Significance test for the mean

Significance test for a mean

The goal of a significance test for a mean is to use sample data to decide whether there’s convincing evidence that the population mean $\mu$ differs from a hypothesized value ${\mu }_0$. In other words, we’re asking:

• Is the difference we see between the sample mean and ${\mu }_0$ something that could reasonably happen just from natural sampling variation?
• Or is the difference large enough that it suggests the true population mean is actually different from ${\mu }_0$?

Procedure for a significance test for a mean

Step 1: State the hypotheses

The null hypothesis always states “no difference” between the population mean and the hypothesized mean:

$$\begin{array}{rl}{H}_0& :\mu ={\mu }_0\end{array}$$

The alternative hypothesis depends on the question being asked:

Definitions

Two-tailed test

The population mean differs from the hypothesized mean in either direction.

$$\begin{array}{rl}{H}_a& :\mu \ne {\mu }_0\end{array}$$

Right-tailed test

The population mean is greater than the hypothesized mean.

$$\begin{array}{rl}{H}_a& :\mu >{\mu }_0\end{array}$$

Left-tailed test

The population mean is less than the hypothesized mean.

$$\begin{array}{rl}{H}_a& :\mu <{\mu }_0\end{array}$$

Step 2: Check the conditions

To conduct a significance test for a mean, check the following:

• Randomness: The sample is random (ideally a simple random sample).
• Independence: The sample size is less than $10\%$ of the population.
• Normality: Either the sample size is greater than $30$ (so the central limit theorem applies) or the population distribution is approximately normal.

Step 3: Calculate the test statistic

Definitions

One-sample t-statistic

A measure of how far the sample mean is from the hypothesized mean, in units of standard error. Used when $\sigma$ is unknown.

$$\begin{array}{rl}t& =\frac{\bar{x}-{\mu }_0}{s/\sqrt{n}}\end{array}$$

where $\bar{x}$ is the sample mean, $s$ is the sample standard deviation, and $n$ is the sample size.

Step 4: Find the p-value

Definitions

P-value

The probability of getting a test statistic at least as extreme as the one observed, assuming $H_0$ is true. In other words, if the true population mean really is ${\mu }_0$, how likely is a t-statistic as extreme as the one observed?

To find the p-value, use a t-distribution table with the correct degrees of freedom:

$$\begin{array}{rl}df& =n-1\end{array}$$

Locate where your t-value falls in the table to approximate the p-value.

If you’re using a TI-84 calculator, press 2nd → VARS, then choose tcdf(lower bound, upper bound, degrees of freedom). This gives a more precise p-value than the table.

Step 5: Make a decision

Use the given significance level. If none is provided, use the common default $\alpha =0.05$.

• If $p<\alpha$, reject $H_0$.
• If $p>\alpha$, fail to reject $H_0$.

Practice problem

Example:

A company claims that its LED light bulbs last $100$ hours on average. A consumer takes a random sample of $36$ lightbulbs and measures their lifetimes, getting a sample mean lifetime of $\bar{x}=103$ and a sample standard deviation of $s=6$. Based on these results, the consumer believes the light bulbs last longer than the company claimed. Use the significance level $\alpha =0.01$.

Part a.

State the hypotheses.

We’re testing whether the mean lifetime is greater than $100$, as the consumer believes.

Solution:

Part b.

Check that the conditions needed for a t-test are met.

There are three conditions to check:

1. Randomness
2. Independence
3. Normality

Solution:

Part c.

Compute the test statistic $t$ and the degrees of freedom $df$.

From the problem:

$$\begin{array}{rl}\bar{x}& =103\\ s& =6\\ n& =36\\ {\mu }_0& =100\end{array}$$

Compute the standard error:

$$\begin{array}{rl}\text{SE}& =\frac{s}{\sqrt{n}}\\ & =\frac{6}{\sqrt{36}}\\ & =\frac{6}{6}\\ & =1\end{array}$$

Compute the test statistic:

$$\begin{array}{rl}t& =\frac{\bar{x}-{\mu }_0}{\text{SE}}\\ & =\frac{103-100}{1}\end{array}$$

Compute the degrees of freedom:

$$\begin{array}{rl}df& =n-1\\ & =36-1\end{array}$$

Solution:

Part d.

Find the p-value and make a decision about the hypotheses. Also state the conclusion in context.

Because this is a right-tailed test, we use the one-tailed p-value.

From the t-table:

$$\begin{array}{rl}{t}_{0.005,df=35}& \approx 2.724\end{array}$$

Compare the test statistic from Part c to this critical value:

$$\begin{array}{rl}3& >2.724\end{array}$$

So the p-value is less than $0.005$, which is also less than $\alpha =0.01$.

Solution:

Part e.

What kind of error is possible in this scenario and how do you know?

Solution:

Part f.

Why might the company have made the claim that the lightbulbs last $100$ hours on average when there is strong evidence that they actually last even longer?

Solution: