Exploratory Data Analysis
1.4. EDA Case Studies
1.4.2. Case Studies
126.96.36.199. Uniform Random Numbers
As a first step in the analysis, common summary statistics are
computed for the data.
Sample size = 500 Mean = 0.5078304 Median = 0.5183650 Minimum = 0.0024900 Maximum = 0.9970800 Range = 0.9945900 Stan. Dev. = 0.2943252Because the graphs of the data indicate the data may not be normally distributed, we also compute two other statistics for the data, the normal PPCC and the uniform PPCC.
Normal PPCC = 0.9771602 Uniform PPCC = 0.9995682The uniform probability plot correlation coefficient (PPCC) value is larger than the normal PPCC value. This is evidence that the uniform distribution fits these data better than does a normal distribution.
One way to quantify a change in location over time is to
fit a straight line
to the data using an index variable as the independent
variable in the regression. For our data, we assume
that data are in sequential run order and that the
data were collected at equally spaced time intervals. In our regression,
we use the index variable X = 1, 2, ..., N, where N is the number
of observations. If there is no significant drift in the location
over time, the slope parameter should be zero.
Coefficient Estimate Stan. Error t-Value B0 0.522923 0.2638E-01 19.82 B1 -0.602478E-04 0.9125E-04 -0.66 Residual Standard Deviation = 0.2944917 Residual Degrees of Freedom = 498The t-value of the slope parameter, -0.66, is smaller than the critical value of t0.975,498 = 1.96. Thus, we conclude that the slope is not different from zero at the 0.05 significance level.
One simple way to detect a change in variation is with a
Bartlett test after dividing the
data set into several equal-sized intervals. However, the Bartlett
test is not robust for non-normality. Since we know this data set is
not approximated well by the normal distribution,
we use the alternative Levene
test. In particular, we use the Levene test based on the median
rather the mean. The choice of the number of intervals is somewhat
arbitrary, although values of four or eight are reasonable. We will divide
our data into four intervals.
H0: σ12 = σ22 = σ32 = σ42 Ha: At least one σi2 is not equal to the others. Test statistic: W = 0.07983 Degrees of freedom: k - 1 = 3 Significance level: α = 0.05 Critical value: Fα,k-1,N-k = 2.623 Critical region: Reject H0 if W > 2.623In this case, the Levene test indicates that the variances are not significantly different in the four intervals.
There are many ways in which data can be non-random. However,
most common forms of non-randomness can be detected with a
few simple tests including the
lag plot shown on
the previous page.
Another check is an autocorrelation plot that shows the autocorrelations for various lags. Confidence bands can be plotted using 95% and 99% confidence levels. Points outside this band indicate statistically significant values (lag 0 is always 1).
The lag 1 autocorrelation, which is generally the one of most interest, is 0.03. The critical values at the 5 % significance level are -0.087 and 0.087. This indicates that the lag 1 autocorrelation is not statistically significant, so there is no evidence of non-randomness.
A common test for randomness is the runs test.
H0: the sequence was produced in a random manner Ha: the sequence was not produced in a random manner Test statistic: Z = 0.2686 Significance level: α = 0.05 Critical value: Z1-α/2 = 1.96 Critical region: Reject H0 if |Z| > 1.96The runs test fails to reject the null hypothesis that the data were produced in a random manner.
Probability plots are a
graphical test of assessing whether a particular distribution provides
an adequate fit to a data set.
A quantitative enhancement to the probability plot is the correlation coefficient of the points on the probability plot, or PPCC. For this data set the PPCC based on a normal distribution is 0.977. Since the PPCC is less than the critical value of 0.987 (this is a tabulated value), the normality assumption is rejected.
Chi-square and Kolmogorov-Smirnov goodness-of-fit tests are alternative methods for assessing distributional adequacy. The Wilk-Shapiro and Anderson-Darling tests can be used to test for normality. The results of the Anderson-Darling test follow.
H0: the data are normally distributed Ha: the data are not normally distributed Adjusted test statistic: A2 = 5.765 Significance level: α = 0.05 Critical value: 0.787 Critical region: Reject H0 if A2 > 0.787The Anderson-Darling test rejects the normality assumption because the value of the test statistic, 5.765, is larger than the critical value of 0.787 at the 0.05 significance level.
Based on the graphical and quantitative analysis, we use the model
95% confidence limit for C = (0.497,0.503)
It is sometimes useful and convenient to summarize the above results
in a report.
Analysis for 500 uniform random numbers 1: Sample Size = 500 2: Location Mean = 0.50783 Standard Deviation of Mean = 0.013163 95% Confidence Interval for Mean = (0.48197,0.533692) Drift with respect to location? = NO 3: Variation Standard Deviation = 0.294326 95% Confidence Interval for SD = (0.277144,0.313796) Drift with respect to variation? (based on Levene's test on quarters of the data) = NO 4: Distribution Normal PPCC = 0.9771602 Normal Anderson-Darling = 5.7198390 Data are Normal? (as tested by Normal PPCC) = NO (as tested by Anderson-Darling) = NO Uniform PPCC = 0.9995683 Uniform Anderson-Darling = 0.9082221 Data are Uniform? (as tested by Uniform PPCC) = YES (as tested by Anderson-Darling) = YES 5: Randomness Autocorrelation = -0.03099 Data are Random? (as measured by autocorrelation) = YES 6: Statistical Control (i.e., no drift in location or scale, data is random, distribution is fixed, here we are testing only for fixed uniform) Data Set is in Statistical Control? = YES