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 assumptionaccording 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 assumptionaccording 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.