Recent questions in Lower Quartile

High school statisticsAnswered question

Nylah Hendrix 2022-07-14

Sample-quartile

I don't know : Is there a sample such that the mean does not lie between the lower and upper quartile? Is there a sample such that the median does not lie between the lower and upper quartile?

I don't know : Is there a sample such that the mean does not lie between the lower and upper quartile? Is there a sample such that the median does not lie between the lower and upper quartile?

High school statisticsAnswered question

rmd1228887e 2022-07-09

Calculating Quartiles Dilemma

When I am calculating the lower quartile and upper quartile, why am I taking (n+1)/4 and 3(n+1)/4, instead of n/4 and 3n/4?

According to me, if total number of observations are n, then that should be counted figuring out the quartiles, isn't it?

When I am calculating the lower quartile and upper quartile, why am I taking (n+1)/4 and 3(n+1)/4, instead of n/4 and 3n/4?

According to me, if total number of observations are n, then that should be counted figuring out the quartiles, isn't it?

High school statisticsAnswered question

Dayanara Terry 2022-07-08

Getting the upper and lower quartiles in data with an even number of observations, or where the quartile lands on a decimal number

I want to draw a box plot, which requires that I know the median, the lower and upper quartiles, and the minimum and maximum values of my data.

I understand that the quartiles are simply the value on certainly "percentile" of the cumulative frequency of the data.

So lower quartile = the value of the observation on the 25th percentile of the data. Now my question (for AQA GCSE prep) is - what if taking 25% of my data ends up in a decimal number, let's say, $3.5$. And my data consists of classes in a grouped frequency table. And two of my classes are:

$class1$ || $2<=h<3.5$

$class2$ || $3.5<=h<5$

So when I take 25% of 3.5 falls in between two classes. Which value should I choose as the lower quartile? Should it be $class1$, or $class2$? Should my rounding of 3.5 be the same as regular rounding is done, i.e. just rounding up to 4 (hence selecting $class2$)? Or should I round choose $class1$ for some reason?

I want to draw a box plot, which requires that I know the median, the lower and upper quartiles, and the minimum and maximum values of my data.

I understand that the quartiles are simply the value on certainly "percentile" of the cumulative frequency of the data.

So lower quartile = the value of the observation on the 25th percentile of the data. Now my question (for AQA GCSE prep) is - what if taking 25% of my data ends up in a decimal number, let's say, $3.5$. And my data consists of classes in a grouped frequency table. And two of my classes are:

$class1$ || $2<=h<3.5$

$class2$ || $3.5<=h<5$

So when I take 25% of 3.5 falls in between two classes. Which value should I choose as the lower quartile? Should it be $class1$, or $class2$? Should my rounding of 3.5 be the same as regular rounding is done, i.e. just rounding up to 4 (hence selecting $class2$)? Or should I round choose $class1$ for some reason?

High school statisticsAnswered question

ttyme411gl 2022-07-07

Boxplot: whiskers and outliers doubt

I have a doubt on boxplot.

I'll expose my knowledge and then my doubt.

$x=\{{x}_{1},{x}_{2}...{x}_{n}\}$: the set of samples

${q}_{1}$,${q}_{3}$: the first and third quartiles

${w}_{l}$,${w}_{u}$: the lower and upper whiskers

$IQR={q}_{3}-{q}_{1}$

box extends from ${q}_{1}$ to ${q}_{3}$

${w}_{l}=max(min(x),{q}_{1}-1.5\cdot IQR)$

${w}_{u}=min(max(x),{q}_{3}+1.5\cdot IQR)$

$outliers=\{{x}_{i}\in x\phantom{\rule{thickmathspace}{0ex}}|\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}{x}_{i}<{w}_{l}\vee {x}_{i}>{w}_{u}\}$

Observations:

$\text{whiskers' distance from box are not symmetric}\phantom{\rule{0ex}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}({w}_{l}=min(x)\vee {w}_{u}=max(x))$

