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Arte-miy333 [17]
3 years ago
10

A wildlife biologist is doing research on chronic wasting disease and its impact on the deer populations in northern Colorado. T

o 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.
Mathematics
1 answer:
jonny [76]3 years ago
4 0

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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(9-8) -2.02 \sqrt{\frac{2^2}{25} +\frac{1^2}{20}}= 0.0743

(9-8) +2.02 \sqrt{\frac{2^2}{25} +\frac{1^2}{20}}= 1.926

And we are 9% confidence that the true mean for the difference of the population means is given by:

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Step-by-step explanation:

For this problem we have the following data given:

\bar X_1 = 9 represent the sample mean for one of the departments

\bar X_2 = 8 represent the sample mean for the other department

n_1 = 25 represent the sample size for the first group

n_2 = 20 represent the sample size for the second group

s_1 = 2 represent the deviation for the first group

s_2 =1 represent the deviation for the second group

Confidence interval

The confidence interval for the difference in the true means is given by:

(\bar X_1 -\bar X_2) \pm t_{\alpha/2} \sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}

The confidence given is 95% or 9.5, then the significance level is \alpha=0.05 and \alpha/2 =0.025. The degrees of freedom are given by:

df=n_1 +n_2 -2= 20+25-2= 43

And the critical value for this case is t_{\alpha/2}=2.02

And replacing we got:

(9-8) -2.02 \sqrt{\frac{2^2}{25} +\frac{1^2}{20}}= 0.0743

(9-8) +2.02 \sqrt{\frac{2^2}{25} +\frac{1^2}{20}}= 1.926

And we are 9% confidence that the true mean for the difference of the population means is given by:

0.0743 \leq \mu_1 -\mu_2 \leq 1.926

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3 years ago
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