9.3 Comparing Two Independent Population Proportions
SLO 4
Demonstrate an understanding of, and ability to use, basic ideas of statistical processes, including hypothesis tests and confidence interval estimation.
Comparing two proportions completes the inference toolkit for two samples. The pooled proportion is the conceptual centre: the null hypothesis asserts the two rates are equal, so the test estimates that single rate from both samples and measures the observed gap against it.
SLO 5
Identify appropriate statistical techniques and use technology-based statistical analysis to describe, interpret, and communicate results.
Interpretation is the harder half here, and the section makes it explicit — a six-figure sample can make a third-of-a-percentage-point difference highly significant, so communicating the result means reporting the size of the gap alongside the p-value.
Learning Objectives
By the end of this section, you will be able to:
- state the conditions that must hold before two independent population proportions can be compared;
- compute the pooled proportion \(p_{c}\) and explain why the test uses it in place of the two separate sample proportions;
- carry out a two-proportion z-test and interpret the result in the context of the claim;
- recognize that pooling is required here even though §9.1 forbade it for means.
Sections 9.1 and 9.2 compared two averages. This section compares two percentages, and the question is just as common: do two medications produce reactions at different rates, do two age groups own electric vehicles at different rates, do two states report the same racial makeup. The logic will be familiar by now — a difference, a standard error, a z-score, a p-value. One step is genuinely new, and it is worth watching for. When the null hypothesis says the two population proportions are equal, it is claiming there is really only one proportion, and the test takes that claim seriously by combining both samples into a single estimate of it.
9.3.1 Conditions and the Pooled Proportion
When conducting a hypothesis test that compares two independent population proportions, the following should be present:
- The two independent samples are simple random samples that are independent of one another.
- The number of successes is at least five and the number of failures is at least five for each of the samples.
- Each population should be at least ten to twenty times the size of its sample. This keeps a population from being over-sampled, which would otherwise distort the result.
Definition 9.3.1 — The success–failure condition is checked on four counts, not two sample sizes.
Generally the null hypothesis states that the two proportions are the same, \(H_{0}: p_{A} = p_{B}\). To conduct the test we use a pooled proportion, \(p_{c}\), which combines both samples into one estimate:
$$ p_{c} = \frac{x_{A} + x_{B}}{n_{A} + n_{B}} $$Definition 9.3.2 — The pooled proportion is one combined pile's rate, not the average of the two sample rates.
The difference of two proportions follows an approximate normal distribution:
$$ P^{\prime}_{A} - P^{\prime}_{B} \sim N\left[0,\ \sqrt{p_{c}(1 - p_{c})\left(\frac{1}{n_{A}} + \frac{1}{n_{B}}\right)}\right] $$Definition 9.3.3 — The difference of two proportions is centred at zero; the sample sizes set its spread.
The test statistic for comparing two independent population proportions is
$$ z = \frac{(p^{\prime}_{A} - p^{\prime}_{B}) - (p_{A} - p_{B})}{\sqrt{p_{c}(1 - p_{c})\left(\frac{1}{n_{A}} + \frac{1}{n_{B}}\right)}} $$The second condition is the one to check first, and it is the reason a proportion test can fail on data that looks perfectly adequate. A sample of 200 with only two successes has plenty of observations but almost no information about the success rate, and the normal approximation underneath this whole method breaks down. Count the successes and the failures separately, in both samples — four numbers, all of which must clear five.
The sample proportions themselves are what you would expect:
$$ p^{\prime}_{A} = \frac{x_{A}}{n_{A}} \qquad \text{and} \qquad p^{\prime}_{B} = \frac{x_{B}}{n_{B}} $$where \(x_{A}\) and \(x_{B}\) are the counts of successes in the two samples.
§9.1 told you never to pool the two sample variances, and now this section pools the two sample proportions. That looks like a contradiction and is not. In §9.1, pooling would have required assuming the two populations have equal standard deviations — an assumption the null hypothesis said nothing about, so you had no right to it. Here the null hypothesis itself asserts that \(p_{A} = p_{B}\), so under the null there is exactly one population proportion, and the best estimate of it uses every observation you collected. You pool when the null gives you permission, and only then.
Under a null hypothesis of equal proportions the term \((p_{A} - p_{B})\) is zero, so the numerator reduces to the observed difference in the two sample proportions.
Notice that \(p_{c}\) appears only in the denominator. The numerator keeps the two sample proportions apart, because the observed difference is the evidence; the denominator merges them, because the null hypothesis is what supplies the standard deviation you are measuring that evidence against.
Comparing two proportions, like comparing two means, is common. If two estimated proportions differ, it may be due to a real difference in the populations or it may be due to chance. A hypothesis test is what separates those two explanations.