${w}_{u}-{q}_{3}<{q}_{1}-{w}_{l}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\nexists {x}_{i}:{x}_{i}\in outliers\wedge {x}_{i}>{w}_{u}$

${w}_{u}-{q}_{3}>{q}_{1}-{w}_{l}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\nexists {x}_{i}:{x}_{i}\in outliers\wedge {x}_{i}<{w}_{l}$

My doubt: if all what I exposed is correct, how do you explain the presence of outliers in this speed of light boxplot (third experiment, lower outliers) and in this plot (see wednesday, lower outliers)?

In the case my reasoning is wrong, please provide a simple numeric counterexample.

I have a doubt on boxplot.

I'll expose my knowledge and then my doubt.

$x=\{{x}_{1},{x}_{2}...{x}_{n}\}$: the set of samples

${q}_{1}$,${q}_{3}$: the first and third quartiles

${w}_{l}$,${w}_{u}$: the lower and upper whiskers

$IQR={q}_{3}-{q}_{1}$

box extends from ${q}_{1}$ to ${q}_{3}$

${w}_{l}=max(min(x),{q}_{1}-1.5\cdot IQR)$

${w}_{u}=min(max(x),{q}_{3}+1.5\cdot IQR)$

$outliers=\{{x}_{i}\in x\phantom{\rule{thickmathspace}{0ex}}|\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}{x}_{i}<{w}_{l}\vee {x}_{i}>{w}_{u}\}$

Observations:

$\text{whiskers' distance from box are not symmetric}\phantom{\rule{0ex}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}({w}_{l}=min(x)\vee {w}_{u}=max(x))$

${w}_{u}-{q}_{3}<{q}_{1}-{w}_{l}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\nexists {x}_{i}:{x}_{i}\in outliers\wedge {x}_{i}>{w}_{u}$

${w}_{u}-{q}_{3}>{q}_{1}-{w}_{l}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\nexists {x}_{i}:{x}_{i}\in outliers\wedge {x}_{i}<{w}_{l}$

My doubt: if all what I exposed is correct, how do you explain the presence of outliers in this speed of light boxplot (third experiment, lower outliers) and in this plot (see wednesday, lower outliers)?

In the case my reasoning is wrong, please provide a simple numeric counterexample.

High school statisticsAnswered question

Jackson Duncan 2022-06-26

How to count $n$th percentile from normally distributed random variable?

I have normally distributed random variable $X\sim \mathcal{N}(100,225)$. How to count $n$th percentile?

In my case I need lower quartile - $x(0.25)$

I have normally distributed random variable $X\sim \mathcal{N}(100,225)$. How to count $n$th percentile?

In my case I need lower quartile - $x(0.25)$

High school statisticsAnswered question

Lucille Cummings 2022-06-26

Is the lower quartile here $18.5$ or $27$?

Take the following list of data:

$4\text{}10\text{}27\text{}27\text{}29\text{}34\text{}34\text{}34\text{}37$

If I remove the median which is $29$ then the left hand side is:

$4\text{}10\text{}27\text{}27$

The median of this side is $\frac{10+27}{2}=18.5$

But if I use the percentile formula:

${P}_{25}=\frac{25}{100}9=2.25$

$2.25$ is not a whole number so we take the next whole number which is $3$

The ${P}_{25}$ is the third value which is $27$

Note: Wikipedia defines the lower quartile as the middle number between the lowest number and the median. In this case it would be $27$

Which of these is the lower quartile, $18.5$, $27$, or somethings else?

