard deviation (S) is the square root of the sample variance, which is √7.60 = 2.76 (rounded to the nearest hundredth)





0.610.840.590.80.481.070.680.33

The given data is presented above. There are four parts to this particular question.

a. Find the range

The range is simply the largest number minus the smallest number in the data set. From the given data the calculation would be 1.07-0.33= 0.74 carats

b. Find the variance i.e. s squared

To find the variance, I copied the data set and pasted it into an excel spreadsheet and used the variance formula under the statistical formula category. Which provided an answer of --> . Variance can also be done manually by first ∑x (summation of x) where x is all of the data points.Next ∑(x)^2 which is each individual data point squared and then added together.

Lastly, the calculation performed is s^2=(∑x^2-∑(x)^2/n)/ n-1....where n= 8 which is the total number of data points

c. Find the standard deviation

The standards deviation represented by "s" is the square root of the variance (s squared). sqrt(0.052)=0.228 carats

d. Which measure of variation best describes the spread of the 8 carat values? Explain.

The standard deviation is the best because it uses all of the values in the calculation, it indicates how spread out the data are, and it is in the same units as the original data.

14 of 23 14 of 23


Very good, note you could use Minitab and explain the steps etc.



Professor,

I've been playing with minitab. I tried to use minitab for a different problem with two sets of data but for some reason, my variables didn't auto-populate when I tried to make a histogram. I will usethis problem and see if it works without any errors.

UPDATE

So I used my numbers in minitab and was able to calculate variance and standard deviation. I first coped the data from mystatlab using the "make tab delimited copy" option. I pasted theinformation in minitab and selected the "Stat" tab --> "Basic Statistics" --> "Graphical Summary". A separate window pops up displaying all the key statistics for the data including a box plot,histogram, confidence interval etc. A screenshot of the window is attached below. From the graphical summary it provided a standard deviation of 0.22910 and a variance of 0.05249. The summarydid not provide the range however it does give the maximum and minimum points for the box plot which can be subtracted (1.07-0.33) to get the range of 0.74. I also used the calculate function toget the range but it once I completed the formula it replaced my data with the answer of 0.74, unsure as to why?

Screen Shot 2015-05-06 at 5.57.29 PM.png


Very cool, thanks for the follow up.


Using "similar problem" my data is as follows:a. x=50, s=5, sample mean = 35

(I couldn't find the sample mean symbol to insert and none of my copy/paste works in here)I'm still working on getting along with the software, I haven't had any success yet so doing my calculations by hand.Answers are: a: 3, z-score within a sample meanb: .25 z-score within a population meanc: 0 z-score within a population meand: -.5 z-score within a sample mean

Whether it is a sample mean or a population mean, in this case is determined by the data provided. The value of the mean is the number of units above or below the mean, for example in questiona. x is 3 standard deviations above the mean. b. x is .25 standard deviations above the mean, c. x is at the mean and d. x is .5 standard deviations below the mean.

RE: Homework Question 8 Ryan Frain 5/6/2015 11:57:13 AM

It$looks$like$we$were$given$a$similar$problem$with$different$data.$I$can$see$that$every$problem$has$been$completed$so$I$am$going$to$add$the$homework$problem$that$is$similar$below:

0.720.780.710.450.40.990.730.6

Find the range of the dataa.

Answer:$I$ordered$the$numbers$from$smallest$to$largest

0.4,$0.45,$0.6,$0.71,$0.72,$0.73,$0.78,$0.99

From$here$I$took$0.99"0.4=$0.59

Find the variance of the data set

Answer:$I$used$the$minitab$to$get$the$answer$it$generated$0.035$carats^2

a.

Find$the$standard$devia5on

Answer:$again$using$the$minitab$the$standard$devia5on$is$0.188$carats$which$is$just$the$square$root$of$the$variance

b.$

Which$measure$of$varia5on$best$describes$the$spread$of$the$8$carat$values?$Explain.

