Question

WILL FAN AND MEDAL!

The table below shows data from a survey about the amount of time students spend doing homework each week. The students were either in college or in high school:

Answer 1

Which of the choices below best describes how to measure the spread of this data?
(Hint: Use the minimum and maximum values to check for outliers.)

a. Both spreads are best described with the IQR.
b. Both spreads are best described with the standard deviation.
c. The college spread is best described by the IQR. The high school spread is best described by the standard deviation.
d. The college spread is best described by the standard deviation. The high school spread is best described by the IQR.

Answer 2

I think it's A? I know it's not B.

Answer 3

I think it may be D, actually. Cause I saw a couple of people say that they got A and C wrong. I put B before and got it wrong, so it must be D? I don't know.

Answer 4

it says use the min and max data to check for outliers

Answer 5

I've already graphed it. I'll attach the box plots.

Answer 6

Don't they both have an outlier?

Answer 7

if the mean > median generally an outlier exists

Answer 8

The median is more accurate than the mean

Answer 9

I'm just having a hard time understanding when to use standard deviation and when to use IQR. I know my textbook says that if the distribution of data is symmetrical, then the best measurement to use is standard deviation. And if the data is asymmetrical, the best measurement is IQR. I don't know if I'm making any sense here, so I'm sorry. I really suck at math.

Answer 10

beuase it takes things in order, so the mean is shifgted upwards or downwards depending on the data

Answer 11

standard deviation is a relative measure (i.e. relative to the mean)

Answer 12

I know, and IQR is relative to the median.

Answer 13

Sorry it took me so long to type, I was looking at my textbook.

Answer 14

IQR , think about it, "quartile" means "quarter" divide into 4 quarters

Answer 15

the lower quartile is the value one way into the data
the upper quartile is the value 3 quarters into the data
the upper quartile minus the lower quartile is the IQR

Answer 16

I keep thinking it's A because the data isn't evenly distributed, and it has outliers. But I remember somebody saying they got that wrong, so I'm just completely stuck.

Answer 17

think about the mean.,. now what is going to upset the mean, only outliers

Answer 18

outliers dont upset the median as much

Answer 19

Sorry, I didn't read what you typed above, it didn't show because my computer froze for a couple minutes. I'm gonna read it now

Answer 20

this is becuase the median will shift as the data extends

Answer 21

So the IQR for the high school would be 5 and the IQR for the college would be 7.5

Answer 22

Okay, so the outliers upset the mean, not the median. So it would be IQR, not standard deviation, right?

Answer 23

take simple examples: use 1, 2, 3, 4 5 what is the mean?

Answer 24

3

Answer 25

correct, now what is the median?

Answer 26

3

Answer 27

correct also, now chose 1, 2, 3, 4 & 13 what is the mean and what is the median

Answer 28

mean: 4.6

median: 3

Answer 29

see what happens to the mean when you have an outlier?

Answer 30

Yeah, like you said before, it messes up the mean.

Answer 31

now you are gettin it, cool!

Answer 32

Well thanks for helping so much :p

Answer 33

now you can look at your table and guess your answer

Answer 34

5-50 average should be about 25

Answer 35

however, the mean is 13.8 which means there must be an outlier somewhere

Answer 36

So it would be A? Because there's also an outlier in the high school plot.

Answer 37

Cause outliers would cause the shape of the data to be asymmetric, and IQR is the best measure of spread for asymmetrical data.

Answer 38

@BPDlkeme234

Answer 39

which shows how to best measure the spread of the data, well you cant use measure of central tendency (mean, median etc) to show spread, you have to use measures of spread (IQR or standard deviation etc)

Answer 40

Once again you need to be able to "see" the data, i.e relative to the mean, to be able to judge which one is best for judging this, this is the whole point of the exercise