A wildlife biologist is doing research on chronic wasting disease and its impact on the deer populations in northern Colorado. To estimate the difference between the proportions of deer with chronic wasting disease in two different regions, a random sample of 200 deer was obtained from one region and a random sample of 197 deer was obtained from the other region. The biologist checked for the following.(200) (0.06) >= 10 (200) (0.94) >=10 (197)(0.086) >=10 (197)(0.914) >= 10Which of the following conditions for inference was the biologist checking? a. The population of deer within each region is approximately normal. b. it is reasonable to generalize from the samples to the populations. c. The samples are independent of each other. d. The observations within each sample are close to independent. e. The sampling distribution of the difference in sample proportions is approximately normal Prisen.

Mathematics College

A wildlife biologist is doing research on chronic wasting disease and its impact on the deer populations in northern Colorado. To estimate the difference between the proportions of deer with chronic wasting disease in two different regions, a random sample of 200 deer was obtained from one region and a random sample of 197 deer was obtained from the other region. The biologist checked for the following.

(200) (0.06) >= 10
(200) (0.94) >=10
(197)(0.086) >=10
(197)(0.914) >= 10

Which of the following conditions for inference was the biologist checking?

a. The population of deer within each region is approximately normal.
b. it is reasonable to generalize from the samples to the populations.
c. The samples are independent of each other.
d. The observations within each sample are close to independent.
e. The sampling distribution of the difference in sample proportions is approximately normal Prisen.

Answers

Answer 1

Answer: The sampling distribution of the difference in sample proportions is approximately normal

Step-by-step explanation:

Given that :

Random sample of deer from one region = 200

p = 0.06

1 - p = 1 - 0.06 = 0.94

Random sample of deer from other region = 197

p = 8.6% = 0.086

1 - p = 1 - 0.086 = 0.914

For an approximately normal distribution :

np ≥ 10

n(1 - p) ≥ 10

(200) (0.06) = 12

(200) (0.94) = 188

(197)(0.086) = 16.942

(197)(0.914) = 180.058

Since all conditions are met :

Then, the sampling distribution of the difference in sample proportion is approximately normal.

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Answer:

a volleyball team with 9 wins and 6 losses

Step-by-step explanation:

Jim’s soccer team has a record of 12 wins and 8 losses.Hence when divded

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Step-by-step explanation:

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Answer:

Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation:

Robert total earnings:

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Answers

Answer:

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Answers

Answer:

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Answers

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Answers

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Answers

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