OPTIMA Statistics

Subsection 5.5.1 Test Statistic for a Single Population Mean

The test statistic for a single mean when using a small sample will be a t-score. This t-score measures how unusual the observed sample is under the assumption of the null hypothesis. The formula is very similar to the one used with large samples with the exception that it is a t-score and not a z-score, and that it explicitly uses the sample standard deviation \(s\) instead of the population standard deviation \(\sigma\text{.}\)

Note that just as in the case of a confidence interval, the test statistic will belong to a certain member of the t-distribution family based on the degrees of freedom. This will become important later on when we actually conduct the hypothesis test. For now, consider the example below.

Subsection 5.5.2 Test Statistic for a Difference Between Means

We may also wish to conduct a hypothesis test for the difference between population means when the sample from one or both of those populations is small, meaning less than 30. To do this, we must make one additional assumption. Just as was done when building confidence intervals for differences using small sample means, we must assume that the variances of the two populations are equal. Recall that this assumption lead us to use a pooled variance estimate, given by:

where \(s_1\) is the standard deviation of the first sample of size \(n_1\) and \(s_2\) is the standard deviation of the second sample of size \(n_2\text{.}\) While we do not expect that \(s_1^2 = s_2^2\text{,}\) we do assume that they both will be close to the same common variance \(\sigma^2\text{,}\) and we therefore use \(\sigma^2 \approx s^2\) as our approximation in the following test statistic computation.

Once again the test statistic we get is a member of the particular t-distribution having \(n_1 + n_2 - 2\) degrees of freedom. This will be important once we start actually conducting the hypothesis test. An example of this computation can be seen in the following problem.

Subsection 5.5.3 The Traditional Test

A traditional hypothesis test using test statistics and critical values drawn from a t-distribution works in much the same way as tests using the normal distribution. Recall that the four steps are:

1. State the null and alternative hypotheses (done).

2. Compute the test statistic (done).

3. Find the rejection region and the critical values that separate it from the acceptance region.

4. Determine if the test statistic is in the rejection region, in which case we reject the null hypothesis, or in the acceptance region, in which case we fail to reject the null hypothesis.

The one catch, as mentioned previously, is that we must choose our critical values from the t-distribution with the appropriate number of degrees of freedom. The value of df will vary with the sample size or sizes with which we are working, but the formulas are the same as those found in Section 4.5 and restated below.

• Single Sample Mean.

  Degrees of freedom is one less than the sample size: \(df = n - 1\text{.}\)

• Difference Between Means.

  Degrees of freedom is two less than the sum of the sample sizes: \(df = n_1 + n_2 - 2\text{.}\)

Let's revisit the two examples we've seen in this lesson so far, finishing them off with a traditional hypothesis test.

You believe that the average lifespan of a supernova is 74 days. Because supernovas are so rare (approximately one every 50 years in our galaxy), you can only find accurate records from 7 supernovas to use in your data. Their average lifespan was 63.8 days with a standard deviation of 12.6 days. Conduct a traditional hypothesis test at the \(\alpha = 0.10\) significance level.

Solution

We have previously seen that the hypotheses are:

\begin{align*} H_0\amp:\ \mu = 74\\ H_A\amp:\ \mu \not= 74 \end{align*}

And the test statistic is:

\begin{equation*} t_{\text{test}} = \frac{63.8 - 74}{12.6/\sqrt{7}} \approx -2.142\text{.} \end{equation*}

Since this is a two-tailed hypothesis test, the rejection region will split the significance level of \(\alpha = 0.10\) into two tails with area \(0.05\) each as shown below.

The number of degrees of freedom is \(7 - 1 = 6\text{.}\) Therefore, from the t-distribution table, the critical values are \(1.943\) for the right tail, and \(-1.943\) for the left tail. Because our test statistic of \(t_\text{test} = -2.142\) is further into the left tail than the critical value -1.943, it is in the rejection region. We therefore must reject the null hypothesis. There is evidence tending towards significance that supernovas do not have an average lifespan of 74 days.

You believe that moon rocks contain less iron than do earth rocks. To test this hypothesis, you collect a sample of 300 rocks from all over the earth and find that they contain an average of 6.7 grams per cubic inch of iron with a standard deviation of 2.9 grams per cubic inch. You only have access to six moon rocks, however, and find that they have an average of 2.4 grams per cubic inch with a standard deviation of 0.35 grams per cubic inch. Conduct a traditional hypothesis test at the \(\alpha = 0.05\) significance level.

Solution

From previous work, the hypotheses are:

\begin{align*} H_0\amp:\ \mu_1 - \mu_2 \geq 0\\ H_A\amp:\ \mu_1 - \mu_2 \lt 0 \end{align*}

Our test statistic is:

\begin{equation*} t_{\text{test}} = \frac{ (2.4-6.7) - 0 }{\sqrt{8.27369\left(\frac{1}{6}+\frac{1}{300}\right)}} \approx -3.626\text{.} \end{equation*}

This is a left-tailed test (note that the alternative hypothesis involves “\(\lt\)”) and therefore our critical value will be negative. To find it, we look up the positive critical value in the t-distribution table that leaves an area of \(0.05\) (the significance level) in the right tail and has

\begin{equation*} df = 300 + 6 - 2 = 304 \end{equation*}

degrees of freedom. This is clearly 30+, so we use the value \(1.645\) from the t-distribution table. To get the left-tail critical value, we change the sign to get -1.645 as shown below.

Since our test statistic of \(t_\text{test} = -3.626\) is much more extreme than the critical value \(-1.645\text{,}\) we reject the null hypothesis. There is statistically significant evidence that the iron content of moon rocks is less than that of rocks on earth.

Subsection 5.5.4 Difficulties of the P-Value Test

What about p-value tests? Up to this point, while we've done both traditional and p-value hypothesis tests, p-value tests have been a little more useful since we find the actual probability of finding a sample at least as extreme as the one we got. This allows those who are reading our work to make up their own minds on what significance level should be used. However, when dealing with t-distributions, we run into a problem. Consider the portion of the t-distribution table show below.

Suppose we wanted to find the p-value for a right-tailed test statistic \(t_\text{test} = 2.316\) with four degrees of freedom. In the normal distribution table, we would look up the z-score for our test statistic and find the probability. But the t-distribution table is much more limited. The t-scores are in the body of the table, and the probabilities for this small selection of t-scores are listed in the column headings.

So, for this example, we would have to look in the \(df = 4\) row and observe that the test statistic of \(2.316\) is between \(2.132\) and \(2.776\text{.}\) Therefore it's p-value is between \(0.05\) and \(0.025\text{,}\) the probabilities that are at the top of those two columns. Notice that the order changed because a larger test statistic will have a smaller p-value. So, we'll reorder this and state that the p-value is between 0.025 and 0.05.

While this is the best we can do with test statistics from t-distributions using tables, we could use a computer to find p-values for any \(t_\text{test}\) statistic. That process depends on the specific computer program being used and is beyond the scope of this text.