9.3.2 Running a Two-Proportion z-Test
Two types of valves are being tested to determine if there is a difference in pressure tolerances. Fifteen out of a random sample of 100 of Valve A cracked under 4,500 psi. Six out of a random sample of 100 of Valve B cracked under 4,500 psi. Test at a 5% level of significance.
Solution — comparing two valve failure rates
Step 1 — Check the conditions. Successes and failures: 15 and 85 for Valve A, 6 and 94 for Valve B. All four counts are at least five, so the method applies.
Step 2 — Write the hypotheses. "Is there a difference" gives no direction, so the test is two-tailed:
$$ H_{0}: p_{A} = p_{B} \qquad H_{a}: p_{A} \neq p_{B} $$Step 3 — Compute the sample and pooled proportions.
$$ p^{\prime}_{A} = \frac{15}{100} = 0.15 \qquad p^{\prime}_{B} = \frac{6}{100} = 0.06 \qquad p_{c} = \frac{15 + 6}{100 + 100} = 0.105 $$Step 4 — Standardize. The standard deviation of the difference is
$$ \sqrt{(0.105)(0.895)\left(\frac{1}{100} + \frac{1}{100}\right)} = 0.0434 $$so \(z = \frac{0.15 - 0.06}{0.0434} = 2.076\).
Step 5 — Decide. The two-tailed p-value is 0.0379. Since \(0.0379 < 0.05\), reject \(H_{0}\).
Answer: At the 5% level of significance, there is sufficient evidence to conclude that the proportions of Valve A and Valve B that crack under 4,500 psi are different. Note that at a 1% level the same data would not have rejected — a reminder that the significance level is a commitment made before the test, not after seeing 0.0379.
Dr. Rafael Ocampo is testing two types of medication for hives to determine if there is a difference in the proportions of adult patient reactions. Twenty out of a random sample of 200 adults given medication A still had hives 30 minutes after taking the medication. Twelve out of another random sample of 200 adults given medication B still had hives 30 minutes after taking the medication. Test at a 1% level of significance.
Solution — a two-tailed test on reaction rates
Step 1 — Name the setting. The problem asks for a difference in proportions, making it a test of two proportions. Let A and B be the subscripts for medication A and medication B, so \(p_{A}\) and \(p_{B}\) are the population proportions of interest.
Step 2 — Name the random variable. \(P^{\prime}_{A} - P^{\prime}_{B}\), the difference in the proportions of adult patients who still had hives after 30 minutes on medication A and on medication B.
Step 3 — Write the hypotheses.
$$ H_{0}: p_{A} = p_{B} \quad \text{or} \quad p_{A} - p_{B} = 0 $$ $$ H_{a}: p_{A} \neq p_{B} \quad \text{or} \quad p_{A} - p_{B} \neq 0 $$The words "is a difference" tell you the test is two-tailed.
Step 4 — Choose the distribution. Since this is a test of two binomial population proportions, the distribution is normal. Pool first:
$$ p_{c} = \frac{x_{A} + x_{B}}{n_{A} + n_{B}} = \frac{20 + 12}{200 + 200} = 0.08 \qquad 1 - p_{c} = 0.92 $$ $$ P^{\prime}_{A} - P^{\prime}_{B} \sim N\left[0,\ \sqrt{(0.08)(0.92)\left(\frac{1}{200} + \frac{1}{200}\right)}\right] $$Step 5 — Compute. The estimated proportions are \(p^{\prime}_{A} = \frac{20}{200} = 0.1\) and \(p^{\prime}_{B} = \frac{12}{200} = 0.06\), so the observed difference is \(0.1 - 0.06 = 0.04\). The standard deviation of the difference is 0.0271, giving \(z = 1.47\). The two-tailed p-value is 0.1404: half of it lies below \(-0.04\) and half above \(0.04\).
Step 6 — Compare and decide. \(\alpha = 0.01\) and the p-value is 0.1404, so \(\alpha <\) p-value. Do not reject \(H_{0}\).
Answer: At a 1% level of significance, from the sample data, there is not sufficient evidence to conclude that there is a difference in the proportions of adult patients who still had hives after 30 minutes on medication A and medication B.
On a TI-83/83+/84/84+: press STAT, arrow over to TESTS, and press 6:2-PropZTest. Enter 20 for x1, 200 for n1, 12 for x2, and 200 for n2. Arrow down to p1: and arrow to the \(\neq p2\) option. Press ENTER, arrow down to Calculate and press ENTER. The calculator returns p = 0.1404 and the test statistic \(z = 1.47\). Running the same instructions but choosing Draw instead of Calculate produces the shaded graph.
