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.

The sampling distribution of the difference in sample proportions is approximately normal because samples and proportions are greater than or equal to 10.

Given :

For dear population 1, the sample size is 200 and proportion is 0.06.

Now, for dear population 2, the sample size is 197 and the proportion is 0.086.

Therefore, from the above calculations, it can be concluded that the sampling distribution of the difference in sample proportions is approximately normal because samples and proportions are greater than or equal to 10.

So, the correct option is e).

For more information, refer to the link given below:

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e. The sampling distribution of the difference in sample proportion is approximately normal.

Step-by-step explanation:

(200) (0.06) >= 10

(200) (0.94) >=10

(197)(0.086) >=10

(197)(0.914) >= 10

We test the sample proportions of it is normal :

Deer population 1:

Sample size, n1 = 200

Proportion, p1 = 0.06

q = 1 - p = 1 - 0.06 = 0.94

n1 * p = 200 * 0.06 = 12

n1 * q = 200 * 0.94 = 188

Deer population 2:

Sample size, n2 = 197

Proportion, p = 0.086

q = 1 - p = 1 - 0.086 = 0.914

n2 * p = 197 * 0.086 = 16.94

n2 * q = 197 * 0.914 = 180.06

Since for samples and proportions ;

n*p and nq ≥ 10 ;

We cm conclude that the sampling distribution of the difference in sample mean is appropriately normal.

16

step-by-step explanation:

2 x 40 = 80

80/5

= 16

stops on an even number.