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.

Complete 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.

(b) What are the mean and standard deviation of the sampling distribution of the difference in sample proportions pˆD−pˆE? Show your work and label each value.

(c) Can it be assumed that the sampling distribution of the difference of the sample proportions pˆD−pˆE is approximately normal? Justify your answer.

(d) Consider your answer in part (a). What is the probability that pˆD−pˆE is greater than the value found in part (a)?

a) 0.1917

b) The mean is 0.1917 and the standard deviation is 0.0914.

c)Yes, the sampling distribution of the difference of the sample proportions pˆD−pˆE is approximately normal

d) 0.1580

Step-by-step explanation:

a) for p`D we have:

= 0.375

For p`E we have:

= 0.1833

Therefore, p`D - p`E, we have:

0.375 - 0.1833

=0.1917

b) The Mean can be calculated as p`D - p`E =

0.375 - 0.1833

=0.1917

For standard deviation:

The mean is 0.1917 and the standard deviation is 0.0914.

c) Yes, the sampling distribution of the difference of the sample proportions pˆD−pˆE is approximately normal, because for normal condition, we have:

i) np ≥ 10

ii) n(1-p) ≥ 10.

From the expressions, we can see the samples satisfy the condition for normality.

d) To get the probability, wen need to find the Z score.

The Z score can be calculated using the formula:

= 1.0029

Therefore,

P(Z > 1.0029) = 1 - P(Z ≤ 1.0029)

From the z distribution table, we have:

P (Z > 1.0029) = 1 - 0.8420 = 0.1580

The probability is 0.1580