Researchers are studying two populations of sea turtles. In population D, 30 percent of the turtles have a shell length greater than 2 feet.

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.

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

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Adalynn 3 weeks 2022-01-03T12:21:25+00:00 1 Answer 0 views 0

Answers ( )

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    2022-01-03T12:22:41+00:00

    Answer:

    a. 0.1917

    b. 0.0914

    d. 0.1580

    Step-by-step explanation:

    (a)

     P^D = \frac{15}{40} = 0.375

     P^E = \frac{11}{60} =0.8133

    Mean, \sigma_{P^D-P^E} = P^D-P^E = 0.375 -0.1833 = 0.1917

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

    Mean,   =  = 0.1917

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

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

    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,

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

    z = \frac{0.1917-0.1}{0.0914} = 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

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