Question

In a certain area of a large city, it is hypothesized that 40 percent of the houses are in a dilapidated condition. A random sample of 75 houses from this section and 90 houses from another section yielded a difference of 0.09. If there is no difference between the two areas in the proportion of dilapidated houses, what is the probability of observing a difference this large or larger?

Answer 1

Using the normal distribution and the central limit theorem, it is found that there is a 0.24 = 24% probability of observing a difference this large or larger.

Normal Probability Distribution

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

• It measures how many standard deviations the measure is from the mean.

• After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

• By the Central Limit Theorem, for a proportion p in a sample of size n, the sampling distribution of sample proportion is approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1 - p)}{n}}$, as long as $np \geq 10$ and $n(1 - p) \geq 10$.

• When two variables are subtracted, the mean is the subtraction of the means, while the standard deviation is the square root of the sum of the variances.

In this problem, for each sample, the mean and standard error are given by:

$\mu_I = 0.4, s_I = \sqrt{\frac{0.4(0.6)}{75}} = 0.0566$

$\mu_{II} = 0.4, s_{II} = \sqrt{\frac{0.4(0.6)}{90}} = 0.0516$

Hence, for the distribution of differences, the mean and the standard error are given by:

$\mu = \mu_I - \mu_{II} = 0.4 - 0.4 = 0$

$s = \sqrt{s_I^2 + s_{II}^2} = \sqrt{0.0566^2 + 0.0516^2} = 0.0766$

The probability of observing a difference this large or larger is given by P(|Z| > Zx), in which Zx is the z-score when X = 0.0766. Hence:

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem:

$Z = \frac{X - \mu}{s}$

$Z = \frac{0.09 - 0}{0.0766}$

Z = 1.175.

Hence the probability is P(|Z| > 1.175), which is 2 multiplied by the p-value of Z = -1.175. Then:

2 x 0.12 = 0.24.

0.24 = 24% probability of observing a difference this large or larger.

To learn more about the normal distribution and the central limit theorem, you can check brainly.com/question/24663213