9.2 Two Population Means with Known Standard Deviations
SLO 4
Demonstrate an understanding of, and ability to use, basic ideas of statistical processes, including hypothesis tests and confidence interval estimation.
Knowing both population standard deviations removes the estimation step, so the difference of sample means is exactly normal and the statistic is a z-score. Seeing the test in its cleanest form makes clear what the Student's t in §9.1 was compensating for.
SLO 5
Identify appropriate statistical techniques and use technology-based statistical analysis to describe, interpret, and communicate results.
The section turns on a single technology-facing check — whether the column says population or sample standard deviation, and whether n₁ + n₂ exceeds 30 — and that check is what decides which rāSHio panel or calculator test you reach for.
Learning Objectives
By the end of this section, you will be able to:
- recognize when a two-sample problem gives you the population standard deviations rather than sample estimates;
- write the sampling distribution of \(\overline{X}_{1} - \overline{X}_{2}\) when both \(\sigma\) values are known;
- carry out a two-sample z-test and interpret its result in context;
- state the condition under which the normal distribution may stand in for the Student's t, and recognize when it does not apply.
Section 9.1 handled the case you will meet almost every time: two independent samples, and no knowledge of how much either population varies. This section handles the other case. When the two population standard deviations are genuinely known, the estimating step disappears, the Student's t is replaced by the normal distribution, and the degrees-of-freedom formula that made §9.1 awkward goes away entirely. Knowing the population standard deviations is not likely in practice — that is worth saying plainly, because it explains why this section is short. It is here partly because the arithmetic is the cleanest version of the two-sample idea, and partly because a second, more common situation borrows the same normal distribution: samples large enough that the Student's t and the normal have nearly converged.
9.2.1 The Sampling Distribution When Both Sigmas Are Known
When both population standard deviations are known and both populations are normal, the difference of the sample means is normally distributed:
$$ \overline{X}_{1} - \overline{X}_{2} \sim N\left[\mu_{1} - \mu_{2},\ \sqrt{\frac{(\sigma_{1})^{2}}{n_{1}} + \frac{(\sigma_{2})^{2}}{n_{2}}}\right] $$and its standard deviation is
$$ \sqrt{\frac{(\sigma_{1})^{2}}{n_{1}} + \frac{(\sigma_{2})^{2}}{n_{2}}} $$Definition 9.2.1 — Sampling distribution of a difference with known standard deviations.
When both population standard deviations are known, the test statistic is a z-score:
$$ z = \frac{(\overline{x}_{1} - \overline{x}_{2}) - (\mu_{1} - \mu_{2})}{\sqrt{\frac{(\sigma_{1})^{2}}{n_{1}} + \frac{(\sigma_{2})^{2}}{n_{2}}}} $$Compare that radical with Definition 9.1.3 and you will see the only difference is the letter. Where §9.1 had \(s_{1}\) and \(s_{2}\) — estimates — this has \(\sigma_{1}\) and \(\sigma_{2}\), the real thing. Because nothing is being estimated, the quantity is a genuine standard deviation rather than a standard error, and the distribution is exactly normal rather than approximately Student's t.
The Student's t exists because of a specific problem: when you divide by an estimated standard deviation, you have introduced a second source of randomness into the denominator, and the resulting ratio has heavier tails than a z-score does. Remove the estimation and the problem removes itself. That is the whole reason this section gets the normal distribution and §9.1 does not — and it is also why the two converge as samples grow, since a standard deviation estimated from a large sample is very nearly the real one.
Under a null hypothesis that the two population means are equal — or that \(\mu_{1} \leq \mu_{2}\), which puts the boundary at equality — the term \((\mu_{1} - \mu_{2})\) is zero and the mean of the distribution is zero.
There are no degrees of freedom to compute here, and no non-integer \(df\) to explain. Once you have the two \(\sigma\) values, the standard deviation of the difference is a single arithmetic step and the p-value comes straight from the normal table or a calculator.
The random variable is the same one as in §9.1, \(\overline{X}_{1} - \overline{X}_{2}\). What changes is that you can write down its distribution exactly instead of approximating it.
