Question

For each data set, which aspect of the data is likely to cause evidence obtained from it to be inconsistent?

Answer 1

Answer:

1.outliers.

2.precision.

3.accuracy.

Explanation:

for future users.

Answer 2

Answer:

DATA SET 1. Outliers

DATA SET 2. Outliers

DATA SET 3: Accuracy

Explanation:

First let's talk about OUTLIERS, ACCURACY and PRECISION

Accuracy refers to the distance of data from the actual or standard value. In this case it is tagged as "correct" value.

Precision refers to the distance of data taken from each other. Or from one data point to the other. So if the data values are close to each other, you can say it is precise. But if they are far from the "correct" value, then you can say that they are precise but inaccurate.

As for outliers, you are looking for abnormal deviation from the set. This could be because of errors like typos, wrong reading or even errors in measurement.

As for case 1:

190, 315, 198, 194, 200, 195, 540, 201, 197, 193

315 and 540 are abnormally high values when you compare them to the others. This could indicate that these are outliers. You can statistically figure it out also by using interquartile range. But you can see that explanation at the end of this.

For case 2:

12, 18, 9, 13, 14, 19, 11, 20, 19, 12

The data points are not close to each other, but they are closer to the correct value individually, so you could say that this is accurate but not precise. Although 9 is low compared to the others, it is still within the accepted value.

For case 3:

50, 49, 48, 51, 52, 50, 50, 49, 51, 53

The data points are far from the correct value, but as you can see they are close to each other. This indicates that the data set is precise, but they are not accurate because they are not close to the correct value.

FURTHER EXPLANATION ON OUTLIERS:

First arrange your data from lowest to highest:

190, 193, 194, 195 ,197, 198, 200, 201 , 315, 540

Find the median (Just look for the value that is in the middle of the data set. Since this is an even number of data, you get the two middle values and divide it by two) :

197 + 198 = 395/2 = 197.5 this is the median of your data

Then get the first quartile and the third

Get the median of the first half:

190, 193, 194, 195, 197

And then the median of the second half:

198, 200, 201 , 315, 540 = 201

Then get the interquartile range (IQR)

IQR = Q3 - Q1 = 201-194 = 7

Then multiply 1.5 by IQR

1.5 x 7 = 10.5

Now add the 10.5 and Q3

201 + 10.5 = 211.5

Anything higher than that is an outlier.

So 315 and 540 are outliers.