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Title: M&M statistics Day and Date: Sept 5 2019, Thursday Course: Chem 223 Name: Nikhila Sampath Partner: Harrison Park Section: AB 5 TA: Sanja Pudar

Sampath 1

Abstract The purpose of this lab was to practice performing statistical analysis on data sets in order to determine if they are similar to each other. This was accomplished by using the F-Test to compare standard deviations and the T-Test to compare means. This was done after ensuring that the data sets had no outliers using Grubbs test. It was determined that though the standard deviation of the red M&Ms was different from that of the blue, green and brown M&Ms there was no significant difference between the mean mass of each color. It was concluded that each color M&M weighed the same.

Introduction The purpose of this lab was to practice performing statistical analysis on data sets in order to be able to compare them. This was done using three tests- namely Grubbs test, F-Test and T-Test. Grubbs test, also known as maximum normed residual test, is used to detect a single outlier in a data set. Gcalculated is determined based on the below mentioned formula and compared to Gtable which is determined based on the number of data points.  Gcalculated =

| Questionable value − X | S

(1)

If Gcalculated > Gtable , the data point can be rejected. However if the data is suspected to have more than one outlier, it is better to use the Tietjen-Moore test or the generalized extreme studentized deviate test1 . The F-Test is used to determine if there is a significant difference in the standard deviation of two sets of data. Fcalculated is determined based on the formula

Sampath 2

Fcalculated =

S12 S22

(2)

where S1 > S2. It is then compared to Ftable which is determined based on the degrees of freedom  which is given by DOF = n -1

(3)

If Fcalculated > Ftable , the standard deviations are significantly different. This test can be a two-tailed test or a one-tailed test. The two-tailed version tests against the alternative that the variances are not equal. The one-tailed version only tests in one direction, that is the variance from the first population is either greater than or less than (but not both) the second population variance2 . The T-Test is used to determine if the mean of two sets of data are significantly different. It is mostly used when the data sets would follow a normal distribution and may have unknown variances. A t-test is used as a hypothesis testing tool, which allows testing of an assumption applicable to a population3 . If the standard deviations are not significantly different, Tcalculated is given by T calculated =

| X 1 −X 2 |

√

S 12 (n1 −1) + S 22 (n2 −1) n1 + n2 −2

(4)

It is then compared to Ttable which depends on the degrees of freedom which is given by DOF = n1 + n2 - 2

(5)

If the standard deviations are significantly different, Tcalculated is given by T calculated =

| X 1 −X 2 |

√

S1 2 + S2 2 n1 + n2

It is then compared to Ttable which depends on the degrees of freedom which is given by

(6)

Sampath 3

DOF =

2

2

( S1 + S2 )2 n1 n2

2 ( S1 ) 2 n1 n1 −1

+

2 ( S2 ) 2 n2 n2 −1

(7)

If Tcalculated > Ttable , the means are significantly different. Based on lab 1, it was predicted that the mass of each bag of M&Ms as well as the mass of each color varied.

Procedure A bag of M&Ms was obtained. The analytical balance was tared without anything on it. Once the reading steadied at zero, the bag was placed on the balance and mass was recorded. Then the bag was removed from the balance and a weigh boat was placed instead. The balance was then tared again with the weigh boat on it. Once the reading steadied at zero, the bag of M&Ms was opened and its contents emptied onto the weigh boat while the wrapper was preserved. The mass of the M&Ms without the wrapper was recorded and the weigh boat was taken off the balance. The balance was tared again without anything on it. Once the reading steadied at zero, the wrapper was placed on the balance and its mass was recorded. After this, the wrapper was taken off the balance and gloves were worn. The M&Ms were then sorted into different weigh boats based on their color. The total number of M&Ms in each color was counted and this number was recorded. Five random M&Ms of each color were chosen. The balance was tared without anything on it and once the reading steadied at zero, the mass of each of the five was measured individually. After that, the mass of all five of them together was measured. This process was repeated for each color. After this, the M&Ms and weigh boats were discarded.

