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.

Question 2
(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)? Show your work

Question

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Respuesta :

yemmy yemmy · Answer 1 · 2020-05-01T06:43:50+02:00

Answer:

a. 0.1917

b. 0.0914

d. 0.1580

Step-by-step explanation:

(a)

[tex]P^D = \frac{15}{40} = 0.375[/tex]

[tex]P^E = \frac{11}{60} =0.8133[/tex]

Mean, [tex]\sigma_{P^D-P^E} = P^D-P^E[/tex] = 0.375 -0.1833 = 0.1917

(b) sample prop ? Show your work and label each value.

Mean, = = 0.1917

Standard deviation = [tex]\sqrt{\frac{P^D(1-P^D)}{N_D} +\frac{P^E(1-P^E)}{N_E} }[/tex]

Standard deviation = [tex]\sqrt{\frac{0.375(1-0.375)}{40} +\frac{0.1833(1 - 0.1833)}{60} }[/tex]

Standard deviation = 0.0914

(c)

Normality condition:

np ≥ 10 and n(1-p) ≥ 10

Both the samples satisfy the normality condition.

(d)

The probability is obtained by calculating the z score,

[tex]z = \frac{(P^D-P^E)-(P^d-P^e)}{\sigma_{P^D - P^E}}[/tex]

[tex]z = \frac{0.1917-0.1}{0.0914}[/tex] = 1.0029

P(z > 1.0029) = 1 - P(z ≤ 1.0029)

The probability is obtained from the z distribution table,

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