# Chapter 6 Introduction to Hypothesis Testing

## 6.1 Module Overview

In this module, we will continue our discussion on statistical inference with a discussion on hypothesis testing. In hypothesis testing, we take a more active approach to our data by asking questions about population parameters and developing a framework to answer those questions. We will root this discussion in confidence intervals before learning about several other approaches to hypothesis testing.

Module Learning Outcomes/Objectives

Test one sample means using

1. confidence intervals.
2. the critical value approach.
3. the p-value approach.

This module’s outcomes correspond to course outcomes (6) apply statistical inference techniques of parameter estimation such as point estimation and confidence interval estimation and (7) apply techniques of testing various statistical hypotheses concerning population parameters.

## 6.2 Logic of Hypothesis Testing

One of our goals with statisticial inference is to make decisions or judgements about the value of a parameter. A confidence interval is a good starting point, but we might also want to ask questions like

• Do cans of soda actually contain 12 oz?
• Is Medicine A better than Medicine B?

A hypothesis is a statement that something is true. A hypothesis test involves two (competing) hypotheses:

1. The null hypothesis, denoted $$H_0$$, is the hypothesis to be tested. This is the “default” assumption.
2. The alternative hypothesis, denoted $$H_A$$ is the alternative to the null.

Note that the subscript 0 is “nought” (pronounced “not”). A hypothesis test helps us decide whether the null hypothesis should be rejected in favor of the alternative.

Example: Cans of soda are labeled with “12 FL OZ.” Is this accurate?

The default, or uninteresting, assumption is that cans of soda contain 12 oz.

• $$H_0$$: the mean volume of soda in a can is 12 oz.
• $$H_A$$: the mean volume of soda in a can is NOT 12 oz.

We can write these hypotheses in words (as above) or in statistical notation. The null specifies a single value of $$\mu$$:

• $$H_0$$: $$\mu=\mu_0$$

We call $$\mu_0$$ the null value. When we run a hypothesis test, $$\mu_0$$ will be replaced by some number. For the soda can example, the null value is 12. We would write $$H_0: \mu = 12$$.

The alternative specifies a range of possible values for $$\mu$$:

• $$H_A$$: $$\mu\ne\mu_0$$. “The mean is different from the null value.”
The Logic of Hypothesis Testing

Take a random sample from the population. If the data area consistent with the null hypothesis, do not reject the null hypothesis. If the data are inconsistent with the null hypothesis and supportive of the alternative hypothesis, reject the null in favor of the alternative.

Example: One way to think about the logic of hypothesis testing is by comparing it to the U.S. court system. In a jury trial, jurors are told to assume the defendant is “innocent until proven guilty.” Innocence is the default assumption, so

• $$H_0$$: the defendant is innocent.
• $$H_A$$: the defendant is guilty.

Like in hypothesis testing, it is not the jury’s job to decide if the defendant is innocent. That should be their default assumption. They are only there to decide if the defendant is guilty or if there is not enough evidence to override that default assumption. The burden of proof lies on the alternative hypothesis.

Notice the careful language in the logic of hypothesis testing: we either reject, or fail to reject, the null hypothesis. We never “accept” a null hypothesis.

### 6.2.1 Decision Errors

• A Type I Error is rejecting the null when it is true. (Null is true, but we conclude null is false.)
• A Type II Error is not rejecting the null when it is false. (Null is false, but we do not conclude it is false.)
$$H_0$$ is
True False
Decision Do not reject $$H_0$$ Correct decision Type II Error
Reject $$H_0$$ Type I Error Correct decision

Example: In our jury trial,

• $$H_0$$: the defendant is innocent.
• $$H_A$$: the defendant is guilty.

A Type I error is concluding guilt when the defendant is innocent. A Type II error is failing to convict when the person is guilty.

How likely are we to make errors? Well, $$P($$Type I Error$$)=\alpha$$, the significance level. (Yes, this is the same $$\alpha$$ we saw in confidence intervals!) For Type II error, $$P($$Type II Error$$)=\beta$$. This is related to the sample size calculation from the previous module, but is otherwise something we don’t have time to cover.

We would like both $$\alpha$$ and $$\beta$$ to be small but, like many other things in statistics, there’s a trade off! For a fixed sample size,

• If we decrease $$\alpha$$, then $$\beta$$ will increase.
• If we increase $$\alpha$$, then $$\beta$$ will decrease.

In practice, we set $$\alpha$$ (as we did in confidence intervals). We can improve $$\beta$$ by increasing sample size. Since resources are finite (we can’t get enormous sample sizes all the time), we will need to consider the consequences of each type of error.