Answer$E:$The$standard$devia5on$is$the$best$because$it$uses$all$of$the$values$in$the$calcula5on,$it$indicates$how$spread$out$the$data$are,$and$it$is$in$the$same$units$as$the$original$data

c.$


Nice job Ryan.


Nice job classmates! I took the shot at this question too, it's somewhat similar to number 7. Yet, I had struggled on question letter b and the letter d.

Similar exercise is below;a) Range =( 1.19 - .44 ) = .75 ; highest number in sample is 1.13 minus the lowest number which is .44.

1.02

15 of 23 15 of 23

0.760.530.640.441.190.830.53

5.94

Start forsolvingletter bquestion-----(step1, get thesum ofthesample)

step 2,get thesquareroot ofeachsamplerange)

1.0404 (this is the

1.02^2)= 1.04040.5776 (.76^2)

0.28090.40960.19361.41610.68890.28094.888

4.888-(5.94)^2/88-1=(4.888-4.41045)/7=.47755

0.068letter b,find thevariance

0.261 std deviation of answer inletter b or of .068, letter c

d) which measure of variation besst describes the spread? Correct answer is letter b or, The Std deviation is the best because it uses of the values in the calculation, it indicates how spreadout the data are, and it is in the same units as the original data.


Ryan,

Good post. Different data set (8):

0.49

0.77

0.72

0.82

0.47

1.02

0.76

0.32

Utilizing the minitab to obtain my data - Statis, basic statistics and input the data. Here are the results:

Variable Mean SE Mean StDev Variance Median

Data 0.6712 0.0802 0.2269 0.0515 0.7400

range was 0.7 - taking 1.02 - 0.32

Variance which I noted incorrectly as 0.510 it is 0.051

Standard Deviation which is the squared root of the variance 0.051 = 0.22693217992544 rounding 3 decimal equally to 0.227

The measure which measure this variation:


RE: Homework Question 8 Arnas Eidukaitis 5/9/2015 10:49:41 PM

Here is my data:0.390.970.570.550.371.190.720.77

a. Find range of dataBiggest data number minus smallest number gives data range, therefore it is as follow: 1.19-0.37=0.82

b. Find the variance of the data setUsed Microsoft Excell formula VAR, therefore it's as follow: 0.08

16 of 23 16 of 23

c. Find the standard deviation of the data setSqrt (variance)=0.283

d.Which measure of variation best describes the spread of the 8 carat values? Explain.




RE: Homework Question 9 Larry Parks 5/5/2015 7:17:55 PM

Compute the Z score

a) z=41-35/3=6/3 z=2b) z=70-69/2=1/2 z=.5c) z=25-25/5=0/5 z=0d) z=24-30/4=-6/4 z=-1.5e-1:4) sample, population, population, samplef-1) 2 standard deviations above the meanf-2) .5 standard deviations above the meanf-3) at the meanf-4) 1.5 standard deviations below the mean


You might lead in with something like... "To compute the z score we subtract the mean from the value in question and then divide by the standard deviation. (No need for fancy symbols).


similar exercise:

1. To compute z-score(standard score), we take the score minus the mean score, divided by the standard deviation.

a. z=(55-45)/5=2b. z=(62-61)/2=0.5c. z=(45-45)/5=0d. z=(44-45)/2=-0.5

2. Based on the formula used, for part a. the z-score locates x within a sample.For part b. the z-score locates x within a population, and for part c. the z-score locates x within a population.For part d. the z-score locates x within a sample.

