A Complete Test of a Mean Fill Volume
In “Writing a Conclusion for a One-Sample t Test,” you learned how to connect a test decision to a claim about a population mean. Now we will carry out the entire test for a bottling line. The question is whether the line’s true mean fill volume is less than its target of 500 milliliters.
The test uses one sample of quantitative measurements: the fill volumes of selected containers. The sample mean \(\bar{x}\) describes the containers measured, while the parameter \(\mu\) describes the population mean named in the question. A one-sample t test uses the sample standard deviation \(s\) to estimate the variability in the population.
A result below 500 in one sample does not, by itself, establish that the population mean is below 500. The test measures how far the sample mean is from 500 relative to its estimated standard error, then evaluates that result under the null hypothesis. As in “The One-Sample t Test Statistic,” the degrees of freedom are \(n-1\), and the p-value must match the direction in \(H_a\).
The Four Steps for This Test
The structure below brings together the hypothesis, condition checks, calculations, and conclusion. In a complete response, explain how the study supports each condition; do not rely on a calculator result alone.
Define \(\mu\) in context, write \(H_0\) and \(H_a\), and state 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 \(t=(\bar{x}-500)/(s/\sqrt{n})\), use \(df=n-1\), and find the area to the left 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 mean fill volume in context.
For a lower-tailed test, values of \(t\) farther below zero are more consistent with the alternative. The p-value is the probability, assuming the true mean is 500 milliliters, of obtaining a sample mean at least as low relative to its standard error as the one observed. It is not the probability that the null hypothesis is true.
Worked Examples
Worked Example: Testing a Production Lot
A fictional beverage plant randomly selects 16 containers from a production lot of 400. Their mean fill volume is \(\bar{x}=498.5\) milliliters, and their sample standard deviation is \(s=3\) milliliters. Assume the population distribution of fill volumes in this lot is approximately Normal. Is there evidence that the true mean fill volume is below 500 milliliters? Use \(\alpha=0.05\).
State. Let \(\mu\) be the true mean fill volume, in milliliters, of containers in this production lot. The hypotheses are \(H_0:\mu=500\) milliliters and \(H_a:\mu<500\) milliliters. Use \(\alpha=0.05\).
Plan and check conditions. Use a one-sample t test for a population mean. The containers were randomly selected, which supports the Random condition. For the 10% condition, \(0.10(400)=40\), and \(16\leq40\), 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. However, the problem states that the population distribution is approximately Normal, supporting the Normal/Large Sample condition. The degrees of freedom will be \(df=16-1=15\).
Do. First calculate the estimated standard error:
The sample mean is 1.5 milliliters below the null value. Dividing that difference by the standard error gives
The alternative is lower-tailed, so the p-value is the area to the left of \(-2.00\) with \(df=15\): \(\operatorname{tcdf}(-1\mathrm{E}99,-2,15)\approx0.0320\), rounded to four decimal places. Since \(0.0320<0.05\), reject \(H_0\).
Conclude. The sample provides convincing evidence that the true mean fill volume of containers in this production lot is less than 500 milliliters.
The test supports a claim about the lot’s population mean, not a claim that every container is underfilled. The sample mean is a little below 500, but the test decision also depends on how large that difference is compared with the estimated sampling variability.
Worked Example: A Small Difference With a Larger P-Value
In another fictional setting, a bottling facility randomly selects 25 containers from a run of 500. The sample mean fill volume is \(\bar{x}=499.6\) milliliters, and the sample standard deviation is \(s=2\) milliliters. Assume the fill volumes in this run follow an approximately Normal distribution. Test whether the true mean is less than 500 milliliters, using \(\alpha=0.05\).
State. Let \(\mu\) be the true mean fill volume, in milliliters, of containers in this run. The hypotheses are \(H_0:\mu=500\) milliliters and \(H_a:\mu<500\) milliliters. Set \(\alpha=0.05\).