Take the following list of data:

$4\text{}10\text{}27\text{}27\text{}29\text{}34\text{}34\text{}34\text{}37$

If I remove the median which is $29$ then the left hand side is:

$4\text{}10\text{}27\text{}27$

The median of this side is $\frac{10+27}{2}=18.5$

But if I use the percentile formula:

${P}_{25}=\frac{25}{100}9=2.25$

$2.25$ is not a whole number so we take the next whole number which is $3$

The ${P}_{25}$ is the third value which is $27$

Note: Wikipedia defines the lower quartile as the middle number between the lowest number and the median. In this case it would be $27$

Which of these is the lower quartile, $18.5$, $27$, or somethings else?

High school statisticsAnswered question

pachaquis3s 2022-06-25

How do I calculate the quartiles for this problem?

I have the following list of numbers, and I'm trying to calculate the quartiles:

2, 4, 4, 5, 7, 7, 7, 7, 7, 7, 8, 8, 9, 9, 9, 9

I'm running into a bit of confusion because the median turns out to be 7. Now I don't know how to demarcate the lower group from the upper group since there are a whole bunch of 7's. Consequently, I don't know how to calculate ${Q}_{1}$ and ${Q}_{3}$

I have the following list of numbers, and I'm trying to calculate the quartiles:

2, 4, 4, 5, 7, 7, 7, 7, 7, 7, 8, 8, 9, 9, 9, 9

I'm running into a bit of confusion because the median turns out to be 7. Now I don't know how to demarcate the lower group from the upper group since there are a whole bunch of 7's. Consequently, I don't know how to calculate ${Q}_{1}$ and ${Q}_{3}$

High school statisticsAnswered question

Emmy Knox 2022-06-21

How to calculate lower & upper quartiles?

I'm sure this has been asked many times before but it's confusing me a lot so hopefully someone can help!

I am given this data set:

$\begin{array}{rrrrrrr}& 0.28& 0.30& 0.42& 0.59& 0.71& 1.67\\ & 1.82& 2.39& 4.71& 4.79& 4.89& 5.00\\ & 5.00& 5.76& 6.09& 6.17& 6.81& 7.12\\ & 7.44& 8.05& 8.59& 8.86& 9.04& 9.78\end{array}$

There are $24$ numbers here. I need to find the lower & upper quartile. I figured that the median is $5.00$

And looking at the answer sheet the Lower Quartile is $1.745$ and the Upper quartile is $7.28$

My question is how do i get to these answers!?

Thank you in advance.

I'm sure this has been asked many times before but it's confusing me a lot so hopefully someone can help!

I am given this data set:

$\begin{array}{rrrrrrr}& 0.28& 0.30& 0.42& 0.59& 0.71& 1.67\\ & 1.82& 2.39& 4.71& 4.79& 4.89& 5.00\\ & 5.00& 5.76& 6.09& 6.17& 6.81& 7.12\\ & 7.44& 8.05& 8.59& 8.86& 9.04& 9.78\end{array}$

There are $24$ numbers here. I need to find the lower & upper quartile. I figured that the median is $5.00$

And looking at the answer sheet the Lower Quartile is $1.745$ and the Upper quartile is $7.28$

My question is how do i get to these answers!?

Thank you in advance.

High school statisticsAnswered question

Santino Bautista 2022-06-19

Identifying the k points in 2D geographic space which are 'most distant' from each other

I have a set of DNA samples from Y plants in a given geographic area. I'm going to be doing DNA sequencing on individuals in this population (and a number of other, separate populations), however due to financial constraints I'm unable to perform sequencing on all Y individuals. I've decided that I can afford to sequence k (out of Y) in each separate area.

I'd like to select the k samples which are farthest apart/most geographically distributed within the Y samples I collected. Samples that are taken from plants in close proximity to each other are more likely to be closely related, and I'd like to sequence what are ultimately the most genetically diverse samples for my later analysis.

So, the question is: given a set of Y points/samples in 2d geographic space (lat/long coordinates), how do I select the k points that are most geographically distributed/distant from each other?

