44. Tropical flowers Different varieties of the tropical flower Heliconia are fertilized by different species of hummingbirds.

Researchers believe that over time, the lengths of the flowers and the forms of the hummingbirds’ beaks have evolved to match each other. Here are data on the lengths in millimeters for random samples of two color varieties of the same species of flower on the island of Dominica:

(a) A Fathom dotplot of the data is shown below. Use the graph to answer these questions:

• Are the centers of the two groups similar or different? Explain
• Are the spreads of the two groups similar or different? Explain.

(b) Construct and interpret a 95% confidence interval for the difference in the mean lengths of these two varieties of flowers.
(c) Does the interval support the researchers’ belief that the two flower varieties have different average lengths? Explain.

To find the centers and spreads of the two groups similar or different from the given data:

Between the dot on the far left and the dot on the far right, the center is nearly in the middle. Then notice that the upper dotplot's center is closer to the right than the left, giving the impression that the centers are different.
The spread is the distance between the farthest left and farthest right dots.
Then notice that the top dotplot's spread appears to be greater than the bottom dotplot's spread, implying that the spread is different.
Hence, the Centers are different and the Spread also different.

A $95\%$ confidence interval for the difference in the mean lengths of these two varieties of flowers

Determine the mean:
${\overline{x}}_1=39.6983$
${\overline{x}}_2=36.18$
Then determine the variance:
$s_1=1.7865$
$s_2=0.9753$
Calculate the degrees of freedom:
$$\begin{gathered} df=\min \left(n_1-1,n_2-1\right) \\ =min(23-1,15-1) \\ =14 \end{gathered}$$

Using table $B$,to find $t^{*}$with $df=14$and $c=95\%$

$t^{*}=2.145$

Since the Confidence interval for${\mu }_1-{\mu }_2$ are:
$$\begin{gathered} \left({\overline{x}}_1-{\overline{x}}_2\right)-t_{\alpha /2}\times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}=(39.6983-36.18)-2.145\times \sqrt{\frac{1.{7865}^2}{23}+\frac{0.{9753}^2}{15}} \\ =2.5538 \end{gathered}$$

Then,

$$\begin{gathered} \left({\overline{x}}_1-{\overline{x}}_2\right)+t_{\alpha /2}\times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}=(39.6983-36.18)+2.145\times \sqrt{\frac{1.{7865}^2}{23}+\frac{0.{9753}^2}{15}} \\ =4.4828 \end{gathered}$$

Hence, the $95\%$confidence interval for the difference in the mean lengths of these two varieties of flowers is$2.5538$and $4.4828$.

To explain the interval support the researchers’ belief that the two flower varieties have different average lengths.

As a result from part (b):
$2.5538,4.4828$
Then notice that the upper dotplot's spread appears to be greater than the lower dotplot's spread, implying that the spreads are different.

Hence, the answer is yes, the interval support the researchers’ belief that the two flower varieties have different average lengths.