Question

Different varieties of the tropical flower Heliconia are fertilized by different species of hummingbirds. Over time, the lengths of the flowers and the form of the hummingbirds' beaks have evolved to match each other. Here are data on the lengths in millimeters of three varieties of these flowers on the island of Dominica. data332.dat

Do a complete analysis that includes description of the data and a significance test to compare the mean lengths of the flowers for the three species. (Round your answers for x to four decimal places, s to three decimal places, and s_(x^^\_) to three decimal places. Round your test statistic to two decimal places. Round your P-value to three decimal places.)

flower type n x^^\_ s s_(x^^\_)

H. bihai

H. caribaea red H.

caribaea yellow

F =

P =

variety length
bihai   49.62
bihai   46.47
bihai   48.14
bihai   47.49
bihai   47.82
bihai   47.48
bihai   48.03
bihai   46.99
bihai   46.57
bihai   51.08
bihai   45.65
bihai   49.78
bihai   49.14
bihai   47.77
bihai   46.91
bihai   47.77
red     37.12
red     41.89
red     38.85
red     39.33
red     40.72
red     40.37
red     41.27
red     41.67
red     39.5
red     41.93
red     41.09
red     42.74
red     40.78
red     39.27
red     40.42
red     41.52
red     40.6
red     38.67
red     37.94
red     41.86
red     37.44
red     42.01
red     40.62
yellow  35.07
yellow  36.23
yellow  34.55
yellow  36.56
yellow  35.33
yellow  36.96
yellow  36.23
yellow  33.76
yellow  34.25
yellow  35.41
yellow  35.25
yellow  37.49
yellow  35.04
yellow  37.48
yellow  37.15

Answer 1

Following is the output of one-way ANOVA analysis:

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance
bihai	16	766.71	47.919375	1.952032917
red	23	927.61	40.33087	2.422090119
yellow	15	536.76	35.784	1.40734
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	1179.03942	2	589.51971	293.9838641	1.01E-28	3.178799
Within Groups	102.269236	51	2.0052791
Total	1281.30866	53

1)

Groups	Count	Average	Sd	se
Bihai	16	47.9194	1.397	0.349
Red	23	40.3309	1.556	0.325
Yellow	15	35.784	1.186	0.306

Here

Hypotheses are:

H0: The mean lengths of the flowers for the three species is same.

Ha: The mean lengths of the flowers for the three species is not same.

The F test statistics:

F = 293.98

The p-value:

p-value = 0.000

Since p-value is less than 0.05 so we reject the null hypothesis.