Plan and check conditions. A one-sample t test is appropriate for the population mean if its conditions are supported. The containers were randomly selected, supporting the Random condition. For the 10% condition, \(0.10(500)=50\), and \(25\leq50\), so independence is reasonable. 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 is
Thus the test statistic is
For the lower-tailed alternative and \(df=24\), the p-value is \(\operatorname{tcdf}(-1\mathrm{E}99,-1,24)\approx0.1636\), rounded to four decimal places. Since \(0.1636>0.05\), fail to reject \(H_0\).
Conclude. The sample does not provide convincing evidence that the true mean fill volume of containers in this run is less than 500 milliliters.
Failing to reject does not show that the true mean is exactly 500 milliliters. It means that this sample does not give convincing evidence for the particular lower-mean claim at the 0.05 significance level.
Worked Example: A Small Sample With Convincing Evidence
A fictional pilot bottling run contains 90 containers. A random sample of 9 containers has mean fill volume \(\bar{x}=499\) milliliters and sample standard deviation \(s=1.5\) milliliters. Assume the population distribution of fill volumes in the run is approximately Normal. Test \(H_0:\mu=500\) against \(H_a:\mu<500\) at \(\alpha=0.05\).
State. Let \(\mu\) be the true mean fill volume, in milliliters, of all containers in this pilot run. The hypotheses are \(H_0:\mu=500\) milliliters and \(H_a:\mu<500\) milliliters, with \(\alpha=0.05\).
Plan and check conditions. Use a one-sample t test. The 9 containers 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 run. 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 p-value is the area to the left of \(-2.00\) for 8 degrees of freedom: \(\operatorname{tcdf}(-1\mathrm{E}99,-2,8)\approx0.0403\), rounded to four decimal places. Since \(0.0403<0.05\), reject \(H_0\).
Conclude. The sample provides convincing evidence that the true mean fill volume of containers in this pilot run is less than 500 milliliters.
This example has fewer observations than the first two, so the stated population shape is especially important: the large-sample route cannot support the t test here. The test’s conclusion is justified in this example because the population is specified to be approximately Normal.
Common Mistakes and AP Exam Tips
A reliable solution keeps the alternative hypothesis, the tail of the p-value, and the final conclusion aligned. Check each of the following before submitting a response.
- Using the wrong tail. Because \(H_a:\mu<500\), find the area to the left of the observed t statistic. A right-tail area answers a different question.
- Using the wrong standard deviation. The one-sample t statistic uses the sample standard deviation \(s\), not a known population standard deviation \(\sigma\). Calculate the standard error as \(s/\sqrt{n}\).
- Using the wrong degrees of freedom. For this one-sample t test, \(df=n-1\). State the degrees of freedom before finding the p-value.
- Checking conditions without evidence. “Random condition met” is not a full check. Say that containers were randomly selected. For the 10% condition, compare \(n\) with 10% of the production lot when sampling without replacement. For the Normal/Large Sample condition, use \(n\geq30\) or relevant evidence about the population or sample shape.
- Changing the alternative after seeing the sample. The direction must come from the question being asked, not from whether \(\bar{x}\) happens to fall below or above 500. Here the question specifies a lower mean, so the alternative is lower-tailed.
- Writing “accept \(H_0\)” or claiming equality. If \(p>\alpha\), say “fail to reject \(H_0\).” The result does not prove that the mean equals 500 milliliters.
- Making a conclusion about individual containers. The parameter \(\mu\) is a population mean. A conclusion about \(\mu\) does not establish that each container is below or above the target.
- Misstating the p-value. Describe it as a probability calculated assuming \(H_0\) is true, for a result at least as extreme in the direction of \(H_a\). 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 a test of \(H_0:\mu=500\) versus \(H_a:\mu<500\), a sample gives \(t=-1.40\) with \(df=19\). Which tail should be used to find the p-value?
- A random sample of 18 containers is drawn without replacement from a lot of 150. Does the 10% condition hold? Show the comparison.
- A bottling sample has \(n=16\), \(\bar{x}=499\), and \(s=2\). Calculate the standard error and the one-sample t statistic for \(H_0:\mu=500\).
- A lower-tailed test has \(p=0.08\). State the decision and write the main idea of the conclusion in the context of mean fill volume.
- Why is “There is a 96% chance that the mean fill volume is below 500 milliliters” not a correct interpretation of a p-value of 0.04?