Question

Researchers are studying two populations of sea turtles. In population D, 30 percent of the turtles have a shell length greater than 2 feet. In population E, 20 percent of the turtles have a shell length greater than 2 feet. From a random sample of 40 turtles selected from D, 15 had a shell length greater than 2 feet. From a random sample of 60 turtles selected from E, 11 had a shell length greater than 2 feet. Let pˆD represent the sample proportion for D, and let pˆE represent the sample proportion for E.
(a) What is the value of the difference pˆD−pˆE? Show your work.

Answer 1

Answer:

$p^D-p^E=0.192$  

Step-by-step explanation:

We are given the following in the question:

Population D:

30% of the turtles have a shell length greater than 2 feet.

Sample size,

$n_D = 40$

Number of turtles that had shell length greater than 2 feet,

$x_D = 15$

Sample proportion:

$p^D=\dfrac{x_D}{n_D} =\dfrac{15}{40} = 0.375$

Population E:

20% of the turtles have a shell length greater than 2 feet.

Sample size,

$n_E = 60$

Number of turtles that had shell length greater than 2 feet,

$x_E = 11$

Sample proportion:

$p^E=\dfrac{x_E}{n_E} =\dfrac{11}{60} = 0.183$

We have to find the difference between the sample proportion.

Difference in sample proportion =

$=p^D-p^E\\=0.375-0.183\\=0.192$