Example We could think about assessing consequences through the jury trial example. Consider two possible charges:

1. Defendant is accused of stealing a loaf of bread. If found guilty, they may face some jail time and will have a criminal record.
2. Defendant is accused of murder. If found guilty, they will have a felony and may spend decades in prison.

Since these are moral questions, I will let you consider the consequences of each type of error. However, keep in mind that we do make scientific decisions that have lasting impacts on people’s lives.

Hypothesis Test Conclusions

• If the null hypothesis is rejected, we say the result is statistically significant. We can interpret this result with:
• At the $$\alpha$$ level of significance, the data provide sufficient evidence to support the alternative hypothesis.
• If the null hypothesis is not rejected, we say the result is not statistically significant. We can interpret this result with:
• At the $$\alpha$$ level of significance, the data do not provide sufficient evidence to support the alternative hypothesis.

Notice that these conclusions are framed in terms of the alternative hypothesis, which is either supported or not supported. We will never conclude the null hypothesis. Finally, when we write these types of conclusions, we will write them in the context of the problem.

## 6.3 Confidence Interval Approach to Hypothesis Testing

We can use a confidence interval to help us weigh the evidence against the null hypothesis. A confidence interval gives us a range of plausible values for $$\mu$$. If the null value is in the interval, then $$\mu_0$$ is a plausible value for $$\mu$$. If the null value is not in the interval, then $$\mu_0$$ is not a plausible value for $$\mu$$.

1. State null and alternative hypotheses.
2. Decide on significance level $$\alpha$$. Check assumptions (decide which confidence interval setting to use).
3. Find the critical value.
4. Compute confidence interval.
5. If the null value is not in the confidence interval, reject the null hypothesis. Otherwise, do not reject.
6. Interpret results in the context of the problem.

Example: Is the average mercury level in dolphin muslces different from $$2.5\mu g/g$$? Test at the 0.05 level of significance. A random sample of $$19$$ dolphins resulted in a mean of $$4.4 \mu g/g$$ and a standard deviation of $$2.3 \mu g/g$$.

1. $$H_0: \mu = 2.5$$ and $$H_A: \mu \ne 2.5$$.
2. Significance level is $$\alpha=0.05$$. The value of $$\sigma$$ is unknown and $$n = 19 < 30$$, so we are in setting 3.
3. For setting 3, the critical value is $$t_{df, \alpha/2}$$. Here, $$df=n-1=18$$ and $$\alpha/2 = 0.025$$:
qt(0.025, 18)
## [1] -2.100922
1. The confidence interval is \begin{align} \bar{x} &\pm t_{df, \alpha/2}\frac{s}{\sqrt{n}} \\ 4.4 &\pm 2.101 \frac{2.3}{\sqrt{19}} \\ 4.4 &\pm 1.109 \end{align} or $$(3.29, 5.51)$$.
2. Since the null value, $$2.5$$, is not in the interval, it is not a plausible value for $$\mu$$ (at the 95% level of confidence). Therefore we reject the null hypothesis.
3. At the 0.05 level of significance, the data provide sufficient evidence to conclude that the true mean mercury level in dolphin muscles is greater than $$2.5\mu g/g$$.

Note: The alternative hypothesis is “not equal to,” but we conclude “greater than” because all of the plausible values in the confidence interval are greater than the null value.

## 6.4 Critical Value Approach to Hypothesis Testing

We learned about critical values when we discussed confidence intervals. Now, we want to use these values directly in a hypothesis test. We will compare these values to a value based on the data, called a test statistic.

Idea: the null is our “default assumption.” If the null is true, how likely are we to observe a sample that looks like the one we have? If our sample is very inconsistent with the null hypothesis, we want to reject the null hypothesis.

### 6.4.1 Test statistics

Test statistics are similar to z- and t-scores: $\text{test statistic} = \frac{\text{point estimate}-\text{null value}}{\text{standard error}}.$

• Setting 1: $$\mu$$ is target parameter, $$X$$ is normal, $$\sigma$$ known

$z = \frac{\bar{x}-\mu_0}{\sigma/\sqrt{n}}$

• Setting 2: $$\mu$$ is target parameter, $$n \ge 30$$, $$\sigma$$ unknown

$z = \frac{\bar{x}-\mu_0}{s/\sqrt{n}}$

• Setting 3: $$\mu$$ is target parameter, $$n < 30$$, $$\sigma$$ unknown

$t = \frac{\bar{x}-\mu_0}{s/\sqrt{n}}$

In fact, they serve a similar function in converting a variable $$\bar{X}$$ into a distribution we can work with easily.

The set of values for the test statistic that cause us to reject $$H_0$$ is the rejection region. The remaining values are the nonrejection region. The value that separates these is the critical value!

