The Sign of t Is Not the Direction of the Test
As in “Worked Example: Two-Sided Test for a Mean,” the one-sample t statistic compares the sample mean with the value stated in the null hypothesis. Its sign tells which side of that null value the sample mean falls on. But the sign alone does not determine the p-value: the alternative hypothesis determines which results count as evidence against \(H_0\).
For example, a positive t statistic means the sample mean is above the null value. That result points in the direction of an upper-tailed alternative, \(H_a:\mu>\mu_0\). It points away from a lower-tailed alternative, \(H_a:\mu<\mu_0\). The test still uses the alternative that was chosen to match the research question—not one selected after looking at the sample.
Recall from “The One-Sample t Test Statistic” that
The denominator is positive, so the sign of \(t\) is the sign of \(\bar{x}-\mu_0\). Under \(H_0\), the t distribution is centered at 0. A positive observed value lies to the right of 0, and a negative one lies to the left.
Match the Alternative to the Tail
A p-value is the probability, assuming \(H_0\) is true, of obtaining a test statistic at least as extreme as the observed statistic in the direction specified by \(H_a\). For a one-sided test, this means using just the tail named by the alternative. The sign tells you where the observed statistic lies within that distribution.
- For \(H_a:\mu<\mu_0\), the area to the left of the observed statistic: \(P(T\leq t)\).
- For \(H_a:\mu>\mu_0\), the area to the right of the observed statistic: \(P(T\geq t)\).
If the statistic points toward the alternative, the observed value is out in the tail that supports the claim, and the p-value can be small. If it points the opposite way, the p-value is the area in the alternative’s tail on the other side of 0. That area can be large. Do not change the tail just because the observed statistic has the opposite sign.
For a two-sided alternative, \(H_a:\mu\ne\mu_0\), both directions count. As explained in the previous tutorial, find the tail area beyond \(|t|\) and double it. That is different from a one-sided test: do not double a one-sided p-value.
| Alternative | Observed t | p-value area | What the sign indicates |
|---|---|---|---|
| \(H_a:\mu<\mu_0\) | Negative | Left of t | Sample mean is below \(\mu_0\), in the alternative’s direction |
| \(H_a:\mu<\mu_0\) | Positive | Left of t | Sample mean is above \(\mu_0\), opposite the alternative |
| \(H_a:\mu>\mu_0\) | Positive | Right of t | Sample mean is above \(\mu_0\), in the alternative’s direction |
| \(H_a:\mu>\mu_0\) | Negative | Right of t | Sample mean is below \(\mu_0\), opposite the alternative |
On a TI-84, the lower-tail area can be found with \(\operatorname{tcdf}(-1\mathrm{E}99,t,df)\), and the upper-tail area with \(\operatorname{tcdf}(t,1\mathrm{E}99,df)\). These large bounds approximate negative and positive infinity. Enter the observed t statistic as it is, including its sign. The calculator does not decide whether the chosen alternative matches the question; you must select the correct tail.
Worked Examples
Worked Example: Testing for a Higher Mean Restart Time
A fictional equipment team randomly selects 10 devices from an inventory of 120. The sample mean restart time is \(\bar{x}=32\) minutes, and the sample standard deviation is \(s=\sqrt{10}\) minutes. Assume restart times in the inventory follow an approximately Normal distribution. At \(\alpha=0.05\), test whether the true mean restart time is greater than 30 minutes.
State. Let \(\mu\) be the true mean restart time, in minutes, for all devices in this inventory. The hypotheses are \(H_0:\mu=30\) minutes and \(H_a:\mu>30\) minutes. Use \(\alpha=0.05\).
Plan and check conditions. Use a one-sample t test for a population mean. The devices were randomly selected, supporting the Random condition. For the 10% condition, \(0.10(120)=12\), and \(10\leq12\), so the sample is no more than 10% of the inventory. Since \(n=10<30\), the large-sample route is not met; the stated approximately Normal population supports the Normal/Large Sample condition. Use \(df=10-1=9\).
Do. First calculate the standard error and t statistic:
The statistic is positive, consistent with the upper-tailed alternative. For \(H_a:\mu>30\), use the area to the right of 2.00, not a doubled area:
The p-value is approximately 0.03828, rounded to five decimal places. Since \(0.03828<0.05\), reject \(H_0\).
Conclude. The sample provides convincing evidence that the true mean restart time for devices in this inventory is greater than 30 minutes.
The positive sign and the upper-tailed alternative point in the same direction. The p-value measures how often, if the mean were 30 minutes, a t statistic of 2.00 or greater would occur.
Worked Example: A Positive t Statistic With a Lower-Tailed Alternative
A fictional clinic is reviewing appointment delays. A random sample of 10 appointments from 150 appointments has a mean delay of \(\bar{x}=10\) minutes and a sample standard deviation of \(s=\sqrt{10}\) minutes. Assume delays follow an approximately Normal distribution. The clinic wants to know whether the true mean delay is below 8 minutes. Test at \(\alpha=0.05\).
State. Let \(\mu\) be the true mean appointment delay, in minutes, for all 150 appointments in the group being reviewed. The hypotheses are \(H_0:\mu=8\) minutes and \(H_a:\mu<8\) minutes. Use \(\alpha=0.05\).