Sampath 4

Data and observations TABLE 1- Data for Yellow M&Ms Total count

10

Mass of #1

0.7933 g

Mass of #2

0.8867 g

Mass of #3

0.8450 g

Mass of #4

0.8432 g

Mass of #5

0.8923 g

Together

4.2607 g

Sum

4.2605 g

Difference

0.0002 g

Mean

0.8521 g

Std Dev

0.04000

Gtable

1.672

Gcalculated for mass #1

1.471

Gcalculated for mass #2

0.8650

Gcalculated for mass #3

0.1775

Gcalculated for mass #4

0.2225

Gcalculated for mass #5

1.005

Sampath 5

TABLE 2- Data for Blue M&Ms Total count

9

Mass of #1

0.8445 g

Mass of #2

0.8943 g

Mass of #3

0.8340 g

Mass of #4

0.8633 g

Mass of #5

0.8214 g

Together

4.2577 g

Sum

4.2575 g

Difference

0.0002 g

Mean

0.8515 g

Std Dev

0.02843

Gtable

1.672

Gcalculated for mass #1

0.2462

Gcalculated for mass #2

1.506

Gcalculated for mass #3

0.6156

Gcalculated for mass #4

0.4151

Gcalculated for mass #5

1.059

Sampath 6

TABLE 3- Data for Red M&Ms Total count

5

Mass of #1

0.8445 g

Mass of #2

0.8461 g

Mass of #3

0.7892 g

Mass of #4

0.8968 g

Mass of #5

1.0096 g

Together

4.3862 g

Sum

4.3862 g

Difference

0g

Mean

0.8772 g

Std Dev

0.08321

Gtable

1.672

Gcalculated for mass #1

0.3935

Gcalculated for mass #2

0.3742

Gcalculated for mass #3

1.058

Gcalculated for mass #4

0.2351

Gcalculated for mass #5

1.591

Sampath 7

TABLE 4- Data for Orange M&Ms Total count

14

Mass of #1

0.8387 g

Mass of #2

0.9154 g

Mass of #3

0.7987 g

Mass of #4

0.8404 g

Mass of #5

0.8275 g

Together

4.2201 g

Sum

4.2207 g

Difference

-0.0006 g

Mean

0.8441 g

Std Dev

0.04320

Gtable

1.672

Gcalculated for mass #1

0.1259

Gcalculated for mass #2

1.650

Gcalculated for mass #3

1.052

Gcalculated for mass #4

0.08658

Gcalculated for mass #5

0.3852

Sampath 8

TABLE 5- Data for Green M&Ms Total count

6

Mass of #1

0.8316 g

Mass of #2

0.8622 g

Mass of #3

0.8534 g

Mass of #4

0.9050 g

Mass of #5

0.8312 g

Together

4.2838 g

Sum

4.2834 g

Difference

0.0004 g

Mean

0.8567 g

Std Dev

0.03022

Gtable

1.672

Gcalculated for mass #1

0.8298

Gcalculated for mass #2

0.1826

Gcalculated for mass #3

0.1085

Gcalculated for mass #4

1.599

Gcalculated for mass #5

0.8430

Sampath 9

TABLE 6- Data for Brown M&Ms Total count

11

Mass of #1

0.8110 g

Mass of #2

0.8701 g

Mass of #3

0.8290 g

Mass of #4

0.8599 g

Mass of #5

0.8849 g

Together

4.2551 g

Sum

4.2549 g

Difference

0.0002 g

Mean

0.8510 g

Std Dev

0.03032

Gtable

1.672

Gcalculated for mass #1

1.319

Gcalculated for mass #2

0.6306

Gcalculated for mass #3

0.7249

Gcalculated for mass #4

0.2942

Gcalculated for mass #5

1.119

Sampath 10

Table 7- Results for F-Test Color combo

S1

S2

Fcalculated

Ftable

Result

Yellow & Blue