3. z-score represents the distance between a given measurement x and the mean, expressed in standard deviations. For a. z-score is 2, therefore the value of x lies 2 standard deviations above the meanFor b. z-score is 0.5, therefore the value of x lies 0.5 standard deviations above the meanFor c. z-score is 0, therefore the value of x lies at the meanFor c. z-score is -0.5, therefore the value of x lies 0.5 standard deviations below the mean

RE: Homework Question 9 Ryan Frain 5/8/2015 11:54:49 AM

For$another$similar$exercise$I$wanted$to$work$this$ques5on$because$the$z"score$is$something$that$I$am$not$as$familiar$with.$So$to$begin$the$data$I$was$given$is:

x= 34, s = 3, x(bar)= 25a. x=79,$µ=$78,$σ=$2b.$µ=$50,$σ=$5,$x=$50c.$s=$6,$x=$26,$x(bar)=$35d.$

z"score=$(x"x(bar))/s$or$(x"µ)/σ

z=3a. z=$0.5b.$z=$0c.$z=$"1.5d.$Asks$which$are$samples$and$which$are$popula5ons.$Parts$a$and$d$are$samples$and$parts$b$and$c$are$popula5ons.e.$Asks$where$x$is$located.$A$is$located$3$above$the$mean,$B$is$located$0.5$above$the$mean,$C$is$at$the$mean,$and$D$is$1.5$below$the$mean.f.$


similar exercise:

1. To compute z-score(standard score), we take the score minus the mean score, divided by the standard deviation.

a. z=(52-40)/6=2b. z=(78-77)/2=.5c. z=(40-40)/5=0d. z=(43-45)/4=-0.5

2. Based on the formula used, for part a. the z-score locates x within a sample.For part b. the z-score locates x within a population, and for part c. the z-score locates x within a population.For part d. the z-score locates x within a sample.

3. z-score represents the distance between a given measurement x and the mean, expressed in standard deviations. For a. z-score is 2, therefore the value of x lies 2 standard deviations above the meanFor b. z-score is 0.5, therefore the value of x lies 0.5 standard deviations above the meanFor c. z-score is 0, therefore the value of x lies at the mean - my original answer on this part was wrong as 16, std deviation below the meanFor c. z-score is -0.5, therefore the value of x lies 0.5 standard deviations below the mean


17 of 23 17 of 23

Modified:5/10/2015 8:42 AM

i like the z score because it makes it really simple to show how far away from the mean the particular data falls also it appears as though it is one of the calculations which seems to be easy todetermine by hand if what you need is already given in the problem. Here is the problem that was given to me in my homework:

a. x=48, s=6, xbar=30, z = (48-30)/6 = 3 - this means that x is 3 standard deviations above the mean in the sample

b. x=69, µ=67, α=4, z = (69-67)/4 = .5 - this means that x is .5 standard deviations above the mean in the population

c. µ =75, α=5, x=75, z = (75-75)/5 = 0 - this means that x is 0 standard deviations above the mean or is the mean of the population

d. s=4, x=40, xbar=50, z = (40-50)/4 = -2.5 - this means that x is 2.5 standard deviations below the mean of the sample.



Compute the z-score corresponding to each of the values of x below.

a.53-45=8/4=2---2 standard deviations above the meanb.86-85/2=.5---.5 standard deviations above the meanc.50-50/5=0---at the meand.24-30/4=-1.5---1.5 standard deviations below the mean

The mean is 0 so if you have a positive number for example two it would be two standard deviations above the mean. If you have a negative number for example -1.5 it would be -1.5 standard deviations below themean.

Sample

Population I knew it was a population if I saw µ(population mean)that symbol in the equation

Population

Sample




A sample data set has a mean of 61 and a standard deviation of 14. Determine whether each of the following sample measurements are outliers.

A. 45 Not an outlier - This would not be an outlier because it is between 61 - 14B. 60 Not an outlier- This would not be an outlier because it is between 61 - 14C. 13 Outlier This is an outlier because it is outside of 61 -14D. 107 Outlier This is an outlier because it is outside of 61 -14

If I understand correctly outliers are outside of other observations, it could be because of an experimental error.


Maybe explain what an outlier is...


An outlier is an observation that lies an abnormal distance from other values. This could be if the observation is smaller than the others in the data set or larger. (McClave 81)

McClave, James T., P. Benson, and Sincich. Statistics for Business and Economics for DeVry University, 11th Edition. Pearson Learning Solutions. VitalBook file.


