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Answer: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.
The sampling distribution of the difference in sample proportions is approximately normal because samples and proportions are greater than or equal to 10.
Given :
- A wildlife biologist is doing research on chronic wasting disease and its impact on the deer populations in northern Colorado.
- A random sample of 200 deer was obtained from one region and a random sample of 197 deer was obtained from the other region.
For dear population 1, the sample size is 200 and proportion is 0.06.
[tex]\rm q_1 = 1-p_1=1-0.06=0.94[/tex]
[tex]\rm n_1p_1=200\times 0.06=12\\[/tex]
[tex]\rm n_1q_1=200\times 0.94 = 188[/tex]
Now, for dear population 2, the sample size is 197 and the proportion is 0.086.
[tex]\rm q_2 = 1-p_2=1-0.086=0.914[/tex]
[tex]\rm n_2p_2=197\times 0.086=16.94[/tex]
[tex]\rm n_2q_2=197\times 0.914 = 180.06[/tex]
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:
https://brainly.com/question/795909
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