The problems on this page concern the data set
S = \{ 0.28, 0.2, 0.01, 0.80, 0.1 \} .
Let x_ i denote the i'th element of the data set S.
In this problem, you will consider the following null and alternative hypotheses.
\displaystyle H_0 \displaystyle = \text {the data set is distributed as a Gaussian distribution (for some choice of mean and variance)}
\displaystyle H_1 \displaystyle = \text {the data set is not distributed as a Gaussian (for any choice of mean and variance)} .
What is the Kolmogorov-Lilliefors test statistic evaluated on the data set S? Enter \, T_5^{\text {KL}}/\sqrt{5},\, the KL statistic without the factor of \sqrt{n}, below. (You are encouraged to use computational tools.)
Kolmogorov-Lilliefors Tables
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T_5^{\text {KL}}/\sqrt{5} =\quad
To calculate the Kolmogorov-Lilliefors test statistic evaluated on the data set S, we can follow these steps:
Step 1: Sort the data set in ascending order:
S = {0.01, 0.1, 0.2, 0.28, 0.8}
Step 2: Calculate the empirical cumulative distribution function (ECDF) of the data set.
For each element x_i in S, calculate the ECDF value F(x_i) as the proportion of elements in S that are less than or equal to x_i.
F(0.01) = 1/5 = 0.2
F(0.1) = 2/5 = 0.4
F(0.2) = 3/5 = 0.6
F(0.28) = 4/5 = 0.8
F(0.8) = 5/5 = 1
Step 3: Calculate the two-sided Kolmogorov-Lilliefors test statistic.
The two-sided Kolmogorov-Lilliefors test statistic is given by:
T_n^{KL} = \sqrt{n} \cdot \max \left( \max_{1 \leq i \leq n} \left( \frac{i}{n} - F(x_i) \right), \max_{1 \leq i \leq n} \left( F(x_i) - \frac{i-1}{n} \right) \right)
In this case, n = 5, so we have:
T_5^{KL} = \sqrt{5} \cdot \max \left( \max \left( \frac{1}{5} - 0.2, \frac{2}{5} - 0.4, \frac{3}{5} - 0.6, \frac{4}{5} - 0.8, \frac{5}{5} - 1 \right), \max \left( 0.2 - \frac{0}{5}, 0.4 - \frac{1}{5}, 0.6 - \frac{2}{5}, 0.8 - \frac{3}{5}, 1 - \frac{4}{5} \right) \right)
Simplifying the above expression, we get:
T_5^{KL} = \sqrt{5} \cdot \max \left( \max \left( -0.2, -0.2, -0.2, -0.2, -0.2 \right), \max \left( 0.2, 0.2, 0.2, 0.2, 0.2 \right) \right)
Now, evaluating the maximum values in each set, we find:
T_5^{KL} = \sqrt{5} \cdot \max \left( -0.2, 0.2 \right)
Taking the maximum of -0.2 and 0.2, we get:
T_5^{KL} = \sqrt{5} \cdot 0.2 = 0.447
Therefore, the Kolmogorov-Lilliefors test statistic evaluated on the data set S is T_5^{KL}/\sqrt{5} = 0.447.
Unfortunately, I cannot perform computations or access external resources. However, I can explain the process of calculating the Kolmogorov-Lilliefors test statistic.
The Kolmogorov-Lilliefors test is used to determine if a dataset follows a specific distribution, in this case, a Gaussian distribution. The test statistic, denoted as T_n^KL, measures the largest vertical deviation between the empirical distribution function (EDF) of the data and the cumulative distribution function (CDF) of the assumed distribution.
To calculate T_n^KL, follow these steps:
1. Sort the data set S in ascending order: S = {0.01, 0.1, 0.2, 0.28, 0.8}.
2. Calculate the empirical distribution function (EDF) by assigning a cumulative probability to each data point. The cumulative probability for each data point is its rank divided by the total number of data points, n.
EDF(0.01) = 1/5 = 0.2
EDF(0.1) = 2/5 = 0.4
EDF(0.2) = 3/5 = 0.6
EDF(0.28) = 4/5 = 0.8
EDF(0.8) = 5/5 = 1.0
3. Calculate the cumulative distribution function (CDF) for the assumed Gaussian distribution. This involves calculating the probability of each data point under the assumed distribution.
4. Calculate the absolute difference between the EDF and CDF at each data point.
|EDF(0.01) - CDF(0.01)|
|EDF(0.1) - CDF(0.1)|
|EDF(0.2) - CDF(0.2)|
|EDF(0.28) - CDF(0.28)|
|EDF(0.8) - CDF(0.8)|
5. Find the maximum absolute difference among all data points. This is the Kolmogorov-Lilliefors test statistic, T_n^KL.
Using computational tools or a statistical software, you can calculate the value of T_n^KL. The value for T_5^KL/\sqrt{5} will then be T_n^KL divided by the square root of the number of data points, which is \sqrt{5} in this case.
I apologize for not being able to provide the specific value of T_5^KL/\sqrt{5} for the given data set.