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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)?
Answer:
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:
[tex] \frac{15}{40} [/tex]
= 0.375
For p`E we have:
[tex] \frac{11}{60} [/tex]
= 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:
[tex] s.d = \sqrt{\frac{p`D (1-p`D)}{N_D} + \frac{p`E(1-p`E)}{N_E}}[/tex]
[tex] s.d = \sqrt{\frac{0.375(1 - 0.375)}{40} + \frac{0.1833(1 - 0.1833)}{60}}= 0.0914[/tex]
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:
[tex] Z = \frac{(p`D - p`E) -(pD - pE)}{s.d}[/tex]
[tex] = \frac{(0.1917) -(0.1)}{0.0914}[/tex]
= 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
The value of the difference pˆD−pˆE is 0.1917. The mean and standard deviation is 0.1917 and 0.0914 respectively and the probability that pˆD−pˆE is greater than the 0.1917 is 0.1580.
Given :
- 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.
a) The value of pˆD is:
[tex]=\dfrac{15}{40}[/tex]
The value of pˆE is:
[tex]=\dfrac{11}{60}[/tex]
So, the value of (pˆD - pˆE) is:
[tex]=\dfrac{15}{40}-\dfrac{11}{60}[/tex]
= 0.375 - 0.1833
= 0.1917
b) Mean is given by the formula:
pˆD - pˆE = 0.1917
For standard deviation using the formula:
[tex]\rm SD =\sqrt{ \dfrac{p\hat{}D(1-p\hat{}D)}{N_D}+\dfrac{p\hat{}E(1-p\hat{}E)}{N_E}}[/tex]
[tex]\rm SD = \sqrt{\dfrac{0.375(1-0.375)}{40}+\dfrac{0.1833(1-0.1833)}{60}}[/tex]
SD = 0.0914
c). Yes, the sampling distribution of the difference of the sample proportions pˆD−pˆE is approximately normal.
d). To determine the probability, first evaluate the z-score.
[tex]\rm Z=\dfrac{(p\hat{}D-p\hat{}E)-(pD-pE)}{SD}[/tex]
[tex]\rm Z = \dfrac{0.1917-0.1}{0.0914}[/tex]
Z = 1.0029
Now, P(Z>1.0029) = 1 - P(Z [tex]\leq[/tex] 1.0029)
= 1 - 0.8420
= 0.1580
For more information, refer to the link given below:
https://brainly.com/question/2561151
