Exercise Answers

Calculate the predicted answer from this past lab and show your work (i.e. How many sets (of 5) and total will result in all heads - or should have occurred).

How does this compare to the results from this week's simulation? Provide the same information as the example above and discuss your conclusions. Explain your answers. Reminder:

Class	# Students who coined	# of total tosses	Results (sets)
Tuesday	10	500	7
Thursday morning	3	300	3
Thursday afternoon	10	500	3

Given:

Equation: Pⁿ
P is the relative proportion of heads (versus not heads) on a fair coin. The number of "events" = n
There are two sides to a coin - 1 head and 1 not head (tail)

1 (head) = 1/2 = 0.5 = P

Answer for Tuesday class:

Coin tosses = 10 X 5 X 10 = 500 tosses

1 head + 1 not head

There will be 5 tosses (events) in which each (=5) will be a head. n = 5
0.5⁵ = 0.031 = 3.1% chance that in 5 tosses the coin will come up all heads
If this is true, the proportion (0.031) can be used to predict how many total (and how many groups of 5) heads will occur:
Multiply the proportion (0.031) by the total number of future coin tosses (500)
0.031 X 500 = 15 or 16 [15.5] heads
16 ¸ 5 = 3 [3.2] sets of 5 heads.
Prediction: 3 sets of 5 out of 500 coin tosses will come up heads
The actual results (7 sets = 7X5 = 35 total heads) were more than the predicted (3 sets = 15 or 16 total heads). The hypothesis “The quarter has one head and one not head” was not supported and fail to reject the null hypothesis. What happened? Is it comforting to accept the null hypothesis that there is not one head on these coins?

Answer for Thursday morning class:

Coin tosses = 10 X 5 X 10 = 500 tosses

1 head + 1 not head

There will be 5 tosses (events) in which each (=5) will be a head. n = 5
0.5⁵ = 0.031 = 3.1% chance that in 5 tosses the coin will come up all heads
If this is true, the proportion (0.031) can be used to predict how many total (and how many groups of 5) heads will occur:
Multiply the proportion (0.031) by the total number of future coin tosses (300)
0.031 X 500 = 9 [9.3] heads
9 ¸ 5 = 2 [1.8] sets of 5 heads.
Prediction: 2 sets of 5 out of 300 coin tosses will come up heads
The actual results (3 sets = 3X5 = 15 total heads) were more than predicted (2 sets = 10 total heads). The hypothesis “The quarter has one head and one not head” was not supported and fail to reject the null hypothesis. What happened? Is it comforting to accept the null hypothesis that there is not one head on these coins? Most likely there were too few samples (tosses) to accurately predict reality.

Answer for Thursday afternoon class:

Coin tosses = 3 X 5 X 20 = 300 tosses

1 head + 1 not head

There will be 5 tosses (events) in which each (=5) will be a head. n = 5
0.5⁵ = 0.031 = 3.1% chance that in 5 tosses the coin will come up all heads
If this is true, the proportion (0.031) can be used to predict how many total (and how many groups of 5) heads will occur:
Multiply the proportion (0.031) by the total number of future coin tosses (500)
0.031 X 500 = 15 or 16 [15.5] heads
16 ¸ 5 = 3 [3.2] sets of 5 heads.
Prediction: 3 sets of 5 out of 300 coin tosses will come up heads
The actual results (3 sets = 3X5 = 15 total heads) were just as predicted (3 sets = 15 or 16 total heads). The hypothesis “The quarter has one head and one not head” was supported and the null hypothesis was rejected.

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