Plan and check conditions. Use a one-sample t test. The 10 appointments were randomly selected, supporting the Random condition. For the 10% condition, \(0.10(150)=15\), and \(10\leq15\), so the sample is no more than 10% of the appointments. Since \(n=10<30\), the large-sample route is not met; the stated approximately Normal population supports the Normal/Large Sample condition. The degrees of freedom are \(df=10-1=9\).
Do. The standard error and statistic are
The statistic is positive, so the sample mean is above 8 minutes. But the alternative is lower-tailed. Use the area to the left of the observed statistic:
The p-value is approximately 0.96172, rounded to five decimal places. Since \(0.96172>0.05\), fail to reject \(H_0\).
Conclude. The sample does not provide convincing evidence that the true mean appointment delay is below 8 minutes.
A p-value near 1 makes sense here: the observed mean is above the null value, opposite the claimed lower direction. The correct p-value is still the left-tail area specified by \(H_a:\mu<8\); using the right tail would test a different claim.
Worked Example: A Negative t Statistic With a Lower-Tailed Alternative
A fictional greenhouse manager randomly selects 10 seedling trays from a shipment of 200. The sample mean time to sprout is \(\bar{x}=38\) hours, with \(s=\sqrt{10}\) hours. Assume sprouting times follow an approximately Normal distribution. At \(\alpha=0.05\), test whether the true mean time to sprout is less than the 40-hour target.
State. Let \(\mu\) be the true mean time to sprout, in hours, for all seedlings in this shipment. The hypotheses are \(H_0:\mu=40\) hours and \(H_a:\mu<40\) hours. Use \(\alpha=0.05\).
Plan and check conditions. Use a one-sample t test. Random selection supports the Random condition. For the 10% condition, \(0.10(200)=20\), and \(10\leq20\), so the sample is no more than 10% of the shipment. Since \(n=10<30\), the large-sample route is not met; the stated approximately Normal population supports the Normal/Large Sample condition. Use \(df=10-1=9\).
Do. Calculate the standard error and t statistic:
The negative statistic agrees with the lower-tailed alternative. Find the area to the left of \(-2.00\):
The p-value is approximately 0.03828, rounded to five decimal places. Since \(0.03828<0.05\), reject \(H_0\).
Conclude. The sample provides convincing evidence that the true mean time to sprout for seedlings in this shipment is less than 40 hours.
The upper-tail area beyond \(+2.00\) is also approximately 0.03828 because the t distribution is symmetric. Here, however, the lower-tail alternative calls for the area below \(-2.00\); symmetry is a check, not a reason to ignore the sign or alternative.
Common Mistakes and AP Exam Tips
A strong response names the alternative, shows the observed statistic with its sign, identifies the corresponding tail, and reports a conclusion in context. The tail comes from the alternative; the sign helps locate the observed statistic and shows whether it points in that direction.
- Choosing a tail from the sign alone. A positive t does not automatically mean use the right tail. For a lower-tailed alternative, even a positive t requires the left-tail area.
- Changing the alternative after seeing the sample. Choose \(H_a\) from the research question before examining the result. A sample mean below the target does not justify switching to a lower-tailed test if the original question was about a higher mean.
- Taking the absolute value for a one-sided test. Using \(|t|\) can send you to the wrong tail. Keep the sign when finding a one-sided p-value. The absolute value is used for the two-sided calculation described in the previous tutorial.
- Doubling a one-sided p-value. A one-sided alternative counts results in one specified direction only. Doubling is for a two-sided test, not for an upper- or lower-tailed test.
- Reporting the area opposite the alternative. For \(H_a:\mu<\mu_0\), calculate the area to the left of \(t\); for \(H_a:\mu>\mu_0\), calculate the area to the right. If the statistic points away from the alternative, the correct p-value may be large.
- Claiming a large p-value proves the null. If \(p>\alpha\), say “fail to reject \(H_0\).” This is not proof that \(\mu=\mu_0\), and it does not establish that the alternative is impossible.
- Leaving out the context in the conclusion. For a small p-value, state that the sample provides convincing evidence for the specified direction of the population mean. For a large p-value, state that it does not provide convincing evidence for that claim.
Check Your Understanding
Use \(\alpha=0.05\). Assume the one-sample t test conditions are satisfied unless a question asks you to discuss them.
- For \(H_0:\mu=25\) versus \(H_a:\mu>25\), the observed statistic is \(t=-1.60\). Which tail area gives the p-value, and does the statistic point toward or away from the alternative?
- For \(H_0:\mu=6\) versus \(H_a:\mu<6\), the observed statistic is \(t=-2.10\) with \(df=12\). Describe the calculator bounds you would use to find the p-value.
- A one-sample test has \(n=14\), \(\bar{x}=71\), and \(s=6\). The null value is \(\mu_0=68\), and the alternative is \(H_a:\mu<68\). Calculate the standard error and t statistic, then state whether the statistic points toward the alternative.
- A lower-tailed test produces a positive t statistic and a p-value of 0.92. At \(\alpha=0.05\), state the decision. Explain why the large p-value is reasonable given the sign.
- Why is it incorrect to double the p-value for a test of \(H_a:\mu>\mu_0\)?