Testing a Claim That a Mean Is Higher
In “Worked Example: Testing a Mean Fill Volume,” you carried out a lower-tailed one-sample t test. This tutorial uses the same test structure for a different claim: that a population mean is higher than a stated value. For example, a battery maker might claim that its rechargeable batteries last more than 10 hours on average.
The direction of the claim determines the alternative hypothesis and the tail used for the p-value. If the claim is that the population mean exceeds a reference value \(\mu_0\), the alternative is \(H_a:\mu>\mu_0\). The p-value is then the area to the right of the observed t statistic. A sample mean above the claimed value is not enough on its own to establish convincing evidence; the difference must also be considered relative to its estimated standard error.
As in “The One-Sample t Test Statistic,” calculate \(t=(\bar{x}-\mu_0)/(s/\sqrt{n})\) and use \(df=n-1\). A positive t statistic means the sample mean is above the null value. The p-value measures how likely it would be, if the null hypothesis were true, to get a t statistic at least as large as the one observed.
The Four Steps for a Right-Tailed t Test
The four-step structure from “Writing the Full Four-Step t Interval Solution” and the mean-test tutorials also applies here: state the hypotheses, plan and check conditions, do the calculations, and conclude in context. The alternative determines which tail to use; it should be chosen from the question, not from the direction of the sample result.
Define \(\mu\) in context, write \(H_0:\mu=\mu_0\) and \(H_a:\mu>\mu_0\), and give the significance level \(\alpha\).
Name the one-sample t test. Check the Random condition, the 10% condition when sampling without replacement, and the Normal/Large Sample condition.
Calculate the t statistic, use \(df=n-1\), and find the area to the right of the observed t statistic.
Compare the p-value with \(\alpha\), state whether you reject or fail to reject \(H_0\), and explain what the evidence says about the population mean in context.
For a right-tailed test, a larger positive t statistic gives a smaller p-value: it places the sample mean farther above the null value in standard-error units. But the same difference between \(\bar{x}\) and \(\mu_0\) can produce different t statistics if the sample size or sample variability differs. The test therefore does not judge the difference in isolation.
Worked Examples
Worked Example: Testing a Battery-Life Claim
A fictional battery company claims that its rechargeable batteries last more than 10 hours on average. A quality-control team randomly selects 16 batteries from a production lot of 300. Their mean life is \(\bar{x}=10.75\) hours and their sample standard deviation is \(s=1.2\) hours. Assume battery life in this lot follows an approximately Normal distribution. At \(\alpha=0.05\), is there evidence that the true mean life exceeds 10 hours?
State. Let \(\mu\) be the true mean battery life, in hours, of all batteries in this production lot. The hypotheses are \(H_0:\mu=10\) hours and \(H_a:\mu>10\) hours. Use \(\alpha=0.05\).
Plan and check conditions. Use a one-sample t test for a population mean. The batteries were randomly selected, supporting the Random condition. For the 10% condition, \(0.10(300)=30\), and \(16\leq30\), so the sample is no more than 10% of the lot and independence is reasonable. Since \(n=16<30\), the large-sample route is not met; the stated approximately Normal population supports the Normal/Large Sample condition. The degrees of freedom are \(df=16-1=15\).
Do. First calculate the estimated standard error:
The sample mean is \(10.75-10=0.75\) hours above the null value. In standard-error units, the test statistic is
Because \(H_a:\mu>10\), use the right-tail area with 15 degrees of freedom. The p-value is \(\operatorname{tcdf}(2.50,1\mathrm{E}99,15)\approx0.01225\), rounded to five decimal places. Since \(0.01225<0.05\), reject \(H_0\).
Conclude. The sample provides convincing evidence that the true mean battery life of batteries in this production lot exceeds 10 hours.
This conclusion is about the lot’s population mean, not a guarantee that every battery lasts more than 10 hours. The p-value is calculated under the assumption that the mean is 10 hours; it is not the probability that the null hypothesis is true.
Worked Example: A Sample Mean Above the Claim but Weak Evidence
In a second fictional test, engineers randomly select 25 batteries from a lot of 600. The sample mean life is \(\bar{x}=10.4\) hours, and the sample standard deviation is \(s=2.5\) hours. Assume battery life in this lot is approximately Normal. Test whether the true mean life exceeds 10 hours, using \(\alpha=0.05\).
State. Let \(\mu\) be the true mean battery life, in hours, of all batteries in this lot. The hypotheses are \(H_0:\mu=10\) hours and \(H_a:\mu>10\) hours, with \(\alpha=0.05\).
Plan and check conditions. Use a one-sample t test. The batteries were randomly selected, supporting the Random condition. For the 10% condition, \(0.10(600)=60\), and \(25\leq60\), so the sample is no more than 10% of the lot. Because \(n=25<30\), the large-sample route is not met; the stated approximately Normal population supports the Normal/Large Sample condition. Use \(df=25-1=24\).
