Question

Weekly Experiment and Discussion - Part 1

• Take 2 coins and flip "together" 50 times
• Tally each set of flips and report the frequencies in the format shown below:

	2nd coin Heads	2nd coin tails
1st coin Heads	# of (H, H)	# of (H, T)
1st coin Tails	# of (T, H)	# of (T, T)

(e.g., if on 12 occasions both coins were heads, there should be a 12 in the top left box)

Weekly Experiment and Discussion - Part 2

A) Given the results of your own experiment, can you conclude the results of each coin were independent? ( Show your supporting work)

B) Using the compiled results ( to be posted on Day #4), can you conclude the results of each coin were independent? ( Show your supporting work)

Note: For each, it is about interpreting the results versus any preconceived notions you may have about this exercise

Answer 1

Answer:

Let's take the case be:-

	2nd coin Heads	2nd Coin Tail
1st coin Heads	13	17
1st coin Tails	8	12

The results of each coin are independent.

Define Event A = 1ST coin will show head

Event B = 2nd coin will show head

Events A & B are said to be independent iff P(A) * P(B) = P(A B)

Here P(A) = n(A)/n(S) = 13/(13+8) = 13/21

P(B) = n(B) / n(S) = 13/(13+ 17) = 13/ 30

P(A B) = n(A B) /n(S) = 13/ 50.

Here P(A) * P(B) = P(A B) = 0.2682 0.26 by probability approach, it is proved that eventa A & B are independent.

Using inferential Statistics we can prove that attributes A & B are independent by Chi Square teat of independence.

To test,

H0 : Attributes A & B are independent.

vs H1 : Attributes A & B are not independent.

It is 2X2 contingency table.

	2nd coin head	2nd coin tail	Total
1st coin head	a = 13	b = 17	a+b = 30
1st coin tail	c = 8	d = 12	c+d = 20
total	a+c = 21	b+d = 29	N = 50

Under H0 , the test statistic is,

  

  

= (50* 400)/ 365400

= 200/3654

= 0.0547

From table the critical value for = 5% at 1 degrees of freedom is 3.841

Since calculated value of is less than that of tabulated value, we accept H0 .

CONCLUSION : The results of each coin were independent.