### Hypothesis Testing

**Null and Alternative Hypotheses**

Hypothesis testing involves testing the difference between a hypothesized value of a population parameter and the estimate of that parameter which is calculated from a sample. If the parameter of interest is the mean of the populations in hypothesis testing, we are essentially determining the magnitude of the difference between the mean of the sample and they hypothesized mean of the population. If the difference is very large, we reject our hypothesis about the population. If the difference is very small, we do not. Below is an overview of this process.

In statistics, the hypothesis to be tested is called the null hypothesis and given the symbol \begin{align*}H_0\end{align*}. The alternative hypothesis is given the symbol \begin{align*}H_a\end{align*}

The null hypothesis defines a specific value of the population parameter that is of interest. Therefore, the null hypothesis always includes the possibility of equality. Consider

\begin{align*}H_0: \mu & = 3.2\\ H_a: \mu & \neq 3.2\end{align*}

In this situation if our sample mean, \begin{align*}\bar{x}\end{align*}, is very different from 3.2 we would reject \begin{align*}H_0\end{align*}. That is, we would reject \begin{align*}H_0\end{align*} if \begin{align*}\bar{x}\end{align*} is much larger than 3.2 or much smaller than 3.2. This is called a 2-*tailed test.* An \begin{align*}\bar{x}\end{align*} that is very unlikely if \begin{align*}H_0\end{align*} is true is considered to be good evidence that the claim \begin{align*}H_0\end{align*} is not true. Consider \begin{align*}H_0: \mu \le 3.2 \ H_a: \mu > 3.2\end{align*}. In this situation we would reject \begin{align*}H_0\end{align*} for very large values of \begin{align*}\bar{x}\end{align*}. This is called a **one tail test .** If, for this test, our data gives \begin{align*}\bar{x}=15\end{align*}, it would be highly unlikely that finding \begin{align*}\bar{x}\end{align*} this different from 3.2 would occur by chance and so we would probably reject the null hypothesis in favor of the alternative hypothesis.

#### Testing the Null Hypothesis

If we were to test the hypothesis that the seniors had a mean SAT score of 1100 our null hypothesis would be that the SAT score would be equal to 1100 or:

\begin{align*}H_0: \mu = 1100\end{align*}

We test the null hypothesis against an alternative hypothesis, which is given the symbol \begin{align*}H_a\end{align*} and includes the outcomes not covered by the null hypothesis. Basically, the alternative hypothesis states that there is a difference between the hypothesized population mean and the sample mean. The alternative hypothesis can be supported only by rejecting the null hypothesis. In our example above about the SAT scores of graduating seniors, our alternative hypothesis would state that there is a difference between the null and alternative hypotheses or:

\begin{align*}H_a: \mu \neq 1100\end{align*}

Let’s take a look at examples and develop a few null and alternative hypotheses.

#### Developing Null and Alternative Hypotheses

1. We have a medicine that is being manufactured and each pill is supposed to have 14 milligrams of the active ingredient. What are our null and alternative hypotheses?

Solution:

\begin{align*}H_0 : \mu &=14\\ H_a : \mu &\neq 14\end{align*}

Our null hypothesis states that the population has a mean equal to 14 milligrams. Our alternative hypothesis states that the population has a mean that is different than 14 milligrams. This is two tailed.

2. The school principal wants to test if it is true what teachers say -- that high school juniors use the computer an average 3.2 hours a day. What are our null and alternative hypotheses?

\begin{align*}H_0: \mu &= 3.2\\ H_a: \mu & \neq 3.2\end{align*}

Our null hypothesis states that the population has a mean equal to 3.2 hours. Our alternative hypothesis states that the population has a mean that differs from 3.2 hours. This is two tailed.

### Example

#### Example 1

eHealthInsurance claims that in 2011, the average monthly premium paid for individual health coverage was $183. Suppose you are suspicious that the average, or mean, cost is actually higher. State the null and alternative hypothesis you would use to test this.

The original claim from eHealthInsurance is that \begin{align*}\mu = 183\end{align*}. Your theory is that \begin{align*}\mu >183\end{align*}. Since the original claim has equality in it, we'll put that in the null hypothesis.

\begin{align*}H_0: \mu &=183\\ H_a: \mu & > 183\end{align*}

### Review

- If the difference between the hypothesized population mean and the mean of the sample is large, we ___ the null hypothesis. If the difference between the hypothesized population mean and the mean of the sample is small, we ___ the null hypothesis.
- At the Chrysler manufacturing plant, there is a part that is supposed to weigh precisely 19 pounds. The engineers take a sample of parts and want to know if they meet the weight specifications. What are our null and alternative hypotheses?

For 3-5, determine whether each of the following is a null or an alternative hypothesis.

- The average weight of golden retriever dogs is the same as the average weight of pit bull dogs.
- The proportion of books in the library that are novels is higher than the proportion of books in the library that are nonfiction.
- The average price of wool coats in San Francisco is lower in the summer than in the winter.

For 6-9, for each of the following write the alternative hypothesis.

- \begin{align*}H_0: p=0.30\end{align*} and the test is two sided.
- \begin{align*}H_0: p=0.35\end{align*} and the test is left sided.
- \begin{align*}H_0: p=0.55\end{align*} and the test is right sided.
- \begin{align*}H_0: \mu=500\end{align*} and the test is two sided.
- Suppose the present success rate in treating a certain type of lung cancer is .75. A research group hopes to demonstrate that the success rate of a new treatment of this cancer is better. Write the null and alternative hypotheses.

### Review (Answers)

To view the Review answers, open this PDF file and look for section 8.1.