As I've explored this a bit, I think the problem I'm really having is how to define 'distance' or 'most distributed'. Some of the metrics I've though of (e.g. maximum average distance between the k points) result in really unintuitive point selection in certain cases (for example, if two points are right next to each other but far away from another cluster of all the other points, the two points will be included even if they're almost on top of each other).

I have a feeling that this is not an uncommon problem, and there must be good answers out there. I'll add that I have access to a large cluster and I can brute force the problem to some extent.

I'd appreciate any and all advice or discussion; this is an interesting theoretical topic to me.

I have a set of DNA samples from Y plants in a given geographic area. I'm going to be doing DNA sequencing on individuals in this population (and a number of other, separate populations), however due to financial constraints I'm unable to perform sequencing on all Y individuals. I've decided that I can afford to sequence k (out of Y) in each separate area.

I'd like to select the k samples which are farthest apart/most geographically distributed within the Y samples I collected. Samples that are taken from plants in close proximity to each other are more likely to be closely related, and I'd like to sequence what are ultimately the most genetically diverse samples for my later analysis.

So, the question is: given a set of Y points/samples in 2d geographic space (lat/long coordinates), how do I select the k points that are most geographically distributed/distant from each other?

As I've explored this a bit, I think the problem I'm really having is how to define 'distance' or 'most distributed'. Some of the metrics I've though of (e.g. maximum average distance between the k points) result in really unintuitive point selection in certain cases (for example, if two points are right next to each other but far away from another cluster of all the other points, the two points will be included even if they're almost on top of each other).

I have a feeling that this is not an uncommon problem, and there must be good answers out there. I'll add that I have access to a large cluster and I can brute force the problem to some extent.

I'd appreciate any and all advice or discussion; this is an interesting theoretical topic to me.

High school statisticsAnswered question

Roland Manning 2022-06-19

Ambiguity between methods of calculating quartiles. Which method is more correct?

I am having confusion between two methods of calculating the first quartiles.

Let me explain by an example:

$A=[1,2,3,4,5]$

The method that I know:

To find the first quartile in the list above I first find the median = $3$ that is the 3rd element.

Now I split the list in the following fashion:

$(1,2)$

$3$

$(4,5)$

Now we take the list $(1,2)$ and we find the media between them that is $(1+2)/2=1.5$

So according to my calculation the first quartile ${Q}_{1}=1.5$

The nearest rank method

$n=\lceil \frac{P}{100}\times N\rceil $

So for the above list the $0.25$ percentile or the first quartile ${Q}_{1}$ will be

$\lceil \frac{25}{100}\times 5\rceil =2$

which is 2nd position.

So which is correct $1.5$ or $2$ ? Or does chosing any of them is fine ?

If I am correct in my original calculation why am I getting a difference between the 2 methods.

I am having confusion between two methods of calculating the first quartiles.

Let me explain by an example:

$A=[1,2,3,4,5]$

The method that I know:

To find the first quartile in the list above I first find the median = $3$ that is the 3rd element.

Now I split the list in the following fashion:

$(1,2)$

$3$

$(4,5)$

Now we take the list $(1,2)$ and we find the media between them that is $(1+2)/2=1.5$

So according to my calculation the first quartile ${Q}_{1}=1.5$

The nearest rank method

$n=\lceil \frac{P}{100}\times N\rceil $

So for the above list the $0.25$ percentile or the first quartile ${Q}_{1}$ will be

$\lceil \frac{25}{100}\times 5\rceil =2$

which is 2nd position.

So which is correct $1.5$ or $2$ ? Or does chosing any of them is fine ?

If I am correct in my original calculation why am I getting a difference between the 2 methods.

