Senske’s First Block AP Senske’s First Block AP StatisticsStatistics

Alesha Seternus and Jenna RorerAlesha Seternus and Jenna Rorer

Objective: Be the first player to reach the Candy Castle by landing on the multi-colored rainbow space at the end of the path.

• Sixty-four cards in a deck:

36 single-colored cards

22 double-colored cards

6 character cards

• The card colors consist of red, orange, yellow, green, blue, and purple

• The characters are Grandma Nutt, Mr. Mint, Jolly, Gingerbread, Lolly, and Princess Frostine

We have chosen the Chi-Squared Test in We have chosen the Chi-Squared Test in order to examine the following order to examine the following

probabilities…probabilities…

Test One: The Probability of Choosing a Single-Colored Card from the Deck

Test Two: The Probability of Choosing a Double-Colored Card from the Deck

Test Three: The Probability of Choosing a Character Card from the Deck

Test One: The Probability of Choosing a Single Test One: The Probability of Choosing a Single Colored Card from the DeckColored Card from the Deck

Hypothesis:

Ho:The observed frequency distribution for picking a single colored card fits the specified distribution

Ha: The observed frequency distribution for picking a single colored card does not fit the specified distribution

Assumptions:

State:

•SRS

•Sample size large enough that all expected values are greater than or equal to 5

Check:

•Assumed

•Refer to chart

Trial Red Purple Yellow Blue Orange Green

1 6 4 3 1 5 54 4 5 4 4 3 23 3 1 1 2 1 14 1 1 1 1 1 15 3 5 0 4 4 36 3 1 5 2 4 57 4 2 5 4 2 18 4 6 11 7 5 39 4 4 2 1 4 410 3 2 2 3 4 511 7 2 4 5 5 312 0 4 3 2 3 313 5 6 4 5 6 614 2 3 5 4 2 215 3 6 4 8 4 816 6 4 1 6 3 417 4 2 4 4 4 418 5 2 4 1 6 519 5 3 3 2 1 120 3 4 2 2 2 321 3 2 5 3 8 522 2 2 3 9 3 823 9 7 8 4 5 524 3 5 6 2 4 325 3 5 3 3 2 326 8 6 6 5 6 627 2 4 3 4 2 328 6 3 7 6 5 529 3 3 3 4 1 330 2 2 1 1 3 2

Trial Expected Value Observed Value

1 23.625 24

2 21.9375 22

3 11.8125 9

4 11.25 6

5 20.8125 19

6 20.25 20

7 18.5625 18

8 31.5 36

9 19.6875 19

10 18.0 19

11 23.625 26

12 14.625 15

13 32.625 32

14 16.875 18

15 31.5 33

16 25.3125 24

17 20.8125 24

18 24.75 23

19 14.625 15

20 14.625 16

21 23.0625 26

22 28.125 27

23 36.5625 28

24 24.75 23

25 16.3125 19

26 35.4375 37

27 20.25 18

28 33.1875 32

29 18.0 17

30 11.8125 11

Test OneTest One:The Probability of Choosing a Single-Colored :The Probability of Choosing a Single-Colored Card from the DeckCard from the Deck

obs-exp)2

= exp(24-23.625)2 + (22-21.9375)2 + . . . 23.625 21.9375

= 8.7155

p( > 8.7155) = 0.9999019051

Conclusion: We fail to reject Ho in favor of Ha because our P-value is greater than alpha (0.05). We have sufficient evidence that the observed

frequency distribution for picking a single colored card fits the specified distribution.

df (k-1) = 29

Test Two: The Probability of Choosing a Double-Test Two: The Probability of Choosing a Double-Colored Card from the DeckColored Card from the Deck

Hypothesis:

Ho:The observed frequency distribution for picking a single colored card fits the specified distribution

Ha: The observed frequency distribution for picking a single colored card does not fit the specified distribution

Assumptions:State:

•SRS

•Sample size large enough that all expected values are greater than or equal to 5

Check:

•Assumed

•Refer to chart

Red Purple Yellow Blue Orange Green

1 3 2

3 2

 21

3 4

 02

5 42            

3 2 2

 22

 13

 32

1 0

1 34            

5  02

 24

51

 24

 3  226  

3       1

2 

7   3 3

 32

 14

   048  3

3      3

2 

9    11

 30

 23

   1110  3

3       3

1 

11    42

 32

 20

   1212  1

4       1

2 

13    41

 42

 41

   3214  1

5       2

3 

15   5  24

 33

  3 016  5

211

     42

 17     3 

5 12

   3318  4

1 14

    3 1

 19  

1   2

2 02

   1220   1 

3    0 

2 

21  43

   14

 22

   2122    7

4     0

1 

23 6 2

   34

 42

   3224   4 

1     3

1 125  2

4  3 

5 14

   26    4

2     2

22 227  2

5   3

3 23

   28    3

6     4

13 329  5

1   2

2 2    

30    3    1  1  0           

   

