1
00:00:05,840 --> 00:00:08,760
Most of us are familiar
with comparing

2
00:00:08,760 --> 00:00:11,040
the average values
between two groups.

3
00:00:11,040 --> 00:00:14,070
For example, the average
teaching evaluation square group

4
00:00:14,070 --> 00:00:15,810
four male instructors compared

5
00:00:15,810 --> 00:00:17,295
with that of female instructors,

6
00:00:17,295 --> 00:00:18,540
and the groups are two,

7
00:00:18,540 --> 00:00:20,670
and we know that such comparisons

8
00:00:20,670 --> 00:00:22,530
are made using the t-test.

9
00:00:22,530 --> 00:00:25,110
But what if you're dealing
with more than two groups?

10
00:00:25,110 --> 00:00:27,540
What if there are three,
four or more groups?

11
00:00:27,540 --> 00:00:29,055
In that particular case,

12
00:00:29,055 --> 00:00:31,975
we would use ANOVA or
analysis of variance,

13
00:00:31,975 --> 00:00:34,160
where our intent or
goal is to compare

14
00:00:34,160 --> 00:00:36,410
the means of more
than two groups.

15
00:00:36,410 --> 00:00:39,665
So in order to accomplish this,

16
00:00:39,665 --> 00:00:44,310
we return back to our
teaching evaluation data.

17
00:00:44,310 --> 00:00:46,235
In that particular case,

18
00:00:46,235 --> 00:00:48,110
we have a variable called age,

19
00:00:48,110 --> 00:00:49,520
where the age of the instructor

20
00:00:49,520 --> 00:00:52,190
is recorded in a number of years.

21
00:00:52,190 --> 00:00:54,880
But we will discretize
this age variable,

22
00:00:54,880 --> 00:00:56,855
that is, we will
create three groups.

23
00:00:56,855 --> 00:00:58,670
So instructors who
are 40 years and

24
00:00:58,670 --> 00:01:01,025
younger we put them in one group,

25
00:01:01,025 --> 00:01:05,040
those between 40 and
56.5 years of age,

26
00:01:05,040 --> 00:01:06,335
they are in another group,

27
00:01:06,335 --> 00:01:08,375
and those who are
57 years and older,

28
00:01:08,375 --> 00:01:09,560
you put them in the third group.

29
00:01:09,560 --> 00:01:12,409
So you have younger instructors,

30
00:01:12,409 --> 00:01:13,580
middle-age instructors and

31
00:01:13,580 --> 00:01:15,965
rather slightly
older instructors,

32
00:01:15,965 --> 00:01:17,705
and the number of observations,

33
00:01:17,705 --> 00:01:20,565
taught by each group,
is reported under

34
00:01:20,565 --> 00:01:22,790
N. What we also have here

35
00:01:22,790 --> 00:01:25,055
is the teaching evaluation
score for each group,

36
00:01:25,055 --> 00:01:26,870
which is not deferring much.

37
00:01:26,870 --> 00:01:28,460
It's pretty much four for

38
00:01:28,460 --> 00:01:30,530
each group and for the
older professors is

39
00:01:30,530 --> 00:01:32,540
slightly less at 3.9 in

40
00:01:32,540 --> 00:01:34,700
that respective standard
deviations of it.

41
00:01:34,700 --> 00:01:36,395
So we have three groups
and let's see what

42
00:01:36,395 --> 00:01:38,270
we are interested
in is to determine

43
00:01:38,270 --> 00:01:41,930
if these three averages for

44
00:01:41,930 --> 00:01:45,250
the three respective
age categories

45
00:01:45,440 --> 00:01:49,360
are statistically the same
or they are different,

46
00:01:49,360 --> 00:01:52,715
so we use the one-way analysis
of variance or ANOVA,

47
00:01:52,715 --> 00:01:55,745
and using the ANOVA we use

48
00:01:55,745 --> 00:01:57,560
the F-distribution to compare

49
00:01:57,560 --> 00:01:59,030
the mean values for
more than two groups.

50
00:01:59,030 --> 00:02:01,430
Our null hypothesis
is that samples in

51
00:02:01,430 --> 00:02:03,320
all groups are drawn from
the same populations

52
00:02:03,320 --> 00:02:04,870
with the same mean values.

53
00:02:04,870 --> 00:02:07,145
We fail to reject
the null hypothesis

54
00:02:07,145 --> 00:02:10,550
if the P-value or
the significance for

55
00:02:10,550 --> 00:02:13,235
the F-test is greater than 0.05

56
00:02:13,235 --> 00:02:17,455
and we then infer equal means.