High school statisticsAnswered question

Dwllane4 2022-06-17

Hypergeometric Distribution

I hate to be that guy who comes here and asks one question, however I've been stuck on this problem for quite some time, and I'm going to assume that I shouldn't even be using the binomial theorem at this point

I was given this question:

"A teacher was asked by her principal to select 7 students at random from her class to take a standardized math test, which will be used to determine how well students at that school are doing with respect to math (and correspondingly, how well the teacher is doing at teaching math). The teacher previously had rank ordered her students on the basis of their performance in her class on math tests, and divided the class into quartiles, such that there were 5 students in the upper quartile and 15 students in the lower three quartiles. When the teacher handed the principal the names of the students she had randomly selected from the class, the number of students from the upper quartile was 5, and the number of students from the lower three quartiles was 2. Is there any statistical evidence that the teacher is biased toward selecting students from the upper quartile?"

The first question given was:

"Compute the probability that, of the students selected, 0 are upper quartile students. Round off to 4 decimal places."

I was able to answer that question with the answer 0.0830, this i was able to answer by doing (15/20)(14/19)(13/18)(12/17)(11/16)(10/15)(9/14), however when it asks a question such as computing the probability that of the students selected, 2 are upper quartile students, I get stuck, I tried using the binomial theorem, but to no avail, perhaps I'm doing something wrong, I can't find a fast enough way to determine all of the sequences that will produce two upper quartile students of the 7 draws.

I hate to be that guy who comes here and asks one question, however I've been stuck on this problem for quite some time, and I'm going to assume that I shouldn't even be using the binomial theorem at this point

I was given this question:

"A teacher was asked by her principal to select 7 students at random from her class to take a standardized math test, which will be used to determine how well students at that school are doing with respect to math (and correspondingly, how well the teacher is doing at teaching math). The teacher previously had rank ordered her students on the basis of their performance in her class on math tests, and divided the class into quartiles, such that there were 5 students in the upper quartile and 15 students in the lower three quartiles. When the teacher handed the principal the names of the students she had randomly selected from the class, the number of students from the upper quartile was 5, and the number of students from the lower three quartiles was 2. Is there any statistical evidence that the teacher is biased toward selecting students from the upper quartile?"

The first question given was:

"Compute the probability that, of the students selected, 0 are upper quartile students. Round off to 4 decimal places."

I was able to answer that question with the answer 0.0830, this i was able to answer by doing (15/20)(14/19)(13/18)(12/17)(11/16)(10/15)(9/14), however when it asks a question such as computing the probability that of the students selected, 2 are upper quartile students, I get stuck, I tried using the binomial theorem, but to no avail, perhaps I'm doing something wrong, I can't find a fast enough way to determine all of the sequences that will produce two upper quartile students of the 7 draws.

High school statisticsAnswered question

Sattelhofsk 2022-06-14

Suppose we have data set: $19,21,22,22,28,31,33,44,50$. Find the interquartile range of this set.

First solution: Firstly, we should find the $75$th percentile of this set. $0,75\cdot 9=6,75$ and rounding up this number to nearest whole we get $7$. So the $75$th percentile of this set is $33$. Secondly, should find the $25$th percentile of this set. $0,25\cdot 9=2,25$ and rounding up this number to nearest whole we get $3$. So the $25$th percentile of this set is $22$. Hence,

$\text{interquartile range}={Q}_{3}-{Q}_{1}=33-22=11.$

Second solution: The median of this set is $28$. First quartile is the median of lower set. Hence ${Q}_{1}={\displaystyle \frac{21+22}{2}}=21.5$, third quartile is the median of upper set. Hence ${Q}_{3}={\displaystyle \frac{33+44}{2}}=38.5$

$\text{interquartile range}={Q}_{3}-{Q}_{1}=38.5-21.5=17.$

Which one is correct? Please explain why one of the solutions is false.

