This template will run an F-test to check if two continuous variables have the same means.
F test compares the means of two continuous variables. In other words it shows if their means were statistically different. We should be careful, while using the F test, because of the strict normality assumption, where strict means approximately normal ditribution is not enough to satisfy that.
The Shapiro-Wilk test, the Lilliefors test and the Anderson-Darling test help us to decide if the above-mentioned assumption can be accepted of the Internet usage for educational purposes (hours per day).
| Method | Statistic | p-value |
|---|---|---|
| Lilliefors (Kolmogorov-Smirnov) normality test | 0.2223 | 2.243e-92 |
| Anderson-Darling normality test | 42.04 | 3.31e-90 |
| Shapiro-Wilk normality test | 0.7985 | 6.366e-28 |
So, the conclusions we can draw with the help of test statistics:
based on Lilliefors test, distribution of Internet usage for educational purposes (hours per day) is not normal
Anderson-Darling test confirms violation of normality assumption
according to Shapiro-Wilk test, the distribution of Internet usage for educational purposes (hours per day) is not normal
As you can see, the applied tests confirm departures from normality.
The Shapiro-Wilk test, the Lilliefors test and the Anderson-Darling test help us to decide if the above-mentioned assumption can be accepted of the Internet usage for educational purposes (hours per day).
| Method | Statistic | p-value |
|---|---|---|
| Lilliefors (Kolmogorov-Smirnov) normality test | 0.17 | 6.193e-54 |
| Anderson-Darling normality test | 32.16 | 1.26e-71 |
| Shapiro-Wilk normality test | 0.8216 | 9.445e-27 |
So, the conclusions we can draw with the help of test statistics:
based on Lilliefors test, distribution of Age is not normal
Anderson-Darling test confirms violation of normality assumption
according to Shapiro-Wilk test, the distribution of Age is not normal
As you can see, the applied tests confirm departures from normality.
In this case it is advisable to run a more robust test, then the F-test.
This template will run an F-test to check if two continuous variables have the same means.
F test compares the means of two continuous variables. In other words it shows if their means were statistically different. We should be careful, while using the F test, because of the strict normality assumption, where strict means approximately normal ditribution is not enough to satisfy that.
Here is the the result of the F test to compare the means of Internet usage for educational purposes (hours per day) and Age.
| Method | Statistic | p-value |
|---|---|---|
| F test to compare two variances | 0.08618 | 3.772e-180 |
We can see from the table (in the p-value coloumn) that there is a significant difference between the means of Internet usage for educational purposes (hours per day) and Age.
This template will run an F-test to check if two continuous variables have the same means.
F test compares the means of two continuous variables. In other words it shows if their means were statistically different. We should be careful, while using the F test, because of the strict normality assumption, where strict means approximately normal ditribution is not enough to satisfy that.
Here is the the result of the F test to compare the means of cyl and drat.
| Method | Statistic | p-value |
|---|---|---|
| F test to compare two variances | 11.16 | 1.461e-09 |
We can see from the table (in the p-value coloumn) that there is a significant difference between the means of cyl and drat.
This report was generated with R (3.0.1) and rapport (0.51) in 0.814 sec on x86_64-unknown-linux-gnu platform.
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