57
00:02:17,455 --> 00:02:20,450
Let's say we are interested
in determining if

58
00:02:20,450 --> 00:02:23,495
the beauty score for
instructors differs by age.

59
00:02:23,495 --> 00:02:25,835
We have three groups, younger,

60
00:02:25,835 --> 00:02:28,295
middle aged, and
older professors.

61
00:02:28,295 --> 00:02:30,170
We have the summary
statistics for

62
00:02:30,170 --> 00:02:32,210
the standardized beauty scores.

63
00:02:32,210 --> 00:02:35,990
We see that there is a
difference as the age goes up,

64
00:02:35,990 --> 00:02:39,140
the average value for the
beauty score goes down.

65
00:02:39,140 --> 00:02:41,600
So let's run an ANOVA to see

66
00:02:41,600 --> 00:02:44,855
if the differences are
statistically significant.

67
00:02:44,855 --> 00:02:47,690
Our null hypothesis will be,

68
00:02:47,690 --> 00:02:50,285
"Mean beauty scores for
instructors don't differ with

69
00:02:50,285 --> 00:02:53,700
age," and the alternative
hypothesis will be,

70
00:02:53,700 --> 00:02:56,095
"At least one of the
means is different."

71
00:02:56,095 --> 00:02:59,315
First, this variable does
not exist in our data.

72
00:02:59,315 --> 00:03:02,210
We will need to group or
been the continuous age data

73
00:03:02,210 --> 00:03:05,285
using the dot loc
function in Pandas.

74
00:03:05,285 --> 00:03:08,360
Then use the F underscore
one wave function in

75
00:03:08,360 --> 00:03:11,945
the Scipy Stats Library to
perform the ANOVA test.

76
00:03:11,945 --> 00:03:15,385
We will then print out the F
statistics and the P-value.

77
00:03:15,385 --> 00:03:17,870
What we can see is
that the P-value is

78
00:03:17,870 --> 00:03:21,995
4.32 times ten raised to the
power of negative eight,

79
00:03:21,995 --> 00:03:24,775
and that is less than 0.05.

80
00:03:24,775 --> 00:03:27,695
We will reject the null
hypothesis as there is

81
00:03:27,695 --> 00:03:29,270
significant evidence that at

82
00:03:29,270 --> 00:03:31,280
least one of the means differ.

83
00:03:31,280 --> 00:03:33,230
If I do the same tests for

84
00:03:33,230 --> 00:03:34,700
the teaching evaluation scores

85
00:03:34,700 --> 00:03:36,440
that we observe for
the three groups,

86
00:03:36,440 --> 00:03:40,040
and we run ANOVA on
these three mean values.

87
00:03:40,040 --> 00:03:43,505
We find out that the
P-value is 0.295,

88
00:03:43,505 --> 00:03:46,145
which is greater than 0.05.

89
00:03:46,145 --> 00:03:50,030
We will fail to reject the
null and infer equal means.

90
00:03:50,030 --> 00:03:52,205
That is, that the three means

91
00:03:52,205 --> 00:03:54,470
are not statistically different.

92
00:03:54,470 --> 00:03:56,450
Here we have the Analysis of

93
00:03:56,450 --> 00:03:58,685
Variants performed
on two samples.

94
00:03:58,685 --> 00:04:00,650
One is the beauty score.

95
00:04:00,650 --> 00:04:03,740
We notice that the difference
in means for beauty scores

96
00:04:03,740 --> 00:04:05,015
between the three groups is

97
00:04:05,015 --> 00:04:07,090
based on the significance value.

98
00:04:07,090 --> 00:04:08,960
This leads us to conclude that

99
00:04:08,960 --> 00:04:10,730
at least one mean is different

100
00:04:10,730 --> 00:04:12,800
and we reject the null hypothesis

101
00:04:12,800 --> 00:04:15,125
that states equal means.

102
00:04:15,125 --> 00:04:17,990
Here, because the
P-value for teaching

103
00:04:17,990 --> 00:04:20,210
evaluation scores
between the three groups

104
00:04:20,210 --> 00:04:22,880
is greater than 0.05,

105
00:04:22,880 --> 00:04:25,730
we fail to reject
the null hypothesis.

106
00:04:25,730 --> 00:04:28,085
We believe that these three means

107
00:04:28,085 --> 00:04:30,690
are statistically equal.