0.04000

0.02843

1.980

6.39

SIMILAR

Yellow & Red

0.08321

0.04000

4.327

6.39

SIMILAR

Yellow & Orange

0.04320

0.04000

1.166

6.39

SIMILAR

Yellow & Green

0.04000

0.03023

1.751

6.39

SIMILAR

Yellow & Brown

0.04000

0.03032

1.740

6.39

SIMILAR

Blue & Red

0.08321

0.02843

8.568

6.39

DIFFERENT

Blue & Orange

0.04320

0.02843

2.309

6.39

SIMILAR

Blue & Green

0.03022

0.02843

1.131

6.39

SIMILAR

Blue & Brown

0.03032

0.02843

1.138

6.39

SIMILAR

Red & Orange

0.08321

0.04320

3.711

6.39

SIMILAR

Red & Green

0.08321

0.03023

7.579

6.39

DIFFERENT

Red & Brown

0.08321

0.03032

7.531

6.39

DIFFERENT

Orange & Green

0.04320

0.03023

2.042

6.39

SIMILAR

Orange & Brown

0.04320

0.03032

2.030

6.39

SIMILAR

Green & Brown

0.03032

0.03023

1.006

6.39

SIMILAR

Sampath 11

Table 8- Results for T-Test Color combo

Result of F-Test

Tcalculated

Ttable

Result of T-Test

Yellow & Blue

SIMILAR

0.03470

2.306

SIMILAR

Yellow & Red

SIMILAR

0.06528

2.306

SIMILAR

Yellow & Orange

SIMILAR

0.04163

2.306

SIMILAR

Yellow & Green

SIMILAR

0.03545

2.306

SIMILAR

Yellow & Brown

SIMILAR

0.03549

2.306

SIMILAR

Blue & Red

DIFFERENT

0.6546

2.571

SIMILAR

Blue & Orange

SIMILAR

0.03656

2.306

SIMILAR

Blue & Green

SIMILAR

0.02934

2.306

SIMILAR

Blue & Brown

SIMILAR

0.02939

2.306

SIMILAR

Red & Orange

SIMILAR

0.06629

2.306

SIMILAR

Red & Green

DIFFERENT

0.5193

2.571

SIMILAR

Red & Brown

DIFFERENT

0.6630

2.571

SIMILAR

Orange & Green

SIMILAR

0.03728

2.306

SIMILAR

Orange & Brown

SIMILAR

0.03732

2.306

SIMILAR

Green & Brown

SIMILAR

0.03027

2.306

SIMILAR

Sampath 12

TABLE 9- General data about M&Ms Self

Group #1

Group #2

Group #3

Group #4

Mass of one bag of M&M

48.1932 g

47.8400 g

48.8400 g

50.7629 g

50.0440 g

Mass of all M&Ms together

47.1422 g

46.8000 g

47.7772 g

49.7019 g

48.9833 g

Number of M&Ms

55

54

56

59

58

Average mass of one M&M

0.8571 g

0.8667 g

0.8532 g

0.8424 g

0.8445 g

Mass of wrapper

1.0163 g

1.0400 g

1.0628 g

1.0481 g

1.0508 g

TABLE 10- Statistical analysis on Table 9 Mean

49.1360 g

Standard Deviation

1.238

Gtable

1.672

Gcalculated for self

0.7618

Gcalculated for group 1

1.047

Gcalculated for group 2

0.2392

Gcalculated for group 3

1.314

Gcalculated for group 4

0.7336

Calculations Mean= X=

M ass 1 + M ass 2 + M ass 3 + M ass 4+M ass 5 + 0.8432 + 0.8923 4.2605 = 0.7933 + 0.8867 + 0.8450 = 5 5 5

= 0.8521 g

Standard Deviation = S =

√

n

∑(xi−X )2 i=1

n−1

= 0.04000

Sampath 13

Gcalculated =

| Questionable value − X | S

=

| 0.7933 − 0.8521 | 0.04000

= 1.471

Gtable = 1.672 ( n = 5) Since Gcalculated < Gtable , the value is not an outlier.