First solution: Firstly, we should find the $75$th percentile of this set. $0,75\cdot 9=6,75$ and rounding up this number to nearest whole we get $7$. So the $75$th percentile of this set is $33$. Secondly, should find the $25$th percentile of this set. $0,25\cdot 9=2,25$ and rounding up this number to nearest whole we get $3$. So the $25$th percentile of this set is $22$. Hence,

$\text{interquartile range}={Q}_{3}-{Q}_{1}=33-22=11.$

Second solution: The median of this set is $28$. First quartile is the median of lower set. Hence ${Q}_{1}={\displaystyle \frac{21+22}{2}}=21.5$, third quartile is the median of upper set. Hence ${Q}_{3}={\displaystyle \frac{33+44}{2}}=38.5$

$\text{interquartile range}={Q}_{3}-{Q}_{1}=38.5-21.5=17.$

Which one is correct? Please explain why one of the solutions is false.

High school statisticsAnswered question

Izabella Ponce 2022-06-10

Standard Deviation after removing outlier.

I am totally new to statistics. I'm learning the basics.

I came upon this question while solving Erwin Kreyszig's exercise on statistics. The problem is simple. It asks to calculate standard deviation after removing outliers from the dataset.

The dataset is as follows: 1, 2, 3, 4, 10. What I did is, I found out qm = 3. Then $q$l $=\frac{1+2}{2}=1.5$ and $q$m $=\frac{4+10}{2}=7$

Now, $IQR=7-1.5=5.5$ and $1.5\ast IQR=8.25$

So, we can say numbers beyond $1.5-5.5=-4$ and $7+5.5=12.5$ will be an outlier.

Since there is no outlier, I found out the Standard Deviation of the set which is 3.53.

But, the answer provided is 1.29 which is different from the standard deviation of the set.

Can anyone help me what I missed?

Also, I have another question - we can see with plain eyes 10 is an outlier. But it is not detected here - why?

I am totally new to statistics. I'm learning the basics.

I came upon this question while solving Erwin Kreyszig's exercise on statistics. The problem is simple. It asks to calculate standard deviation after removing outliers from the dataset.

The dataset is as follows: 1, 2, 3, 4, 10. What I did is, I found out qm = 3. Then $q$l $=\frac{1+2}{2}=1.5$ and $q$m $=\frac{4+10}{2}=7$

Now, $IQR=7-1.5=5.5$ and $1.5\ast IQR=8.25$

So, we can say numbers beyond $1.5-5.5=-4$ and $7+5.5=12.5$ will be an outlier.

Since there is no outlier, I found out the Standard Deviation of the set which is 3.53.

But, the answer provided is 1.29 which is different from the standard deviation of the set.

Can anyone help me what I missed?

Also, I have another question - we can see with plain eyes 10 is an outlier. But it is not detected here - why?

High school statisticsAnswered question

Petrovcic2x 2022-06-06

Quartiles and appending elements to an end of a stack

"The company at which Mark is employed has 80 employees, each of whom has a different salary. Mark’s salary of $43,700 is the second-highest salary in the ﬁrst quartile of the 80 salaries. If the company were to hire 8 new employees at salaries that are less than the lowest of the 80 salaries, what would Mark’s salary be with respect to the quartiles of the 88 salaries at the company, assuming no other changes in the salaries?"

The answer is

"The fifth-lowest salary in the second quartile."

This doesn't make sense.

Imagine the employees ranked in order from highest salary to lowest salary. If we add 8 new employees who have salaries lower than the pre-existing 80, then we append 8 salaries to the bottom of the list. How does Mark's position in the queue change? Shouldn't his position stay the same?

"The company at which Mark is employed has 80 employees, each of whom has a different salary. Mark’s salary of $43,700 is the second-highest salary in the ﬁrst quartile of the 80 salaries. If the company were to hire 8 new employees at salaries that are less than the lowest of the 80 salaries, what would Mark’s salary be with respect to the quartiles of the 88 salaries at the company, assuming no other changes in the salaries?"

The answer is

"The fifth-lowest salary in the second quartile."

This doesn't make sense.

Imagine the employees ranked in order from highest salary to lowest salary. If we add 8 new employees who have salaries lower than the pre-existing 80, then we append 8 salaries to the bottom of the list. How does Mark's position in the queue change? Shouldn't his position stay the same?

