Part 1:
The data is about the distance between the coin and the wall which is changing each time of flipping a coin. The data recoded of the 40 trails of flipping a coin. Each time recorded the distance in the continuous scale. We can check this data for which distribution it follows so that we make the conclusion or expect the next value of the distance between coin and wall. This will help us to model the data then we can predict the future values of the distance.
The table of the data as follows
Distance Inches Freq
0.125 1
0.25 1
0.5 1
1.25 3
1.5 1
2.0625 1
2.125 1
2.5 1
3.0625 1
3.125 1
3.25 1
3.35 1
3.5 1
4.0625 2
4.25 1
5.0625 1
5.125 1
5.25 1
5.5 2
6 1
6.0625 2
6.25 1
6.5 1
8 1
8.125 1
8.25 1
8.5 3
9.25 1
12.125 1
12.25 1
15.125 1
18.25 1
18.5 1
The above table where collected data table. In that some of the values came repeatedly the most is 8.5 is 3 times and 5.5, 4.0625 are 2 times. So most of the time distance was not same. The distances vary trail to trail not certain numbers. We can find this data about which distribution it is following so it will be useful to know the pattern of the outcomes.
Histogram of the Distance where it measured in inches as follows
The above histogram says that the data has right tailed that mean most of the data in the left side of the distribution. So the distribution is not following normal we can identify this from the above plot. The distance from the coin to wall is in the most of the case is less than 10 inches and sometimes only it is far from the wall.
Part 2:
The parameter of interest is distance in inches.
Chi square test:
Null Hypothesis: The data follows normal distribution
Alternative Hypothesis: The data not following normal distribution
The output of the chi square test from the R programming as follows
Chi-squared test for given probabilities
X-squared = 134.62, df = 39, p-value = 1.91e-12
The p value is less than the significance value that is 0.05 so we reject the null hypothesis and we can accept the alternative hypothesis which means that the data not following normal distribution.
Kolmograv smirnov test:
The output of the kolmograv test as follows
Null Hypothesis: The data follows normal distribution
Alternative Hypothesis: The data not following normal distribution
One-sample Kolmogorov-Smirnov test
D = 7.1863, p-value < 2.2e-16 Alternative hypothesis: two-sided The p value is less than the significance value that is 0.05 so we reject the null hypothesis and we can accept the alternative hypothesis which means that the data not following normal distribution. Part 3: Simulation process: Chi square test: The output from R programming of the chi square test as follows Chi-squared test for given probabilities X-squared=267.24, df=99, p-value < 2.2e-16 Kolmograv smirnov test: The output of the test as follows One-sample Kolmogorov-Smirnov test D=7.8748, p-value < 2.2e-16 Alternative hypothesis: two-sided In the above two tests the p value is less than significance level that is 0.05. So here also we reject the null hypothesis means that accept the alternative hypothesis that is the data not following required distribution with the specified parameter values. Part 4: The simulation data is generated for 100 sample values. The result from the simulation data of the two tests same from the original data. But there is an assumption that if the sample is large then the resulting data distribution follows normal distribution. In this case the large sample data also not following normal distribution. Simulation will be helpful to get the data as to be satisfied with the assumptions of the models. So it can be useful to make model with accurately. The simulation also not modeled correctly to the data.
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