Defined as how many standard deviations away from the mean?

RE: Homework Question 10 Jolene Whitmore 5/7/2015 7:31:58 PM

If it is 3 or more deviations away from the mean, then it is an outlier. Thought I would add my two cents. LOL


Thank you for the two cents!!


Similar Problem...

Sample data has a mean of 68 and a standard deviation of 14. Determine if the following are outliers: 58,121,69,4.

Adding and subtracting 14 from 68 results in 82 and 54. Numbers between these values are NOT outliers. Numbers greater than 82 and less than 54 are. Hence 58 and 69 are not outliers; they are within onestandard deviation. 121 and 4 are outliers since they fall outside of the range of 1 standard deviation.

Professor and classmates, How does this homework question relate to 6 sigma and quality management? Stated in a different way; when looking at variation in a product or process, why would 6 sigma be acceptable? At first glance callinganything beyond 1 standard deviation an outlier or "defective" makes sense. Less variation in the final end product... correct?

18 of 23 18 of 23


I had a similar problem as well....

Sample data has a mean of 70 and a standard deviation of 14. Determine whether each of the following sample measurements are outliers.

a. 17b. 65c. 49d. 87

In order to solve this problem the first step is to determine three standard deviations to the right (above) and to the left (below) the mean. Three standard deviations is above and below the mean isstandard for a normal distribution (general assumption unless otherwise noted in a given problem).

Therefore, three standard deviations to the right are the following:

70+14= 841. 84+14=982. 98+14=1123.

Therefore, three standard deviations to the left are the following:

70-14= 561. 56-14=422. 42-14=283.

Now that we know the three standard deviations above and below the mean of 70 we can return to the original question of potential outliers. An outlier in this case will fall either below 28 (to the left ofthe mean) or above 112 (to the right of the mean). Based on this information the answers are listed below:

a. 17 is an outlierb. 65 not an outlierc. 49 not an outlierd. 87 not an outlier

Charles in regards to your question: "when looking at variation in a product or process, why would 6 sigma be acceptable? At first glance calling anything beyond 1 standard deviation an outlier or "defective"makes sense. Less variation in the final end product... correct?"

I have taken a few six sigma courses and done research pertaining to the subject. In the instance of variation in a product or process 6 sigma would be acceptable because the focus is minimizing thedefects in the actual product where the closer you are the achieving six sigma the less variance or defects you have. Therefore, achieving six sigma is actually equal to 99.9% which is the tail ends of adistribution which each section divided amongst three percentiles of 68%, 95% and 99.7%.


Charles, six sigma is a little different animal... However, I will tell you after taking this class, you have to the statistical tools to go into it very deeply. It is really not that complicated.


Similar problem:A sample data set has a mean of 75 and a standard deviation of 15. Determine whether each of the following sample measurements are outliers.

a. 132b. 23c. 5d. 54

According to view an example description (by clicking on view an example on the right side panel), observations with z-scores greater than 3 in an absolute value are considered outliers. For some highly skewed datasets, observations with z-scores greater than 2 in absolute value may be outliers.

With that in mind, we can computer the following:a. (132-75)/15=3.83.8 is greater than 3 in absolute value, therefore the measurement 132 is an outlierb. (23-75)/15=-3.46-3.46 is greater than 3 in absolute value, therefore the measurement 23 is an outlierc. (5-75)/15=-4.67-4.67 is greater than 3 in absolute value, therefore the measurement 5 is an outlierd. (54-75)/15=-1.4-1.4 is less than 3 in absolute value, therefore the measurement 54 is not an outlier


Very good here, everyone please note that I don't respond to every single post (it would make for a big mess). I try to find examples of good posts to serve as examples.


The formula I used that was very helpful to determine outliers was: z= x - m/ s

z= z-scoresx= the number providedm= means= standard deviation

If the z-scores is lesser than 3 then it it would not be considered an outlier, but if it is greater than 3 then it will be considered an outlier.An outlier is an inconsistent or unusual number within the data collected. The common causes consist of invalid information recorded, misclassified measurement meaning a number from a different population, andrare chance event.There are actually two ways to determine these outliers, which are by box plots and z-scores. This example required outliers to be determine by z-scores.