Do. The standard error and test statistic are
The right-tail p-value for \(t=0.80\) with 24 degrees of freedom is \(\operatorname{tcdf}(0.80,1\mathrm{E}99,24)\approx0.2158\), rounded to four decimal places. Since \(0.2158>0.05\), fail to reject \(H_0\).
Conclude. The sample does not provide convincing evidence that the true mean battery life in this lot exceeds 10 hours.
The sample mean is above 10 hours, but the difference is small relative to the estimated standard error. Failing to reject \(H_0\) does not establish that the population mean is exactly 10 hours or that it is not higher; it means this sample does not provide convincing evidence for the higher-mean claim at the 0.05 level.
Worked Example: A Small Sample With Convincing Evidence
A fictional pilot lot contains 90 batteries. A random sample of 9 batteries has mean life \(\bar{x}=10.4\) hours and sample standard deviation \(s=0.6\) hours. Assume the battery-life distribution for this lot is approximately Normal. Test whether the true mean life exceeds 10 hours at \(\alpha=0.05\).
State. Let \(\mu\) be the true mean battery life, in hours, of all batteries in this pilot lot. The hypotheses are \(H_0:\mu=10\) hours and \(H_a:\mu>10\) hours. Set \(\alpha=0.05\).
Plan and check conditions. Use a one-sample t test. The 9 batteries were randomly selected, supporting the Random condition. For the 10% condition, \(0.10(90)=9\), and \(9\leq9\), so the sample is no more than 10% of the lot. Since \(n=9<30\), the large-sample route is not met; the stated approximately Normal population supports the Normal/Large Sample condition. The degrees of freedom are \(df=9-1=8\).
Do. Calculate the standard error and test statistic:
The right-tail p-value is \(\operatorname{tcdf}(2.00,1\mathrm{E}99,8)\approx0.0403\), rounded to four decimal places. Since \(0.0403<0.05\), reject \(H_0\).
Conclude. The sample provides convincing evidence that the true mean battery life in this pilot lot exceeds 10 hours.
This sample is small, so the Normal/Large Sample condition depends on the stated population shape rather than the \(n\geq30\) route. The condition checks are part of why the inference is appropriate; a calculator’s p-value alone does not establish that.
Common Mistakes and AP Exam Tips
A complete response keeps the population parameter, alternative hypothesis, p-value tail, and conclusion aligned. These details are especially important when the question asks whether a mean is higher.
- Using a lower-tailed p-value. For \(H_a:\mu>10\), find the area to the right of the observed t statistic. The left-tail area addresses a different claim.
- Choosing the alternative from the sample result. The direction comes from the question being tested. Do not switch to a higher-mean alternative only because \(\bar{x}\) happens to be above \(\mu_0\).
- Using \(s\) instead of the standard error. The denominator in the test statistic is \(s/\sqrt{n}\), not \(s\). The standard error measures the estimated variability of the sample mean.
- Forgetting degrees of freedom. For a one-sample t test, use \(df=n-1\), not \(n\). The p-value depends on the degrees of freedom as well as the t statistic.
- Skipping condition evidence. Name the condition and cite the relevant study detail. For example, say that the batteries were randomly selected, show the 10% comparison when sampling without replacement, and explain how the sample size or population shape supports the Normal/Large Sample condition.
- Writing “accept the null” or claiming there is no effect. When \(p>\alpha\), say “fail to reject \(H_0\).” This does not prove the mean equals the claimed value.
- Overstating the conclusion. The inference concerns a population mean, not every individual battery. A full-credit conclusion identifies the population and measurement, then describes the evidence for the higher-mean claim.
- Misinterpreting the p-value. A p-value is calculated assuming \(H_0\) is true. For a right-tailed test, it is the probability of a t statistic at least as large as the observed one. It is not the probability that \(H_0\) is true.
Check Your Understanding
Use \(\alpha=0.05\) for each question. Assume any unstated t test conditions are supported unless the question asks you to evaluate them.
- For \(H_0:\mu=10\) versus \(H_a:\mu>10\), a sample produces \(t=1.75\) with \(df=14\). Which tail should be used to find the p-value?
- A random sample of 20 batteries is drawn without replacement from a lot of 180. Does the 10% condition hold? Show the comparison.
- A battery sample has \(n=16\), \(\bar{x}=10.6\) hours, and \(s=1.2\) hours. Calculate the standard error and the test statistic for \(H_0:\mu=10\).
- A right-tailed test of a higher mean has \(p=0.08\). State the decision and write the main idea of the conclusion in context.
- Why is “There is a 92% chance that the mean battery life exceeds 10 hours” not a correct interpretation of a p-value of 0.08?