• Random Samples are better than Convenience Sampling
  • A random sample
  • A stratified sample
  • A cluster sample

• Central Limit Theorem
  • Let \(x\) be a random variable.
  • Select a sample of size \(n\) from the distribution of \(x\)
  • Even if \(x\) is not normally distributed, the mean of \(x\) is!
  • The sampling distribution is normally distributed
  • The mean is equal to \(\overline{x}=\frac{1}{n}\sum_{k=1}^nx_k\)
  • The standard deviation is equal to \(s_{\overline{x}}=\frac{s}{\sqrt{n}}\)
  • Note as \(n\rightarrow\infty\) the sample standard deviation goes to zero, \(s_{\overline{x}}\rightarrow0\)

Suppose a random sample of size n = 49 is selected from a population with mean \(\mu\) = 8 and standard deviation \(\sigma\) = 7.

What is the probability that the sample mean is between 7.8 and 8.2?

You can answer this question by taking the following steps

1. Change the x values into z values.
2. Find the cumulative probability for each Z value.
3. Subtract the two probability values

Step 1 Let's recall the z-score formula for the sample mean \[z=\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}\]

Now, let's write the probability we want to find \[Pr(7.8\leq\overline{x}\leq8.2)\]

The z-score for the lower bound is (7.8-8)/1/1=-0.2

The z-score for the upper bound is (8.2-8)/1/1=0.2

\[Pr(-0.2\leq z\leq 0.2)\]

Step 2

Use either a table or excel to find the \(Pr(z\leq -0.2)\) and \(Pr(z\leq 0.2)\)

\(Pr(z\leq -0.2)=0.421\) and \(Pr(z\leq 0.2)=0.579\)

Step 3

We need to subract the two probabilities.

\(Pr(z\leq 0.2)-Pr(z\leq -0.2)=0.579 - 0.421 = 0.158\)

We can do the same thing with proportions.

Suppose that the population proportion of voters that support gay marriage is 60 percent, what is the probability that a sample of 400 voters will yield a proportion equal to or less than 50 percent?

i.e. if \(\pi=0.60\) and \(n=400\), what is the \(Pr(p \leq 0.5)\)

In the case of proportions, we need to use a different z-score. \[z=\frac{p-\pi}{\sqrt{\frac{\pi(1-\pi)}{n}}}\]

The standard deviation of the proportion is \(\sigma_{p}=\sqrt{\frac{\pi(1-\pi)}{n}}=\sqrt{\frac{0.6(0.4)}{400}}=0.0245\)

The z-score is equal to \(\frac{0.5-0.6}{.0245}=-4.08\)

\(Pr(z<-4.08)=.00002\)