The numbers given were: 73, 94, 116, 118Mean: 72Standard deviation: 13

Work:

73: z= 73-72/ 13 = 1/13= .08 (rounded)= .08<3, therefore 73 is not an outlier because the z-score is less than 3.94: z= 94-72/ 13 = 22/13= 1.69 = 1.69<3, therefore 94 is not an outlier because the z-score is less than 3.116: z= 116-72/ 13 = 44/13= 3.38 = 3.38> 3, therefore 116 is an outlier because the z-score is more than 3.118: z= 118-72/13 = 46/13= 3.54 = 3.54> 3, therefore 118 is an outlier because the z-score is more than 3.

McClave, J., Benson, P., & Sincich, T. (2011). Statistics for Business and Economics (11th ed.). Boston, MA: Pearson Education.


19 of 23 19 of 23

Given information: 102, 70, 67, 62 mean 63 and standard deviation of 10102 is an outlier; greater than 3 times outside the standard deviation (63 mean, deviation 53-73)70, 67 & 62 are all not outliers (within deviation range)According to our text outliers are usually due to:1. The measurement is observed, recorded, or entered into the computer incorrectly.2. The measurement comes from a different population.3. The measurement is correct but represents a rare (chance) event.If outliers can likely be caused by error, non-population data or rarities this means that most data is consistent. If you have outliers in your data set it would be abest practice to review the raw data and the sources of the data. This is related to data integrity; if you have error in your data you work to ensure integrity ofdata collection, data entry and interpretation.

RE: Homework Question 10 Eric Deckwa 5/9/2015 4:50:16 PM

Sometimes, unusual or inconsistent measurements occur in a data set that is being observed. An observation that is unusually large or small relative to the data values we want to describe is called an outlier.

To determine if the following sample measurements with a mean of 67 and standard deviation of 11 are outliers, I take the measurement, subtract the mean and then divide by the standard deviation. the absoluteresult greater than a z-score of 3 are outliers.

A) Measurement of 19: (19-67)-15 = -4.36 abs = 4.36>3; this is an outlier

B) Measurement of 79: (79-67)-15 = 1.09 < 3; not an outlier

c) Measurement of 128: (128-67)-15 = 4.06>3; this is an outlier

d)Measurement of 76: (76-67)/15 = .6<3; not an outlier



RE: Homework Question 11 Kelly Taylor 5/5/2015 5:14:05 PM

a. Find the approximate 25th percentile for the PASI score before treatment.- The approximate 25th percentile is 20 as this is where the bottom line of the box is plotted - the Lower Quartile is the 25th Percentile.Find the approximate median for the PASI scores before treatment.- The approximate median is 30 as this is where the line is plotted within the box, also known as the middle quartile.Find the approximate 75th percentile for the PASI scores before treatment.- The approximate 75th percentile is 44 as this is where the upper line of the box is plotted, also called the upper quartile.

b. Find the approximate 25th percentile for the PASI score after treatment.- The approximate 25th percentile is 4 as this is where the bottom line of the box is plotted - the Lower Quartile is the 25th Percentile.Find the approximate median for the PASI scores after treatment.- The approximate median is 6 as this is where the line is plotted within the box, also known as the middle quartile.Find the approximate 75th percentile for the PASI scores after treatment.- The approximate 75th percentile is 8 as this is where the upper line of the box is plotted, also called the upper quartile.

According to the text, a box plot is based on the interquartile range (IQR) or the distance that is between the upper and the lower quartiles or 25th and 27th percentiles.

c. Comment on the effectiveness of ichthyotherapy in treating psoriasis.- The ichthyotherapy is effective as the PASI scores as a whole after the treatment are much lower than prior to treatment.


Excellent Kelly


I'm awarding Kelly a top shelf prize for being the only one to have worked this problem thus far... Collect attached Pink Pig...