Fcalculated =

S12 S22

( S1 > S2) =

0.04000 0.02843

= 1.980

Degree of Freedom for S1 = n1 -1 = 5 - 1 = 4 Ftable = 6.39 ( n1 = n1 = 5 ) Since Fcalculated < Ftable , the standard deviations are similar. If Fcalculated > Ftable , the standard deviations are different.

If standard deviations are similar, T calculated =

| X 1 −X 2 |

√

S 1 2(n1 −1) + S 2 2(n2 −1) n1 + n2 −2

=

| 0.8521 − 0.8515 |

√

((0.04000 2)×4) + ((0.028432 )×4) 8

=

0.03470 Degrees of Freedom = n1 + n2 - 2 = 8 Ttable (n1 = n2 = 5) = 2.306

If standard deviations are different, T calculated = Degrees of Freedom =

2 S22 2 ( S1 + n2 ) n1 S12 2 S22 ( n1 ) ( n2 ) 2 n1 −1 + n2 −1

=

| X 1 −X 2 | 2 2 S1 + S2 n1 + n2

√

2 0.028422 2 ( 0.0832 + 5 ) 5 0.028422 2 0.08322 2 ( ) ( ) 5 5 + 4 4

=

| 0.8772 − 0.8515 |

√

0.0832 2 + 0.02842 2 5+5

= 0.6546

= 4.921 ≈ 5

Ttable (n1 = n2 = 5) = 2.571 Since Tcalculated < Ttable , the means are similar. If Tcalculated > Ttable , the means are different.

Sampath 14

Discussion In Lab 1, it was speculated that the mean mass for each color was different and that the different color M&Ms had different masses. However, the data from Table 8 disproves this. Though the standard deviation of the red M&Ms was different from that of the blue, green and brown M&Ms (see Table 7), there was no significant difference between the mean mass of each color. In light of this new evidence, it can be said that all colors of M&Ms weigh the same. Each of the 5 mass readings obtained for each color was tested and it was determined that there were no outliers. Based on the count of each color per bag, it was noticed that the color distribution was being controlled. When asked about this, an M&M representative claimed that the number of M&Ms produced in each color was the same and the number of M&Ms of each color were chosen based on weight to ensure that each bag weighed the same. Though the truthfulness of this explanation was initially doubted, it was later determined that this can be considered a plausible explanation since as seen from Table 10, the mass of each bag was similar as the standard deviation was 1.238 and none of the data points could be considered an outlier. Another potential reason for the controlled color distribution could be that it was economical to reduce the number of M&Ms of certain colors to save on food dye costs. It could also be possible that certain colors were found to be more aesthetically attractive than others.

Conclusion In this lab, the various sets of data obtained in lab 1 was statistically analyzed using the F-Test, T-Test and Grubbs Test. It was determined that there were no outliers in the data using Grubbs Test, indicating that the data obtained was precise. Though there were some differences

Sampath 15

in standard deviations of the mass of each color observed while performing the F-Test, the T-Test indicated that the mean mass of each color was the same, disproving the hypothesis that the mass of each color differed.

Reference 1. Itl.nist.gov. (2019). 1.3.5.17.1. Grubbs' Test for Outliers. [online] Available at: https://www.itl.nist.gov/div898/handbook/eda/section3/eda35h1.htm [Accessed 4 Oct. 2019]. 2. People.richland.edu. (2019). Stats: F-Test . [online] Available at: https://people.richland.edu/james/lecture/m170/ch13-f.html [Accessed 4 Oct. 2019]. 3. Investopedia. (2019). T-Test Definition . [online] Available at: https://www.investopedia.com/terms/t/t-test.asp [Accessed 4 Oct. 2019]....