High school statisticsAnswered question

patzeriap0 2022-05-29

Where should I use median instead of average?

Is there a general law, or rule of thumb, or rationale, when to use median and when average?

Although I know the difference and how they are computed, when I try to translate to simple English I would say in both cases that they both are a value that justly and fairly represents a big group of values of a certain category.

Examples:

Grades across different subjects for a single student.

Grades in one subject across students in one class.

Time to close a ticket per worker of a support desk.

Jail time given by a judge for a certain crime.

Lap time in a 10 laps run for a certain runner.

Monthly income per household in a given neighborhood.

So what should I use in each case above? And what is the general rule. And as a side question, are there other types of aggregate functions other than median and average that relate to this?

Is there a general law, or rule of thumb, or rationale, when to use median and when average?

Although I know the difference and how they are computed, when I try to translate to simple English I would say in both cases that they both are a value that justly and fairly represents a big group of values of a certain category.

Examples:

Grades across different subjects for a single student.

Grades in one subject across students in one class.

Time to close a ticket per worker of a support desk.

Jail time given by a judge for a certain crime.

Lap time in a 10 laps run for a certain runner.

Monthly income per household in a given neighborhood.

So what should I use in each case above? And what is the general rule. And as a side question, are there other types of aggregate functions other than median and average that relate to this?

High school statisticsAnswered question

Liberty Mack 2022-05-26

Hypothesis Testing help

Really have no idea where to start :(

In an experiment comparing two weight-loss regimes A and B 20 test subjects were matched into 10 pairs so that within each pair the subjects were as similar as possible. Then A was randomly allocated to one of the subjects in each pair, and then B allocated to the other. The number of kilograms lost for each person is obtained and then the ordered A−B differences for each pair (in kg) are given below in the object d, together with some summary statistics:

sort(d) [1] -0.9 -0.3 0.2 0.4 0.6 1.2 1.4 3.3 3.5 4.3

mean(d) [1] 1.37

sd(d) [1] 1.755025

By specifying and checking (with a boxplot) an appropriate normality assumption perform a formal hypothesis test of H0:“regimes the same” against H1:“regimes not the same”.

Any help is appreciated!

Really have no idea where to start :(

In an experiment comparing two weight-loss regimes A and B 20 test subjects were matched into 10 pairs so that within each pair the subjects were as similar as possible. Then A was randomly allocated to one of the subjects in each pair, and then B allocated to the other. The number of kilograms lost for each person is obtained and then the ordered A−B differences for each pair (in kg) are given below in the object d, together with some summary statistics:

sort(d) [1] -0.9 -0.3 0.2 0.4 0.6 1.2 1.4 3.3 3.5 4.3

mean(d) [1] 1.37

sd(d) [1] 1.755025

By specifying and checking (with a boxplot) an appropriate normality assumption perform a formal hypothesis test of H0:“regimes the same” against H1:“regimes not the same”.

Any help is appreciated!

High school statisticsAnswered question

Jaime Coleman 2022-05-09

What is the lower quartile of the set of data?

I came across this problem asking the lower quartile of the ungrouped data. My answer is 3, but other references say it should be 2.5. Here's the data:

1, 1, 2, 2, 3, 3, 4, 4, 6, 7, 8, 10, 11, 14, 15, 20, 22

What do you think?

I came across this problem asking the lower quartile of the ungrouped data. My answer is 3, but other references say it should be 2.5. Here's the data:

1, 1, 2, 2, 3, 3, 4, 4, 6, 7, 8, 10, 11, 14, 15, 20, 22

What do you think?

High school statisticsAnswered question

dresu9dnjn 2022-05-09

Calculating the position of the median

I know this is a simple question, but I cannot find a straight answer anywhere.