9.2.2 Running a Two-Sample z-Test
The means of the number of revolutions per minute of two competing engines are to be compared. Thirty engines of each type are randomly assigned to be tested. Both populations have normal distributions. The table shows the result. Do the data indicate that Engine 2 has higher RPM than Engine 1? Test at a 5% level of significance.
| Engine | Sample Mean Number of RPM | Population Standard Deviation |
|---|---|---|
| 1 | 1,500 | 50 |
| 2 | 1,600 | 60 |
Solution — comparing two engines at the 5% level
Step 1 — Identify the setting. Two independent samples, two population means, and the table hands you population standard deviations. This is the known-\(\sigma\) case, so the distribution is normal and the test statistic is a z-score.
Step 2 — Write the hypotheses. The claim under investigation is that Engine 2 has the higher RPM, so that direction is the alternative. With subscript 2 for Engine 2:
$$ H_{0}: \mu_{2} \leq \mu_{1} \qquad H_{a}: \mu_{2} > \mu_{1} $$This is a right-tailed test.
Step 3 — Find the standard deviation of the difference.
$$ \sqrt{\frac{50^{2}}{30} + \frac{60^{2}}{30}} = \sqrt{83.33 + 120} = 14.26 $$Step 4 — Standardize. \(z = \frac{1600 - 1500}{14.26} = 7.01\).
Step 5 — Decide. A z-score of 7.01 puts essentially no area in the right tail: the p-value is about \(1.2 \times 10^{-12}\), far below 0.05. Reject \(H_{0}\).
Answer: At the 5% level of significance, there is sufficient evidence to conclude that Engine 2 has a higher mean RPM than Engine 1. Note how decisive this is compared with the examples in §9.1 — a 100-RPM gap against a standard deviation of the difference near 14 is an enormous effect, and no realistic significance level would fail to detect it.
The mean lasting time of two competing floor waxes is to be compared. Twenty floors are randomly assigned to test each wax. Both populations have normal distributions. The data are recorded below. Does the data indicate that wax 1 is more effective than wax 2? Test at a 5% level of significance.
| Wax | Sample Mean Number of Months Floor Wax Lasts | Population Standard Deviation |
|---|---|---|
| 1 | 3 | 0.33 |
| 2 | 2.9 | 0.36 |
Solution — a right-tailed z-test on wax durability
Step 1 — Name the setting. This is a test of two independent groups, two population means, population standard deviations known.
Step 2 — Name the random variable. \(\overline{X}_{1} - \overline{X}_{2}\), the difference in the mean number of months the competing floor waxes last.
Step 3 — Write the hypotheses.
$$ H_{0}: \mu_{1} \leq \mu_{2} \qquad H_{a}: \mu_{1} > \mu_{2} $$The words "is more effective" say that wax 1 lasts longer than wax 2, on average. "Longer" is a \(>\) symbol and goes into \(H_{a}\), so this is a right-tailed test.
Step 4 — Choose the distribution. The population standard deviations are known, so the distribution is normal:
$$ \overline{X}_{1} - \overline{X}_{2} \sim N\left(0,\ \sqrt{\frac{0.33^{2}}{20} + \frac{0.36^{2}}{20}}\right) $$Since \(H_{0}\) says \(\mu_{1} \leq \mu_{2}\), the boundary case is \(\mu_{1} - \mu_{2} = 0\), so the mean of the normal distribution is zero.
Step 5 — Compute. The observed difference is \(\overline{x}_{1} - \overline{x}_{2} = 3 - 2.9 = 0.1\). The standard deviation of the difference is 0.1092, giving \(z = 0.9157\) and a right-tail p-value of 0.1799.
Step 6 — Compare and decide. \(\alpha = 0.05\) and the p-value is 0.1799, so \(\alpha <\) p-value. Do not reject \(H_{0}\).
Answer: At the 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the mean time wax 1 lasts is longer than the mean time wax 2 lasts — that is, that wax 1 is more effective.
On a TI-83/83+/84/84+: press STAT, arrow over to TESTS, and press 3:2-SampZTest. Arrow over to Stats and press ENTER. Enter .33 for \(\sigma_{1}\), .36 for \(\sigma_{2}\), 3 for the first sample mean, 20 for n1, 2.9 for the second sample mean, and 20 for n2. Arrow down to \(\mu_{1}\): and arrow to the \(>\mu_{2}\) option. Press ENTER, arrow down to Calculate and press ENTER. The calculator returns p = 0.1799 and the test statistic \(z = 0.9157\). Running the same instructions but choosing Draw instead of Calculate produces the shaded graph.