Trial Expected Value Observed Value

1 14.438 17

2 13.406 15

3 7.2188 10

4 6.875 12

5 12.719 14

6 12.375 14

7 11.344 12

8 19.25 18

9 12.031 12

10 11.0 11

11 14.438 14

12 8.937 8

13 19.938 21

14 10.313 9

15 19.25 21

16 15.468 17

17 12.719 12

18 15.125 18

19 8.9375 9

20 8.9375 8

21 14.094 14

22 17.188 17

23 22.344 21

24 15.125 17

25 9.9688 9

26 21.656 21

27 12.375 13

28 20.281 21

29 11.0 13

30 7.2188 8

Double Color Card Probability Scatterplot

0

5

10

15

20

25

Game Number

Fre

qu

ency

expected

observed

Test Two: The Probability of Choosing a Double-Test Two: The Probability of Choosing a Double-Colored Card from the DeckColored Card from the Deck

(obs-exp)2 = (17-14.438)2 + (15-13.406)2 + … exp 14.438 13.406

= 8.26496

p( > 8.26496) = 0.9999441532

Conclusion: We fail to reject Ho in favor of Ha because our P-value is greater than alpha (0.05). We have sufficient evidence that the observed frequency distribution for picking a single colored card fits the

specified distribution

df (k-1) = 29

Test Three: The Probability of Choosing a Test Three: The Probability of Choosing a Character Card from the DeckCharacter Card from the Deck

Hypothesis:

Ho:The observed frequency distribution for picking a character card fits the specified distribution

Ha: The observed frequency distribution for picking a character card does not fit the specified distribution

Assumptions:State:

•SRS

•Sample size large enough that all expected values are greater than or equal to 5

Check:

•Assumed

•Refer to chart

Trial Gingerbread

Mr. Mint Jolly Princess Frostine

Gramma Nutt

Lolly

1 1 0 0 0 1 0

2 0 0 1 0 1 0

3 0 0 1 0 1 0

4 0 0 0 1 1 0

5 1 0 1 1 0 1

6 1 0 0 0 1 0

7 1 0 0 1 0 1

8 1 1 1 0 0 0

9 0 1 1 1 1 1

10 1 0 0 0 0 0

11 0 0 0 1 1 0

12 0 0 1 0 0 1

13 1 1 1 1 0 1

14 0 0 0 0 1 1

15 0 1 1 0 0 1

16 0 1 1 1 1 0

17 1 0 0 0 1 1

18 0 0 0 0 1 1

19 0 1 1 0 1 0

20 0 0 0 0 1 1

21 0 1 1 1 0 0

22 0 1 1 2 1 1

23 0 1 1 1 2 1

24 0 1 1 0 1 1

25 0 0 0 0 0 1

26 1 1 1 2 1 0

27 1 1 1 0 1 1

28 1 0 0 2 1 1

29 1 0 0 0 0 1

30 0 0 0 2 0 0

Trial Expected Value Observed Value

1 3.9375 1

2 3.365635 2

3 1.96875 2

4 1.875 2

5 3.46875 4

6 3.375 2

7 3.09375 3

8 5.25 2

9 3.28125 5

10 3.0 2

11 3.9375 2

12 2.4375 3

13 5.4375 5

14 2.8125 3

15 5.25 2

16 4.21875 4

17 3.46875 3

18 4.125 3

19 2.4375 2

20 2.4375 2

21 3.84375 2

22 4.6875 6

23 6.09375 6

24 4.125 4

25 2.71875 1

26 5.90625 6

27 3.375 5

28 5.53125 6

29 3.0 2

30 1.9688 2

Character Card Probability Scatterplot

0

2

4

6

8

Game Number

Fre

qu

en

cy

expected

observed

Test Three: The Probability of Choosing a Test Three: The Probability of Choosing a Character Card from the DeckCharacter Card from the Deck

= (obs-exp)2 = (1-3.9375)2 + (2-3.65625)2 + … exp 3.9375 3.6525

p( >14.0230261) = 0.99126

=14.0230261

Conclusion: We fail to reject Ho in favor Ha because our P-value is greater than alpha (0.05). We have sufficient evidence that the observed frequency

distribution for picking a character card fits the specified distribution.

df (k-1) = 29

•Personal Opinions/ Conclusions

•Bias/Error

Our experiment was conducted through random samplings of the 64 cards (no bias)

An example of a bias experiment would be if we had arranged or drawn the cards in a specific order or pattern as to predict/control the outcomes.

If the 30 trials happened to be played by separate groups, all groups had to collect data under identical conditions.

We have come to the conclusion that the probabilities of picking either a single-colored, double-colored, or character card is similar to the expected values.