Question

Consider a population of people who suffer occasionally from migraine headache. Suppose 62% of those people get some relief from taking ibuprofen (true proportion). To investigate the effectiveness of ibuprofen, a random sample of 100 people is obtained at random from this population. A. (4 pts.) Determine the sampling distribution of sample proportion. Also, find the mean and standard deviation of the sampling distribution.

Answer 1

Answer:

The sampling distribution of sample proportion is approximately normal, with mean 0.62 and standard deviation 0.0485.

Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem estabilishes 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 $s = \sqrt{\frac{p(1-p)}{n}}$

62% of those people get some relief from taking ibuprofen (true proportion).

This means that $p = 0.62$

Sample of 100

This means that $n = 100$

A. (4 pts.) Determine the sampling distribution of sample proportion. Also, find the mean and standard deviation of the sampling distribution.

By the Central Limit Theorem, it is approximately normal with

Mean $\mu = p = 0.62$

Standard deviation $s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.62*0.38}{100}} = 0.0485$