Steps:

1. State the null and alternative hypotheses.
2. Determine the significance level $$\alpha$$. Check assumptions (decide which setting to use).
3. Compute the value of the test statistic.
4. Determine the critical values.
5. If the test statistic is in the rejection region, reject the null hypothesis. Otherwise, do not reject.
6. Interpret results.

Example: Is the average mercury level in dolphin muslces different from $$2.5\mu g/g$$? Test at the 0.05 level of significance. A random sample of $$19$$ dolphins resulted in a mean of $$4.4 \mu g/g$$ and a standard deviation of $$2.3 \mu g/g$$.

1. $$H_0: \mu = 2.5$$ and $$H_A: \mu \ne 2.5$$.
2. Significance level is $$\alpha=0.05$$. The value of $$\sigma$$ is unknown and $$n = 19 < 30$$, so we are in setting 3.
3. The test statistic is \begin{align} t &= \frac{\bar{x}-\mu_0}{s/\sqrt{n}} \\ &= \frac{4.4-2.5}{2.3/\sqrt{19}} \\ &= 3.601 \end{align}
4. The critical value is $$t_{df, \alpha/2}$$. Here, $$df=n-1=18$$ and $$\alpha/2 = 0.025$$:
qt(0.025, 18)
## [1] -2.100922
1. The test statistic is in the rejection region, so we will reject the null hypothesis:

1. At the 0.05 level of significance, the data provide sufficient evidence to conclude that the true mean mercury level in dolphin muscles is greater than $$2.5\mu g/g$$.

Notice that this is the same conclusion we came to when we used the confidence interval approach. These approaches are exactly equivalent!

## 6.5 P-Value Approach to Hypothesis Testing

If the null hypothesis is true, what is the probability of getting a random sample that is as inconsistent with the null hypothesis as the random sample we got? This probability is called the p-value?

Example: Is the average mercury level in dolphin muslces different from $$2.5\mu g/g$$? Test at the 0.05 level of significance. A random sample of $$19$$ dolphins resulted in a mean of $$4.4 \mu g/g$$ and a standard deviation of $$2.3 \mu g/g$$.

Probability of a sample as inconsistent as our sample is $$P(t_{df} \text{ is as extreme as the test statistic})$$. Consider $P(t_{18} > 3.6) = 0.001$ but we want to think about the probability of being “as extreme” in either direction (either tail), so $\text{p-value} = 2P(t_{18}>3.6) = 0.002$

If $$\text{p-value} < \alpha$$, reject the null hypothesis. Otherwise, do not reject.

### 6.5.1 P-Values

• Setting 1: $$\mu$$ is target parameter, $$X$$ is normal, $$\sigma$$ known $2P(Z > |z|)$ where $$z$$ is the test statistic.

• Setting 2: $$\mu$$ is target parameter, $$n \ge 30$$, $$\sigma$$ unknown $2P(Z > |z|)$ where $$z$$ is the test statistic.

• Setting 3: $$\mu$$ is target parameter, $$n < 30$$, $$\sigma$$ unknown $2P(t_{df} > |t|)$ where $$t$$ is the test statistic.

Note: $$|a|$$ is the “absolute value” of $$a$$. The absolute value takes a number and throws away the sign, so $$|2|=2$$ and $$|-3|=3$$.

Steps:

1. State the null and alternative hypotheses.
2. Determine the significance level $$\alpha$$. Check assumptions (decide which setting to use).
3. Compute the value of the test statistic.
4. Determine the p-value.
5. If $$\text{p-value} < \alpha$$, reject the null hypothesis. Otherwise, do not reject.
6. Interpret results.

We often use p-values instead of the critical value approach because they are meaningful on their own (they have a direct interpretation).

Example: For the dolphins,

1. $$H_0: \mu = 2.5$$ and $$H_A: \mu \ne 2.5$$.
2. Significance level is $$\alpha=0.05$$. The value of $$\sigma$$ is unknown and $$n = 19 < 30$$, so we are in setting 3.
3. The test statistic is \begin{align} t &= \frac{\bar{x}-\mu_0}{s/\sqrt{n}} \\ &= \frac{4.4-2.5}{2.3/\sqrt{19}} \\ &= 3.601 \end{align}
4. The p-value is $2P(t_{df} > |t|) - 2P(t_{18} > 3.601) = 0.002$
5. Since $$\text{p-value}=0.002 < \alpha=0.05$$, reject the null hypothesis.
6. At the 0.05 level of significance, the data provide sufficient evidence to conclude that the true mean mercury level in dolphin muscles is greater than $$2.5\mu g/g$$.

As before, this is the same conclusion we came to when we used the confidence interval and critical value approaches. All of these approaches are exactly equivalent.