Topic 7.5 Notes – Carrying Out a Test for a Population Mean

What a One-Sample t-Test for a Mean Does

A one-sample t-test is used to test a claim about a population mean $\mu$ when the population standard deviation $\sigma$ is unknown.

That “σ unknown” part is why we use t instead of z. We estimate variability using the sample standard deviation $s$, which adds extra uncertainty. The t-distribution accounts for that.

You’ll always see:

• Parameter: $\mu$ (true population mean)
• Null hypothesis: $H_0: \mu = \mu_0$
• Alternative hypothesis:
  • $H_a: \mu > \mu_0$
  • $H_a: \mu < \mu_0$
  • $H_a: \mu \ne \mu_0$

The test statistic measures how many standard errors the sample mean is from the hypothesized mean.

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

with
$$\text{df} = n - 1$$

You don’t have to memorize this formula. You can rebuild it from
(statistic − parameter) / standard error.

Calculating the Test Statistic

Let’s walk through a quick example.

A fitness company claims its protein bars have a mean of 20 grams of protein. A random sample of 16 bars has:

• $\bar{x} = 18.8$
• $s = 2.4$

We test $H_0: \mu = 20$ vs. $H_a: \mu < 20$.

1. Standard error
   $$\text{SE} = \frac{s}{\sqrt{n}} = \frac{2.4}{\sqrt{16}} = \frac{2.4}{4} = 0.6$$
2. Test statistic
   $$t = \frac{18.8 - 20}{0.6} = \frac{-1.2}{0.6} = -2.0$$
3. Degrees of freedom
   $$\text{df} = 16 - 1 = 15$$

Interpretation of $t = -2.0$:
The sample mean is 2 standard errors below the claimed mean.

Larger absolute values of $t$ mean stronger evidence against $H_0$.

Finding the p-Value

The p-value is the probability, assuming $H_0$ is true, of getting a test statistic as extreme or more extreme than the one observed.

For our test statistic $t = -2.0$ with df = 15, that means finding the area to the left of −2.0 under a $t$-distribution.

Left-tailed p-value for $t = -2.0$, df = 15

Because this is left-tailed, we use the shaded area to the left of −2.0 with df = 15 as the p-value.

You can:

• Use a t-table (round df down if needed).
• Use a calculator: STAT → TESTS → T-Test.

Always report:

• $t = -2.0$
• df = 15
• p-value (from technology or table)

Two-tailed test? Double the one-tail area.
One-tailed test? Don’t double it. That mistake costs points.

Interpreting the p-Value

A correct interpretation sounds like this:

The p-value is the probability of getting a sample mean of 18.8 grams or less, assuming the true mean protein content is 20 grams.

Two things must be there:

1. Assuming the null is true
2. In context

Never say:

• “The probability that the null is true is…”
• “There is a 3% chance the mean is 20.”

The parameter is fixed. The sample result varies.

Making and Justifying a Decision

You compare the p-value to $\alpha$.

• If p ≤ α, reject $H_0$.
• If p > α, fail to reject $H_0$.

Suppose $p = 0.032$ and $\alpha = 0.05$.

Since $0.032 < 0.05$, we reject $H_0$.

Conclusion in context:

There is convincing evidence at the 0.05 level that the true mean protein content of these bars is less than 20 grams.

Notice how the final sentence answers the research question about the population.

Matched Pairs t-Test

If data are paired (before/after, twins, same person twice), reduce the data to differences first.

Example: blood pressure before and after medication.

1. Compute each person’s difference.
2. Let $\bar{d}$ = mean difference, $s_d$ = SD of differences.
3. Test $H_0: \mu_d = 0$.
4. Use $$t = \frac{\bar{d} - 0}{s_d / \sqrt{n}}$$ with df = $n - 1$, where $n$ is the number of pairs.

Conditions apply to the differences, not the original measurements. That detail shows up a lot on FRQs.