PinkPig.jpg


I have used a similar set of data for my analysis.

Find the approximate 25th percentile, Median and 75th percentile for the Psoriasis Area Severity Index (PASI) score before treatment – Base Line. A screen print of the data I was presented with is in the attached power point. For the baseline, it shows a minimum value of 2 and a maximum of 40. The Interquartile range (coloredarea of the box) appears to be from 16 to 30, with a median of 20.Based on that information, the approximate 25th percentile is 15 or 16 (it is kind of hard for me to tell). I chose 16 as the answer and it was considered correct – a closer inspectionwith the graph maximized probably shows 15 as the 25th percentile. I am sticking with 16 since the system told me 16 was correct.The line representing the median is at 20.The approximate 75th percentile is 30 as that is the top of the IQR.

Find the approximate 25th percentile, Median and 75th percentile for the PASI score after 3 weeks of treatment.

For the after treatment PASI score it shows a minimum value of 1 and a maximum of 21. The Interquartile range (colored area of the box) appears to be from 4 to 8 with a median ofjust under 6.Based on that information, the approximate 25th percentile is 4. The line representing the median is just under 6, but I went with 6 and the system was ok with that. The approximate75th percentile is 8 as that is the top of the IQR.Based on the PASI scores presented, the ichthyotherapy treatment is extremely effective. The IRQ shows that the majority of the patients had a much lower PASI after treatment. Theseverity index not only shows that the median is 20 before and 6 after, but it also shows the maximum indicator before treatment of 40 and after treatment of 21. While the extremevalues are not as statistically significant, for the patient rated at 40 prior to the treatment, even if they had the highest post treatment rating, they had an approximate 50%improvement.

week 1 q11.pptx

RE: Homework Question 11 Jolene Whitmore 5/9/2015 8:15:23 AM

This is where I had the most trouble figuring out why the treatment was effective, even though I guessed correctly. I understood how to get the correct numeric answers in each graph and where eachquartile laid on the Y-axis, but interpreting was not inherently evident to me. After reading your post it helped me to really put that question in perspective. The example from MyStatLab said the treatment iseffective because even the 75th percentile, after treatment, is less than the 25th percentile before. It still did not click for me, until you wrote in your post "While the extreme values are not asstatistically significant, for the patient rated at 40 prior to the treatment, even if they had the highest post treatment rating, they had an approximate 50% improvement."

20 of 23 20 of 23

Thanks for the great post.


Very good, thanks Dennis.


What helped me solved this type of question, a thru c is that I have I click the icon for the table. I had it open the whole time as I tried to answer each question, from 25%, median and to the 75% 9a to b), and for letter c, iuse the the same table from before treatment compared to the table after treatment. When the percentage with diseases shows lower, it goes to show the treatment is good or effective. I initially check the firstchoice, so I got it right.




The question contained a chart of data and four scatter graphs, one of which was correct. I copied the chart data to Minitab and created a scatter graph. My scatter graph matched option 2. This was the correct answer.

The second part to the question was 'Is there any trend to the data?'

The data contained two variables related to 18 sit in events, - the number of days of the sit in and the number of arrests for each event.

The scatter graph displayed the fact that the number of arrests did not show any significant trend when compared to the number of days of the sit in. Shorter sit in’s had some arrests, medium length sit ins had zero,and longer ones had a few. There seemed to be no obvious correlation between the length of the sit in and the number of arrests.


Dennis, I too used minitab to solve this problem. Simply copying the data using the default option didn't work for me, I had to use space delimiters to get the variables and data to import correctly. Once Ihad the data in Minitab graphing was easy. Select "Graph" then "Scatterplot" then "Simple". This will open a new window where you select which columns are the X and Y variables. After selecting yourvariables just click "OK" and Minitab generates the graph. You could also generate the graph manually. This wouldn't be too difficult with the limited data set given for this problem, but Minitab is so muchfaster.


