Reading Degrees of Freedom From 2-SampTTest Output

Why Degrees of Freedom Matter in Two-Sample t Procedures

In “The Two-Sample t Test Statistic,” you calculated how far an observed difference in sample means is from the null difference, in estimated standard-error units. To find a p-value for that statistic—or a critical value for an interval—you also need a reference t distribution. The degrees of freedom, abbreviated \(df\), identify which t distribution to use.

For an unpooled two-sample t procedure, technology commonly calculates the Welch degrees of freedom from both sample sizes and standard deviations. The result may be a decimal. A conservative alternative uses the smaller of \(n_1-1\) and \(n_2-1\). Both approaches account for estimating variability from samples; they differ in how they choose the reference t distribution.

Definition: Degrees of freedom for a two-sample t procedure determine the t distribution used to find a p-value or a critical value. The calculator’s Welch value uses both groups’ sample sizes and standard deviations; the conservative value is \(\min(n_1-1,n_2-1)\).

The Welch formula was introduced in “Computing Two-Sample Degrees of Freedom With the Welch Formula.” In terms of the separate variance contributions \(a=s_1^2/n_1\) and \(b=s_2^2/n_2\), it is:

df_{\text{Welch}} = \frac{(a+b)^2} {\frac{a^2}{n_1-1}+\frac{b^2}{n_2-1}}

You usually do not need to calculate this formula by hand when using a calculator. In a 2-SampTTest with pooled variances turned off, read the \(df\) reported with the test statistic and p-value. That decimal is the Welch value. Do not confuse it with the sample sizes, the test statistic, or the conservative value.

For the conservative approach, find each sample’s degrees of freedom, \(n_1-1\) and \(n_2-1\), then take the smaller. It does not use the sample standard deviations. The conservative choice is a practical way to use one t distribution without calculating Welch’s formula.

Comparing the Two Choices

A smaller \(df\) gives a t distribution with heavier tails. For a confidence interval at a fixed confidence level, this generally means a larger critical value, so the interval is wider when the center and standard error stay the same. For a fixed absolute t statistic, a smaller \(df\) gives a larger two-sided p-value.

For a one-sided test, the direction of the statistic relative to the alternative matters. When the statistic is in the direction of the alternative, a smaller \(df\) usually gives a larger p-value. When it is in the opposite direction, a smaller \(df\) gives a smaller p-value. Either way, degrees of freedom do not change the calculated t statistic or standard error; they affect the reference distribution used to assess it.

Key distinction: The calculator’s Welch \(df\) is based on the two samples’ sizes and standard deviations. The conservative \(df=\min(n_1-1,n_2-1)\) uses only the sample sizes. Use the method required for the task, and do not report one value while using the other to find a p-value or critical value.

When the sample sizes and standard deviations are similar, the Welch value may be much larger than the conservative value. When one variance contribution is especially influential, Welch’s value can be closer to the smaller sample degrees of freedom. The examples show why it is useful to identify which method produced the reported \(df\).

Worked Examples

Worked Example: Reading Test Output for Battery Life

A hypothetical study takes independent random samples of rechargeable batteries from two large product populations. The response is battery life, in hours. Group 1 has \(n_1=12\), \(\bar{x}_1=48\), and \(s_1=6\); group 2 has \(n_2=20\), \(\bar{x}_2=44\), and \(s_2=4\). A 2-SampTTest with pooled variances turned off reports \(t=2.052\), \(df=16.951\), and a two-sided p-value of about \(0.056\).

State: Let \(\mu_1\) be the true mean battery life for the first product population and \(\mu_2\) the true mean for the second. Test \(H_0:\mu_1-\mu_2=0\) against \(H_a:\mu_1-\mu_2\ne0\).

Plan: The response is quantitative, and the samples contain different batteries, so the groups are not paired. The samples were randomly selected from large populations, with each sample less than 10% of its population; the sampling design supports treating observations within each group as independent. Because both sample sizes are below 30, plots should show no strong skewness or outliers. Suppose inspection finds no such features. Use an unpooled two-sample t test.

Do: The standard error and test statistic are:

SE=\sqrt{\frac{6^2}{12}+\frac{4^2}{20}} =\sqrt{3+0.8} =\sqrt{3.8} \approx 1.9494\text{ hours}

t=\frac{48-44}{1.9494}\approx 2.052

The calculator’s \(df=16.951\) is its Welch value. For the conservative approach:

df_{\text{conservative}} =\min(12-1,20-1) =\min(11,19) =11

Using the smaller conservative \(df\) gives a two-sided p-value of about \(0.064\), compared with the calculator’s Welch p-value of about \(0.056\). These values are approximate and rounded. Both are above \(0.05\).

Conclude: At the 5% significance level, fail to reject \(H_0\) using either method. The samples do not provide convincing evidence that the two product populations have different mean battery life. The difference between the two p-values illustrates the effect of the reference distribution; the observed means, standard error, and t statistic are unchanged.

Worked Example: Comparing Degrees of Freedom for a Mean Interval

In a hypothetical environmental comparison, independent random samples estimate the daily amount of a particular airborne particle near two types of parks. The units are micrograms per cubic meter. For the first type, \(n_1=9\), \(\bar{x}_1=7.8\), and \(s_1=3\). For the second, \(n_2=16\), \(\bar{x}_2=6.0\), and \(s_2=6\). Assume each sample is less than 10% of its target population, and that the data plots show no strong skewness or outliers.

