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
and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
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
and standard deviation
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:
![p_1 = \frac{69}{210} = 0.3286](https://tex.z-dn.net/?f=p_1%20%3D%20%5Cfrac%7B69%7D%7B210%7D%20%3D%200.3286)
![s_1 = \sqrt{\frac{0.3286*0.6714}{210}} = 0.0324](https://tex.z-dn.net/?f=s_1%20%3D%20%5Csqrt%7B%5Cfrac%7B0.3286%2A0.6714%7D%7B210%7D%7D%20%3D%200.0324)
Indian Wells:
22 out of 162, so:
![p_2 = \frac{22}{162} = 0.1358](https://tex.z-dn.net/?f=p_2%20%3D%20%5Cfrac%7B22%7D%7B162%7D%20%3D%200.1358)
![s_2 = \sqrt{\frac{0.1358*0.8642}{162}} = 0.0269](https://tex.z-dn.net/?f=s_2%20%3D%20%5Csqrt%7B%5Cfrac%7B0.1358%2A0.8642%7D%7B162%7D%7D%20%3D%200.0269)
Distribution of the difference:
![p = p_1 - p_2 = 0.3286 - 0.1358 = 0.1928](https://tex.z-dn.net/?f=p%20%3D%20p_1%20-%20p_2%20%3D%200.3286%20-%200.1358%20%3D%200.1928)
![s = \sqrt{s_1^2+s_2^2} = \sqrt{0.0324^2 + 0.0269^2} = 0.0421](https://tex.z-dn.net/?f=s%20%3D%20%5Csqrt%7Bs_1%5E2%2Bs_2%5E2%7D%20%3D%20%5Csqrt%7B0.0324%5E2%20%2B%200.0269%5E2%7D%20%3D%200.0421)
a. Find a 99% confidence interval for p1 -p2.
The confidence interval is:
![p \pm zs](https://tex.z-dn.net/?f=p%20%5Cpm%20zs)
In which
z is the z-score that has a p-value of
.
99% confidence level
So
, z is the value of Z that has a p-value of
, so
.
The lower bound of the interval is:
![p - zs = 0.1928 - 2.575*0.0421 = 0.0844](https://tex.z-dn.net/?f=p%20-%20zs%20%3D%200.1928%20-%202.575%2A0.0421%20%3D%200.0844)
The upper bound of the interval is:
![p + zs = 0.1928 + 2.575*0.0421 = 0.3012](https://tex.z-dn.net/?f=p%20%2B%20zs%20%3D%200.1928%20%2B%202.575%2A0.0421%20%3D%200.3012)
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.