Question

Consider the following data sets.

1. Find the mean, median, and range for each of the two data sets.
2. Give the five-number summary and draw a boxplot for each of the two data sets.
3. Find the standard deviation for each of the two data sets.
4. Apply the range rule of thumb to estimate the standard deviation of each of the two data sets. How well does the rule work in each case? Briefly discuss why it does or does not work well.
5. Based on all your results, compare and discuss the two data sets in terms of their center and variation.

Researchers at the Pennsylvania State University conducted experiments with poplar trees in which one group of trees was given fertilizer and irrigation while the other group was given no treatment. The weights in kilograms of trees in the two groups are as follows:

No treatment:	0.15	0.02	0.16	0.37	0.22
Treatment:	2.03	0.27	0.92	1.07	2.38

Answer 1

a) For no treatment dataset

Mean = (0.15 + 0.02 + 0.16 + 0.37 + 0.22)/5 = 0.92/5 = 0.184

Median = 0.16

Range = 0.37 - 0.02 = 0.35

For treatment dataset

Mean = (2.03 + 0.27 + 0.92 + 1.07 + 2.38)/5 = 6.67/5 = 1.334

Median = 1.07

Range = 2.38 - 0.27 = 2.11

b) 5-number summary for No treatment dataset

Minimum Value = 0.02

First Quartile = 0.085

Median = 0.16

Third Quartile = 0.295

maximum Value = 0.37

The boxplot for the same is

5-number summary for treatment data

Minimum Value = 0.27

First Quartile = 0.595

Median = 1.07

Third Quartile = 2.205

maximum Value = 2.38

c) The standard deviation for the no-treatment dataset is

The standard deviation for the trearment dataset is

d) The range rule of thumb suggests that the range is generally 4 times the standard deviation.

Range for no-treatment dateset is = 0.35

Estimated standard deviation is = 0.35/4 = 0.0875

Actual standard deviation is = 0.1136

Ratio = 0.35/0.1136 = 3.081

Rule works fairly good in this case.

Range for treatment dateset is = 2.11

Estimated standard deviation is = 2.11/4 = 0.5275

Actual standard deviation is = 0.768

Ratio = 2.11/0.768 = 2.747

Rule doesn't hold well in this case.

The reason that the rule doesn't hold well in these cases is because of the small dataset size.

e) Mean and standard deviation for no treatment data is 0.184 and 0.1136

The mean and standard deviation for the treatment data is 1.334 and 0.768.

The mean weight for trees undergoing treatment is much more than that of trees under no treatment.

The variation of weight of trees undergoing treatment is also more than that of trees under no treatment.

This implies that the treatment is affective.

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