Try it in rāSHio
The tell that sends you to a different panel is the column heading: Population Standard Deviation rather than Sample Standard Deviation. In rāSHio choose Stats → Z Stats and enter this example's values — 3 and 0.33 with 20 for wax 1, 2.9 and 0.36 with 20 for wax 2 — then set the alternative to greater-than. It returns \(z = 0.9157\) and the p-value 0.1799 without any degrees of freedom to set, which is the practical difference between this section and §9.1. If a problem gives you sample standard deviations instead, the Z Stats panel is the wrong one and Stats → T Stats is the right one.
Figure 9.2.1 — The rāSHio Z Stats panel: Stats → Z Stats. The walkthrough opens the panel on its one-sample setting and reads the z-statistic and p-value straight off the results card. For this example set Samples to 2 first, which gives you a second block for wax 2; the panel then returns z and p the same way, with no degrees of freedom to configure. Its demonstration numbers are not this example's floor-wax data.
9.2.3 The Large-Sample Shortcut and Its Limit
When the population standard deviations are unknown but the sum of the sample sizes exceeds 30 — that is, \(n_{1} + n_{2} > 30\) — the normal distribution may be used to approximate the Student's t for a difference of two means, with the sample standard deviations standing in for \(\sigma_{1}\) and \(\sigma_{2}\).
Definition 9.2.3 — Normal approximation to the Student's t.
Two cautions belong with this definition, and the second is the one that catches people.
First, it is a permission, not an instruction. As §9.1 said, use the Student's t whenever you can; the approximation was a convenience from the era of printed tables and buys nothing when a calculator is in front of you.
Second, the condition is on the sum of the sample sizes, and it is easy to satisfy it in one problem and miss it in the next that looks identical. Example 9.2.2 and Try It Now 9.2.2 below are deliberately the same question about the ages of two groups of people. One qualifies for the approximation and one does not.
There is a second route to the normal distribution, and it does not require knowing either \(\sigma\).
The average age of 10 professors selected randomly at university A is 46.672 years with a standard deviation of 8.53. The average age of 10 professors selected randomly at university B is 47.531 years with a standard deviation of 7.83. Does the data indicate that university A has older professors than university B, on average? Test at a 5% level of significance.
Solution — why this one is not a z-test
Step 1 — Check the condition before choosing a distribution. This looks like Example 9.2.2 and it is not. The standard deviations are again sample standard deviations, but here \(n_{1} + n_{2} = 10 + 10 = 20\), which is not greater than 30. The normal approximation of Definition 9.2.3 is unavailable, so this is a Student's t problem and belongs to §9.1's method.
Step 2 — Write the hypotheses. The claim is that university A's professors are older, so with subscript A for university A:
$$ H_{0}: \mu_{A} \leq \mu_{B} \qquad H_{a}: \mu_{A} > \mu_{B} $$This is a right-tailed test.
Step 3 — Standardize with a t. The standard error is
$$ \sqrt{\frac{8.53^{2}}{10} + \frac{7.83^{2}}{10}} = 3.662 $$so \(t = \frac{46.672 - 47.531}{3.662} = -0.235\), with \(df \approx 17.87\) from the Aspin–Welch formula.
Step 4 — Find the p-value. The right-tail area beyond \(t = -0.235\) is 0.4086.
Step 5 — Decide. Since \(0.4086 > 0.05\), do not reject \(H_{0}\).
Answer: At the 5% level of significance, there is not sufficient evidence to conclude that university A has older professors than university B. Notice that the sample actually points the other way — university A's mean age is the lower of the two — so a right-tailed test was never going to find support for the claim, and the p-value above 0.4 records that.
The lesson. Nothing about the wording of this problem tells you which distribution to use. Only the sample sizes do, and the check is one addition: \(n_{1} + n_{2} > 30\) or not. Get it wrong here and you would have reported a z-test; the conclusion happens to survive, but that is luck, not method.
