Question

On the Navajo Reservation, a random sample of 210 permanent dwellings in the Fort Defiance region showed that 69 were traditional Navajo hogans. In the Indian Wells region, a random sample of 162 permanent dwellings showed that 22 were traditional hogans. Let p1 be the population proportion of all traditional hogans in the Fort Defiance region, and let p2 be the population proportion of all traditional hogans in the Indian Wells region.
Required:
a. Find a 99% confidence interval for p 1 - P2.
b. Examine the confidence interval and comment on its meaning. Does it include numbers that are all positive?

Answer 1

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 $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

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 $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

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:

$p_1 = \frac{69}{210} = 0.3286$

$s_1 = \sqrt{\frac{0.3286*0.6714}{210}} = 0.0324$

Indian Wells:

22 out of 162, so:

$p_2 = \frac{22}{162} = 0.1358$

$s_2 = \sqrt{\frac{0.1358*0.8642}{162}} = 0.0269$

Distribution of the difference:

$p = p_1 - p_2 = 0.3286 - 0.1358 = 0.1928$

$s = \sqrt{s_1^2+s_2^2} = \sqrt{0.0324^2 + 0.0269^2} = 0.0421$

a. Find a 99% confidence interval for p1 -p2.

The confidence interval is:

$p \pm zs$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$.

99% confidence level

So $\alpha = 0.01$, z is the value of Z that has a p-value of $1 - \frac{0.01}{2} = 0.995$, so $Z = 2.575$.

The lower bound of the interval is:

$p - zs = 0.1928 - 2.575*0.0421 = 0.0844$

The upper bound of the interval is:

$p + zs = 0.1928 + 2.575*0.0421 = 0.3012$

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.