The standard error for a two-sample t interval is:

SE=\sqrt{\frac{3^2}{9}+\frac{6^2}{16}} =\sqrt{1+2.25} =\sqrt{3.25} \approx 1.8028

The observed difference is \(7.8-6.0=1.8\) micrograms per cubic meter. The calculator’s Welch \(df\) can be read from the 2-SampTInt output. To verify its value, set \(a=3^2/9=1\) and \(b=6^2/16=2.25\):

df_{\text{Welch}} = \frac{(1+2.25)^2}{\frac{1^2}{9-1}+\frac{2.25^2}{16-1}} = \frac{10.5625}{0.125+0.3375} \approx 22.838

The conservative value is \(\min(9-1,16-1)=8\). At 95% confidence, the critical value is about \(2.069\) using Welch \(df\) and \(2.306\) using conservative \(df=8\). The intervals are therefore:

\text{Welch: }1.8\pm 2.069(1.8028) \approx (-1.93,\ 5.53)

\text{Conservative: }1.8\pm 2.306(1.8028) \approx (-2.36,\ 5.96)

Both intervals estimate the same parameter, \(\mu_1-\mu_2\), in micrograms per cubic meter. The conservative interval is wider because it uses the smaller degrees of freedom and thus the larger critical value. Both intervals include zero, so both leave zero as a plausible value for the difference in population means.

Worked Example: When Welch Degrees of Freedom Are Relatively Large

Suppose two independent random samples compare mean weekly practice time for members of two community music programs. Time is measured in hours. Program A has \(n_1=15\), \(\bar{x}_1=6.4\), and \(s_1=5\); Program B has \(n_2=18\), \(\bar{x}_2=3.1\), and \(s_2=5\). Assume the samples are each less than 10% of their target populations and the data have no strong skewness or outliers.

The standard error and t statistic for testing a difference of zero are:

SE=\sqrt{\frac{5^2}{15}+\frac{5^2}{18}} =\sqrt{1.6667+1.3889} \approx 1.7480\text{ hours}

t=\frac{6.4-3.1}{1.7480} =\frac{3.3}{1.7480} \approx 1.888

For the Welch value, \(a=25/15=1.6667\) and \(b=25/18=1.3889\). Substitution gives:

df_{\text{Welch}} = \frac{(1.6667+1.3889)^2} {\frac{1.6667^2}{14}+\frac{1.3889^2}{17}} \approx 29.94

The conservative value is \(\min(14,17)=14\). Here, the sample standard deviations are equal, so the variance contributions are relatively similar and Welch’s value is close to \(n_1+n_2-2=31\). The conservative value is still 14. This comparison does not mean the conservative value is a different test statistic; both choices use \(t\approx1.888\), but refer to different t distributions when finding a p-value or critical value.

How to Report Degrees of Freedom Clearly

For calculator output, identify the value as the Welch degrees of freedom and use it with the reported test statistic and p-value, or with the critical value for an interval. For the conservative approach, show how the smaller of the two sample degrees of freedom was selected. Match the method used to the result you report.

Degrees of freedom are not measurement units. A report such as “\(df=16.951\)” is appropriate for a Welch result, even though it is not a whole number. The conservative result is generally a whole number because it is one of the two sample sizes minus one.

The choice of degrees of freedom does not replace the conditions for inference. As in “Conditions for a Two-Sample t Interval,” and the corresponding checks for a test, consider the design, independence, the 10% condition for sampling without replacement, and whether the sample data support using a t procedure. Having calculator output does not by itself establish those conditions.

Common Mistakes and AP Exam Tips

Calling the calculator’s decimal value an error: Welch degrees of freedom can be noninteger. Read the displayed value as given rather than rounding it to a whole number unless a task specifically directs you to do so.
Using the larger sample’s degrees of freedom for the conservative method: Calculate both \(n_1-1\) and \(n_2-1\), then use the smaller.
Mixing methods: Do not quote the calculator’s Welch p-value while saying you used the conservative degrees of freedom. Use a p-value or critical value that matches the stated method.
Assuming a smaller \(df\) always means a larger one-sided p-value: That is not true without considering the direction. For a fixed statistic in the direction of the one-sided alternative, smaller \(df\) usually gives a larger p-value; for a statistic in the opposite direction, it gives a smaller p-value. A smaller \(df\) gives a larger two-sided p-value for a fixed absolute statistic.
Claiming the degrees of freedom change the observed t statistic: The t statistic is calculated from the sample means and standard error. The degrees of freedom select the reference distribution used afterward.
Forgetting the effect on an interval: At a fixed confidence level and standard error, the smaller conservative \(df\) generally produces a larger \(t^*\) and a wider interval. It does not change the interval’s center.
Skipping context in the conclusion: A full-credit conclusion connects the decision to the population means and response variable, not merely to \(df\), \(t\), or a calculator display.

A clear response names the method, reports the matching degrees of freedom, and then interprets the p-value or interval in context. When comparing approaches, say specifically whether the difference is in the reference distribution, the p-value, or the interval’s critical value and width.

Key takeaway: Read the calculator’s unpooled two-sample output as Welch degrees of freedom. The conservative alternative is \(\min(n_1-1,n_2-1)\). Match the degrees of freedom to the p-value or critical value you use, and remember that degrees of freedom affect the reference distribution—not the sample means, standard error, or t statistic.

Check Your Understanding

For each question, distinguish the degrees of freedom from the test statistic and keep the method clear.

A two-sample test has sample sizes \(n_1=13\) and \(n_2=21\). What is the conservative degrees of freedom?
A calculator reports \(df=18.6\) for an unpooled two-sample t test. What method does this value represent, and should it be rounded simply because it is a decimal?
For the same fixed absolute t statistic, which approach gives the larger two-sided p-value: Welch \(df=18.6\) or conservative \(df=12\)?
At a fixed confidence level, what generally happens to the critical value and interval width when you use a smaller degrees of freedom?
Does changing the degrees of freedom change the calculated two-sample t statistic? Explain what it changes instead.