An interested citizen wanted to know if Democratic U.S. senators are older than Republican U.S. senators, on average. During a certain year, the mean age of 30 randomly selected Republican senators was 61.675 years with a standard deviation of 10.17 years. The mean age of 30 randomly selected Democratic senators was 61.704 years with a standard deviation of 9.55 years. Do the data indicate that Democratic senators are older than Republican senators, on average? Test at a 5% level of significance.
Solution — the normal approximation with 60 observations
Step 1 — Name the setting. This is a test of two independent groups, two population means. The population standard deviations are unknown — 10.17 and 9.55 are sample standard deviations. But \(n_{1} + n_{2} = 30 + 30 = 60\), which is greater than 30, so the normal approximation to the Student's t is available. Let subscript 1 be Democratic senators and subscript 2 Republican senators.
Step 2 — Name the random variable. \(\overline{X}_{1} - \overline{X}_{2}\), the difference in the mean age of Democratic and Republican U.S. senators.
Step 3 — Write the hypotheses.
$$ H_{0}: \mu_{1} \leq \mu_{2} \quad \text{or} \quad \mu_{1} - \mu_{2} \leq 0 $$ $$ H_{a}: \mu_{1} > \mu_{2} \quad \text{or} \quad \mu_{1} - \mu_{2} > 0 $$The words "older than" translate to a \(>\) symbol and go into \(H_{a}\), so this is a right-tailed test.
Step 4 — Choose the distribution.
$$ \overline{X}_{1} - \overline{X}_{2} \sim N\left[0,\ \sqrt{\frac{(9.55)^{2}}{30} + \frac{(10.17)^{2}}{30}}\right] $$Since \(\mu_{1} \leq \mu_{2}\) gives \(\mu_{1} - \mu_{2} \leq 0\), the mean for the normal distribution is zero.
Step 5 — Compute. The observed difference is \(61.704 - 61.675 = 0.029\) years — about ten days. The standard deviation of the difference is 2.547, so \(z = 0.0114\) and the right-tail p-value is 0.4955.
Step 6 — Decide. \(\alpha = 0.05\) and the p-value is 0.4955, so \(\alpha <\) p-value. Do not reject \(H_{0}\).
Answer: At the 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the mean age of Democratic senators is greater than the mean age of the Republican senators. A p-value just under 0.5 is what you should expect from a difference this small: the two groups are, as far as this sample can tell, the same age.
Try it in rāSHio
Try It Now 9.2.2 and Example 9.2.2 ask the same question about ages, and only \(n_{1} + n_{2}\) decides which panel is allowed to answer it. For the senators, \(30 + 30 = 60\) clears 30, so open rāSHio, choose Stats → Z Stats, and enter the sample standard deviations 9.55 and 10.17 where the panel asks for σ — it returns \(z = 0.0114\) and the p-value 0.4955. For the professors, \(10 + 10 = 20\) does not clear 30, so the same figures belong in Stats → T Stats instead — the panel §9.1 walks through — which returns \(t = -0.235\) on \(df \approx 17.87\) and the p-value 0.4086. Run both and you have done the check the section is really about: one addition, and it picks the panel for you.
Problem Set 9.2
Use the following information to answer the next five exercises. The mean speeds of fastball pitches from two different baseball pitchers are to be compared. A sample of 14 fastball pitches is measured from each pitcher. The populations have normal distributions. The table below shows the result. Scouts believe that Rodriguez pitches a speedier fastball.
| Pitcher | Sample Mean Speed of Pitches (mph) | Population Standard Deviation |
|---|---|---|
| Wesley | 86 | 3 |
| Rodriguez | 91 | 7 |
Problem 1. What is the random variable?
Problem 1 Solution
Step 1 — Identify the setting: Two independent samples of 14 pitches each, and the table gives population standard deviations, not sample estimates. This is the known-\(\sigma\) two-sample case, so the object of interest is the difference of the two sample means.
Answer: \(\overline{X}_{1} - \overline{X}_{2}\), the difference in mean speeds of the fastball pitches of the two pitchers.
Problem 2. State the null and alternative hypotheses.
Problem 2 Solution
Step 1 — Assign subscripts: Let subscript 1 be Wesley and subscript 2 be Rodriguez, matching the table.
