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

answer:

it is general formula that is used to write value to its excet position

a11 a12 a13

a21 a22 a23

it is the example of order 2 by 3 matrix in this

a11 means value occur in 1 st row 1st column

a12 means value occurs in 1st row 2nd column

a13 means value occur in 1st row and 3rd column

a21 means value occurs in 2nd row and 1st column

a22 means value occurs in 2nd row and in 2nd column

a23 means value occurs in 2nd row and in 3rd column

step-by-step explanation:

1