When calculating the medium of listed data, the formula is (n+1)/2. My Statistics teacher said for grouped data the position of the medium is n/2. However, this seems contradictory as for discrete grouped data, the data could be written in a list if the original values were known. Therefore two different values for the median are found.

I know this question is probably going to get flagged as it has already been asked, however, it has never been answered. I find it frustrating that such a fundamental concept in Statistics, what is supposed to be precise and never subjective has a wishy-washy answer.

Rant over, I think the position should be (n+1)/2 for discrete data and n/2 for continuous data. Moreover, this then leads into the obvious question of how we should calculate quartiles as well.

It would be nice if everyone could agree on a certain method for calculating the medium as all my textbooks are saying different things.

Someone please dispell the confusion and this statistical mess.

I know this is a simple question, but I cannot find a straight answer anywhere.

When calculating the medium of listed data, the formula is (n+1)/2. My Statistics teacher said for grouped data the position of the medium is n/2. However, this seems contradictory as for discrete grouped data, the data could be written in a list if the original values were known. Therefore two different values for the median are found.

I know this question is probably going to get flagged as it has already been asked, however, it has never been answered. I find it frustrating that such a fundamental concept in Statistics, what is supposed to be precise and never subjective has a wishy-washy answer.

Rant over, I think the position should be (n+1)/2 for discrete data and n/2 for continuous data. Moreover, this then leads into the obvious question of how we should calculate quartiles as well.

It would be nice if everyone could agree on a certain method for calculating the medium as all my textbooks are saying different things.

Someone please dispell the confusion and this statistical mess.

High school statisticsAnswered question

spazzter08dyk2n 2022-05-08

calculating upper and lower quartiles

The number of goals scored by a football team during a months worth of matches is recorderd.

They scored $0$ goals $4$ times, $1$ goal $6$ six times and $2$ goals $3$ times. Calculate the lower and upper quartiles.

Would it simply be this?: lower quartile is the $3.5$th value which is zero, and upper quartile is the $10.5$th value which is $2$?

The number of goals scored by a football team during a months worth of matches is recorderd.

They scored $0$ goals $4$ times, $1$ goal $6$ six times and $2$ goals $3$ times. Calculate the lower and upper quartiles.

Would it simply be this?: lower quartile is the $3.5$th value which is zero, and upper quartile is the $10.5$th value which is $2$?

High school statisticsAnswered question

iyiswad9k 2022-05-02

Probability and Statistics (Quartiles)

Q: Find the upper and lower quartiles of the random variable.

$f(x)=\frac{1}{x\mathrm{ln}(1.5)}$

for $4\le x\le 6$

I set $F(x)=0.25$ and $F(x)=0.75$ to denote $Q1$ and $Q3$

This is what I got for the answers:

For

$F(X)=0.25,x=9.86$

For

$F(X)=0.75,x=3.28$

However, these answers are wrong as compared with my textbook. It should be:

For

$F(X)=0.25,x=4.43$

For

$F(X)=0.75,x=5.42$

Please help me. I don't know what I'm doing wrong.

Q: Find the upper and lower quartiles of the random variable.

$f(x)=\frac{1}{x\mathrm{ln}(1.5)}$

for $4\le x\le 6$

I set $F(x)=0.25$ and $F(x)=0.75$ to denote $Q1$ and $Q3$

This is what I got for the answers:

For

$F(X)=0.25,x=9.86$

For

$F(X)=0.75,x=3.28$

However, these answers are wrong as compared with my textbook. It should be:

For

$F(X)=0.25,x=4.43$

For

$F(X)=0.75,x=5.42$

Please help me. I don't know what I'm doing wrong.

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When you start with the calculations and offer lower quartile word problems based on statistical data for your social sciences course or a school project under the supervision of your teacher, the lower quartile will help you to find the value that is under 25% when you are working with the information that is arranged in increasing order. You’ll be able to find the range of your spread and the leaning of the data towards one side. You can take a look at the lower quartile math definition and examples that can be found among the questions that we offer.