Step 2 — Translate the claim: "Rodriguez pitches a speedier fastball" says \(\mu_{2} > \mu_{1}\), or equivalently \(\mu_{1} < \mu_{2}\). That belongs in \(H_{a}\).
Answer:
$$ H_{0}: \mu_{1} \geq \mu_{2} \qquad H_{a}: \mu_{1} < \mu_{2} $$This is a left-tailed test.
Problem 3. What is the test statistic?
Problem 3 Solution
Step 1 — Find the standard deviation of the difference:
$$ \sqrt{\frac{3^{2}}{14} + \frac{7^{2}}{14}} = \sqrt{\frac{9}{14} + \frac{49}{14}} = \sqrt{4.1429} = 2.0354 $$Step 2 — Standardize: \(z = \dfrac{86 - 91}{2.0354} = -2.4565\).
Answer: \(z \approx -2.46\).
Problem 4. What is the p-value?
Problem 4 Solution
Step 1 — Identify the tail: \(H_{a}: \mu_{1} < \mu_{2}\) is left-tailed, so the p-value is the area to the left of \(z = -2.4565\) under the standard normal curve.
Answer: p ≈ 0.0070.
Problem 5. At the 1% significance level, what is your conclusion?
Problem 5 Solution
Step 1 — Compare p-value to \(\alpha\): \(\alpha = 0.01\) and the p-value is 0.0070, so p-value \(< \alpha\). Reject \(H_{0}\).
Answer: At the 1% significance level, there is sufficient evidence to conclude that the mean speed of Rodriguez's fastball is faster than Wesley's.
Use the following information to answer the next five exercises. A researcher is testing the effects of plant food on plant growth. Nine plants have been given the plant food. Another nine plants have not been given the plant food. The heights of the plants are recorded after eight weeks. The populations have normal distributions. The table below is the result. The researcher thinks the food makes the plants grow taller.
| Plant Group | Sample Mean Height of Plants (inches) | Population Standard Deviation |
|---|---|---|
| Food | 16 | 2.5 |
| No food | 14 | 1.5 |
Problem 6. Is the population standard deviation known or unknown?
Problem 6 Solution
Step 1 — Read the table: The column is headed "Population Standard Deviation," so 2.5 and 1.5 are \(\sigma\) values handed to you directly, not estimates computed from the samples.
Answer: Known. \(\sigma_{\text{Food}} = 2.5\) and \(\sigma_{\text{No Food}} = 1.5\) are population standard deviations, so this is the known-\(\sigma\) two-sample z-test.
Problem 7. State the null and alternative hypotheses.
Problem 7 Solution
Step 1 — Assign subscripts: Let subscript 1 be Food and subscript 2 be No Food.
Step 2 — Translate the claim: "The food makes the plants grow taller" says \(\mu_{1} > \mu_{2}\). That belongs in \(H_{a}\).
Answer:
$$ H_{0}: \mu_{1} \leq \mu_{2} \qquad H_{a}: \mu_{1} > \mu_{2} $$Problem 8. What is the p-value?
Problem 8 Solution
Step 1 — Find the standard deviation of the difference:
$$ \sqrt{\frac{2.5^{2}}{9} + \frac{1.5^{2}}{9}} = \sqrt{\frac{6.25}{9} + \frac{2.25}{9}} = \sqrt{0.9444} = 0.9718 $$Step 2 — Standardize: \(z = \dfrac{16 - 14}{0.9718} = 2.0580\).
Step 3 — Find the right-tail area: \(H_{a}: \mu_{1} > \mu_{2}\) is right-tailed, so the p-value is the area beyond \(z = 2.0580\).
Answer: p ≈ 0.0198.
Problem 9. Draw the graph of the p-value.
Problem 9 Solution
Step 1 — Choose the curve: This is a right-tailed test, so sketch a standard normal curve centered at 0 — the null value of \(\mu_{1} - \mu_{2}\).
Step 2 — Locate the observed difference: The observed sample difference is \(16 - 14 = 2\) inches, which standardizes to \(z = 2.0580\).
Step 3 — Shade the tail: Shade the region under the curve to the right of \(z = 2.0580\). That shaded area is the p-value.
