Q. 46 Different varieties of the tropi... [FREE SOLUTION]

91影视

The Practice of Statistics for AP Examination

Daren Starnes, Josh Tabor

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 837 pages

ISBN: 9781319113339

Chapter 10: Q. 46 (page 667)

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鈥� 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 dot-plot of the data is shown, along with summary statistics. Write a few sentences comparing the distributions.
b. Construct and interpret a $95 %$ confidence interval for the difference in the true mean lengths of these two varieties of flowers.

Short Answer

Expert verified

Part a. There appear to be no outliers in either distribution because there are no dots that appear to be unusually far from the other dots in the dot plot.

Part b. There is $95 %$ confidence that the mean length of H. caribaea red is between $(2.5538, 4.4822)$ millimeters higher than the mean length of H. caribaea yellow.

Step by step solution

Part a. Step 1. Explanation

A dot plot of the data is shown in the question along with the summary statistics. Thus, on comparing the two distribution we see that:

Shape: The distribution of red is skewed to the right because most of the dots lie near the left of the dot plot and the distribution of yellow is skewed to the left because most of the dots lie near the right of the dot plot.

Center: The center for the red distribution is higher than the center for the yellow because most of the dots in the dot plot of red lie to the right of most of the dots in the dot plot of yellow and the mean and the median are both greater for the red.

Spread: The distribution of red is more variable than the distribution of yellow because the dot plot of the red is wider than the dot plot of the yellow and the standard deviation of the red is higher than the standard deviation of yellow.

Unusual features: There appear to be no outliers in either distribution because there are no dots that appear to be unusually far from the other dots in the dot plot.

Part b. Step 1. Given information

Given:

${\overset{炉}{x}}_{1} = 39.698 {\overset{炉}{x}}_{2} = 36.18 n_{1} = 23 n_{2} = 15 s_{1} = 1.786 s_{2} = 0.975 c = 0.95$

Part b. Step 2. Calculation

Now we will be calculating the t-value for this we need to find out the degree of freedom. Thus, the degree of freedom will be:

$d f = m i n (n_{1} - 1, n_{2} - 1) = m i n (23 - 1, 15 - 1) = 14$

Then the t-value will be as:

$t_{伪 / 2} = 2.145$

Thus the confidence interval be:

$({\overset{炉}{x}}_{1} - {\overset{炉}{x}}_{2}) - t_{伪 / 2} 脳 \sqrt{\frac{s_{1}^{2}}{n_{1}} + \frac{s_{2}^{2}}{n_{2}}} = (39.698 - 36.18) - 2.145 脳 \sqrt{\frac{1 . 786^{2}}{23} + \frac{0 . 975^{2}}{15}} = 3.518 - 0.942 = 2.5538 ({\overset{炉}{x}}_{1} - {\overset{炉}{x}}_{2}) + t_{伪 / 2} 脳 \sqrt{\frac{s_{1}^{2}}{n_{1}} + \frac{s_{2}^{2}}{n_{2}}} = (39.698 - 36.18) + 2.145 脳 \sqrt{\frac{1 . 786^{2}}{23} + \frac{0 . 975^{2}}{15}} = 3.518 + 0.9642 = 4.4822$

Thus, we conclude that there is $95 %$ confidence that the mean length of H. caribaea red is between $(2.5538, 4.4822)$ millimeters higher than the mean length of H. caribaea yellow.