Charles did a good job with a "quality response" to a fellow student here...


Charles and Dennis, like both of you I used the tool of minitab. It is really helpful, it comes handy on having the data infront of you .I click the space delimiters to get the import properly otherwise it will give me weird numbers, have you tried choosing other options?:) The, I did the next steps from graph to "simple" in a newwindow. Clicking "ok" as simple as that will give me the correct answer, for the graph. Then, I got the right answers on how the dots should look like and pick the right answer. Viewing the samplebefore answering this type of question, always helps me.


I like this.... This is one time a png or jpg of the graph might be helpful (HOWEVER YOU DON'T HAVE TO INCLUDE ONE)

No Posts Below Here Professor Heard 4/30/2015 10:15:21 AM

No posts other than additional information I might add should be below this point.

Examples of What I DO NOT WANT Professor Heard 5/4/2015 7:18:00 PM

I will post some examples of what I DO NOT Want as a response to this post.

RE: Examples of What I DO NOT WANT Professor Heard 5/4/2015 7:18:42 PM

I did number 4 it is attached.

Lazy Daisy

(No, you did not "do" number 4, you took a picture of the answers and did not explain anything... I do not want "pictures" of the answers.)

Number4.jpg


I did number 5, the answer was Mean is 15.64, Median is 15.00 and Mode is 14.00.

In A Hurry Murray

(Murray did not show us the data to verify his calculations, he did not explain what he did to get these (Minitab steps, or formulas, etc.), and did not give us a brief description of what each was

21 of 23 21 of 23

(mean is the average of the data, median is the center etc.)


An example of commenting on a student's post that is NOT what I want....

Daisy,

Nice Job, I got the same thing.

Getting By Bob



Murray,

You seem to know what you are doing. I am confused.

Will Wake Up in Week Seven and Wonder Why I'm in a Hole Harry



Hey Everybody,

What is Minitab and Where is it?

Week 6 and Didn't Bother Billy

RE: Paul Sandel Week one Paul Sandel 5/6/2015 7:35:19 PM


A: 90-100 28 0.07B: 80-89 96 0.24C: 65-79 156 0.39D: 50-64 64 0.16F: Below 50 56 0.14Total 400 1

These questions were simple for me. To figure out the relative frequency you take the frequency divided by 400. To find the frequency you add up all of the known numbers and subtract from the total.

RE: Paul Sandel Week one Professor Heard 5/6/2015 8:18:18 PM

Hey Paul, make sure you put this in the correct place...

Thanks,

Brent

RE: Paul Sandel Week one Paul Sandel 5/8/2015 6:11:11 PM

I did, my mistake.

RE: Paul Sandel Week one Professor Heard 5/9/2015 7:46:00 PM

No Biggie...

RE: HOMEWORK QUESTION 1 Mohamed Ketat 5/7/2015 12:01:47 AM

Modified:5/10/2015 3:57 PM

Grade on Business Statistics Exam Frequency Relative FrequencyA: 90-100 24 0.06B: 80-89 88 .22C: 65-79 172 .43D: 50-64 60 .15F: Below 50 56 .14Total 400 1

To figure out the relative frequency you take the frequency divided by 400. To find the frequency you add up all of the known numbers and subtract from the total.

22 of 23 22 of 23

RE: HOMEWORK QUESTION 1 Professor Heard 5/7/2015 8:09:49 AM

Please hit respond under the appropriate question...

Thanks,

Brent

StatCrunch Professor Heard 5/9/2015 7:47:49 PM

I get questions about StatCrunch all the time. I don't mind if you use it, but in the lectures I show you how to use Minitab (I have to). If you have extra time, I have a presentation where I answered a few undergraduatequestions with StatCrunch just for the fun of it. You might like it... See attached.

Playing with Statcrunch Part 1.pptx

23 of 23 23 of 23

Date post:	12-Jan-2016
Category:	Documents
Upload:	kristian-martin
View:	21 times
Download:	0 times

Week 1 Discussions

Documents