Answer: A right-tailed normal curve centered at 0, with the region to the right of \(z = 2.0580\) (the standardized form of the observed 2-inch difference) shaded; the shaded area is 0.0198.
Problem 10. At the 1% significance level, what is your conclusion?
Problem 10 Solution
Step 1 — Compare p-value to \(\alpha\): \(\alpha = 0.01\) and the p-value is 0.0198, so p-value \(> \alpha\). Do not reject \(H_{0}\).
Answer: At the 1% significance level, there is not sufficient evidence to conclude that the plant food makes the plants grow taller.
Use the following information to answer the next five exercises. Two metal alloys are being considered as material for ball bearings. The mean melting point of the two alloys is to be compared. Fifteen pieces of each metal are being tested. Both populations have normal distributions. The table below is the result. It is believed that Alloy Zeta has a different melting point.
| Sample Mean Melting Temperature (°F) | Population Standard Deviation | |
|---|---|---|
| Alloy Gamma | 800 | 95 |
| Alloy Zeta | 900 | 105 |
Problem 11. State the null and alternative hypotheses.
Problem 11 Solution
Step 1 — Assign subscripts: Let subscript 1 be Alloy Gamma and subscript 2 be Alloy Zeta.
Step 2 — Translate the claim: "Alloy Zeta has a different melting point" makes no claim about direction, only that the means are unequal.
Answer:
$$ H_{0}: \mu_{1} = \mu_{2} \qquad H_{a}: \mu_{1} \neq \mu_{2} $$Problem 12. Is this a right-, left-, or two-tailed test?
Problem 12 Solution
Step 1 — Read \(H_{a}\): The alternative hypothesis uses \(\neq\), which splits the rejection region between both tails of the distribution.
Answer: Two-tailed test.
Problem 13. What is the p-value?
Problem 13 Solution
Step 1 — Find the standard deviation of the difference:
$$ \sqrt{\frac{95^{2}}{15} + \frac{105^{2}}{15}} = \sqrt{601.6667 + 735} = \sqrt{1336.6667} = 36.5605 $$Step 2 — Standardize: \(z = \dfrac{800 - 900}{36.5605} = -2.7352\).
Step 3 — Find the two-tailed area: \(H_{a}: \mu_{1} \neq \mu_{2}\) is two-tailed, so the p-value doubles the area beyond \(|z| = 2.7352\).
Answer: p ≈ 0.0062.
Problem 14. Draw the graph of the p-value.
Problem 14 Solution
Step 1 — Choose the curve: This is a two-tailed test, so sketch a standard normal curve centered at 0.
Step 2 — Locate the observed difference: The observed sample difference is \(800 - 900 = -100^{\circ}\text{F}\), which standardizes to \(z = -2.7352\).
Step 3 — Shade both tails: Shade the region to the left of \(z = -2.7352\) and the mirror region to the right of \(z = 2.7352\). Together those two regions are the p-value.
Answer: A two-tailed normal curve centered at 0, with the regions beyond \(z = -2.7352\) and \(z = 2.7352\) both shaded; the combined shaded area is 0.0062.
Problem 15. At the 1% significance level, what is your conclusion?
Problem 15 Solution
Step 1 — Compare p-value to \(\alpha\): \(\alpha = 0.01\) and the p-value is 0.0062, so p-value \(< \alpha\). Reject \(H_{0}\).
Answer: Reject the null hypothesis. The data support that the melting point for Alloy Zeta is different from the melting point of Alloy Gamma.
Key Terms
known population standard deviation — the case in which \(\sigma_{1}\) and \(\sigma_{2}\) are given rather than estimated, so the difference of sample means is exactly normal and the test statistic is a z-score.
standard deviation of a difference — \(\sqrt{\frac{(\sigma_{1})^{2}}{n_{1}} + \frac{(\sigma_{2})^{2}}{n_{2}}}\); the same shape as the standard error in §9.1, but built from population values, so nothing is being estimated.
two-sample z-test — the test of two independent population means when both population standard deviations are known; it has no degrees of freedom to compute.
normal approximation to the Student's t — the permission to use the normal distribution for a difference of two means when \(n_{1} + n_{2} > 30\), even though the population standard deviations are unknown; a convenience, not a preference.