Use minitab to obtain the following: thank you PLEASE ANSWER QUESTION #2 BOTH a.
ID: 3331970 • Letter: U
Question
Use minitab to obtain the following: thank you PLEASE ANSWER QUESTION #2 BOTH a. & b.
Instructions : Clearly label each problem and each part of a problem on your paper. Make sure and print information and explain in detail on any questions or comments.
1. ) Simulate flipping a fair coin by performing the following operations.
a. ) Randomly generate a column of data (c1) with 2500 entries { CalcRandom DataInteger }. Minimum number is 0 and maximum number is 1.
b. ) Randomly generate a column of data (c2) with 4500 entries { CalcRandom DataInteger }. Minimum number is 0 and maximum number is 1
c. ) Randomly generate a column of data (c3) with 9500 entries { CalcRandom DataInteger }. Minimum number is 0 and maximum number is 1.
2. ) For the data generated in number 1, perform the following items.
**a. ) Find the empirical probabilities for c1, c2, and c3 { StatTablesTally Individual Variables }. Be sure to check both counts (Frequency) and percents (relative frequency).
**b. ) In which sample did the empirical probabilities, most closely match the theoretical probability of 0.5 ? Why ?
Explanation / Answer
As directed, I used Minitab to solve this problem.
(1) In the first question of this problem, we have to simulate flipping of a fair coin. As a result, I randomly generated 3 columns of binomial data (that is, either 0 or 1) containing 2500, 4500 and 9500 rows of entries, which means that the fair coin was tossed 2500, 4500 and 9500 times respectively.
In Minitab, I used the following procedure to generate the numbers in the 3 columns : Calc -> Random Data -> Integer.
The following code was generated, hence.
MTB > Random 2500 c1;
SUBC> Integer 0 1.
MTB > Random 4500 c2;
SUBC> Integer 0 1.
MTB > Random 9500 c3;
SUBC> Integer 0 1.
(2)
(a) In the second question of this problem, we are to find the empirical probabilities for the 3 columns of data - we name it c1, c2 and c3 for our convenience. We followed the following Minitab procedure to find the empirical probabilities and made sure to calculate both the counts (Frequency) and percents (relative frequency) : Stat -> Tables -> Tally Individual Variables.
We applied it on the 3 columns of data and the following output was received.
MTB > Tally C1;
SUBC> Counts;
SUBC> Percents.
Tally for Discrete Variables: C1
C1 Count Percent
0 1299 51.96
1 1201 48.04
N= 2500
MTB > Tally C2;
SUBC> Counts;
SUBC> Percents.
Tally for Discrete Variables: C2
C2 Count Percent
0 2237 49.71
1 2263 50.29
N= 4500
MTB > Tally C3;
SUBC> Counts;
SUBC> Percents.
Tally for Discrete Variables: C3
C3 Count Percent
0 4741 49.91
1 4759 50.09
N= 9500
(b) From the Minitab output pasted above, we are to judge which sample's empirical probability matches the theoretical probability, that is, 0.50 or 50%.
The answer is the 3rd sample. It has an empirical probability of 50.09% and is the closest to the theoretical probability among the 3 samples considered in this problem.
We know that empirical probability is the probability of an event occuring when an experiment was conducted. It is calculated as the number of event occurences divided by the total number of trials of the experiment.
Here, the 3rd sample's empirical probability is closer to the theoretical probability due to the fact that if we increase the number of trials of an experiment, the empirical (or experimental) probability of an event tends towards the theoretical probability. This is often called the Law of Large Numbers. This establishes the fact that a large sample size gives us a better accuracy than a small sample size.
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