Respuesta :
Answer:
a) The 99% confidence interval for the difference of proportions is (0.0844, 0.3012).
b) We are 99% sure that the true difference in proportions is between 0.0844 and 0.3012. Since all values are positive, there is significant evidence at the 1 - 0.99 = 0.01 significance level to conclude that the proportion is the Fort Defiance region is higher than in the Indian Wells region.
Step-by-step explanation:
Before finding the confidence interval, we need to understand the central limit theorem and subtraction of normal variables.
Central Limit Theorem
The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]
Subtraction between normal variables:
When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.
Fort Defiance:
69 out of 210, so:
[tex]p_1 = \frac{69}{210} = 0.3286[/tex]
[tex]s_1 = \sqrt{\frac{0.3286*0.6714}{210}} = 0.0324[/tex]
Indian Wells:
22 out of 162, so:
[tex]p_2 = \frac{22}{162} = 0.1358[/tex]
[tex]s_2 = \sqrt{\frac{0.1358*0.8642}{162}} = 0.0269[/tex]
Distribution of the difference:
[tex]p = p_1 - p_2 = 0.3286 - 0.1358 = 0.1928[/tex]
[tex]s = \sqrt{s_1^2+s_2^2} = \sqrt{0.0324^2 + 0.0269^2} = 0.0421[/tex]
a. Find a 99% confidence interval for p1 -p2.
The confidence interval is:
[tex]p \pm zs[/tex]
In which
z is the z-score that has a p-value of [tex]1 - \frac{\alpha}{2}[/tex].
99% confidence level
So [tex]\alpha = 0.01[/tex], z is the value of Z that has a p-value of [tex]1 - \frac{0.01}{2} = 0.995[/tex], so [tex]Z = 2.575[/tex].
The lower bound of the interval is:
[tex]p - zs = 0.1928 - 2.575*0.0421 = 0.0844[/tex]
The upper bound of the interval is:
[tex]p + zs = 0.1928 + 2.575*0.0421 = 0.3012[/tex]
The 99% confidence interval for the difference of proportions is (0.0844, 0.3012).
Question b:
We are 99% sure that the true difference in proportions is between 0.0844 and 0.3012. Since all values are positive, there is significant evidence at the 1 - 0.99 = 0.01 significance level to conclude that the proportion is the Fort Defiance region is higher than in the Indian Wells region.