Try it in rāSHio
A proportion test needs a different panel from a mean, even though the logic is identical. In rāSHio choose Stats → Prop Stats and enter this example's counts — 20 successes out of 200 for medication A, 12 out of 200 for medication B — then set the alternative to not-equal. It returns \(z = 1.47\) and the two-tailed p-value 0.1404. Two things it does for you are the two easiest to lose by hand: it forms the pooled proportion \(p_{c}\) itself, and because you chose not-equal it doubles the tail area without being asked. The clip below runs this example end to end: Samples set to two, 20 of 200 against 12 of 200, and the same z and p-value off the results card.
Figure 9.3.1 — The rāSHio Prop Stats panel: Stats → Prop Stats. The walkthrough switches Samples to two and runs this example's own counts — 20 of 200 against 12 of 200 — returning z = 1.4744 and the two-tailed p-value 0.1404. The pooled proportion never appears as a row of its own; it is already inside the standard error the card reports as 0.0271.
9.3.3 One-Tailed Comparisons
When the claim points in a direction, the test becomes one-tailed and the order in which you subtract starts to matter. The two examples below are both directional, and one of them ends up rejecting while the other does not.
A survey was conducted about tea as a favored beverage. The data collected are summarized below. Is the proportion of men favoring tea more than the proportion of women favoring tea? Test at a 1% level of significance.
| Men | Women | |
|---|---|---|
| Favor tea | 16 | 18 |
| Total surveyed | 230 | 218 |
Solution — a right-tailed test that was never going to reject
Step 1 — Check the conditions. Successes and failures: 16 and 214 for men, 18 and 200 for women. All at least five.
Step 2 — Write the hypotheses. The claim is that men favor tea at a higher rate, so with M for men and W for women:
$$ H_{0}: p_{M} \leq p_{W} \qquad H_{a}: p_{M} > p_{W} $$This is a right-tailed test.
Step 3 — Compute the proportions.
$$ p^{\prime}_{M} = \frac{16}{230} = 0.0696 \qquad p^{\prime}_{W} = \frac{18}{218} = 0.0826 \qquad p_{c} = \frac{16 + 18}{230 + 218} = 0.0759 $$Step 4 — Standardize. The standard deviation of the difference is 0.0250, so
$$ z = \frac{0.0696 - 0.0826}{0.0250} = -0.52 $$Step 5 — Decide. For a right-tailed test the p-value is the area above \(z = -0.52\), which is 0.6983. Since \(0.6983 > 0.01\), do not reject \(H_{0}\).
Answer: At the 1% level of significance, there is not sufficient evidence to conclude that the proportion of men favoring tea is more than the proportion of women favoring tea.
Read the sign before you calculate. The sample proportion for men, 0.0696, is lower than the one for women, 0.0826 — the data point the opposite way from the claim. A right-tailed test on data that leans left cannot produce a small p-value, and a p-value near 0.7 is the signature of exactly that situation. Recognizing it early saves you the arithmetic.
A research study was conducted about gender differences in the use of seat belts in motor vehicles. The researcher believed that the proportion of women not wearing seat belts is less than the proportion of men not wearing seat belts. The data collected represent a random sample of U.S. adults and are summarized below. Is the proportion of women not wearing seat belts less than the proportion of men not wearing seat belts? Test at a 1% level of significance.
| Men | Women | |
|---|---|---|
| Does not wear seat belts | 183 | 156 |
| Total number surveyed | 2,231 | 2,169 |
Solution — a left-tailed test on a large sample
Step 1 — Name the setting. This is a test of two population proportions. Let M and F be the subscripts for men and women, so \(p_{M}\) and \(p_{F}\) are the population proportions of interest.
Step 2 — Name the random variable. \(P^{\prime}_{F} - P^{\prime}_{M}\), the difference in the proportions of women and men who do not wear seat belts.
Step 3 — Write the hypotheses.
$$ H_{0}: p_{F} = p_{M} \quad \text{or} \quad p_{F} - p_{M} = 0 $$ $$ H_{a}: p_{F} < p_{M} \quad \text{or} \quad p_{F} - p_{M} < 0 $$The words "less than" tell you the test is left-tailed. Because the researcher's claim is about women being lower, the difference is written with \(p_{F}\) first.
Step 4 — Choose the distribution. Since this is a test of two population proportions, the distribution is normal. Pool:
$$ p_{c} = \frac{x_{F} + x_{M}}{n_{F} + n_{M}} = \frac{156 + 183}{2169 + 2231} = 0.077 \qquad 1 - p_{c} = 0.923 $$ $$ P^{\prime}_{F} - P^{\prime}_{M} \sim N\left(0,\ \sqrt{(0.077)(0.923)\left(\frac{1}{2169} + \frac{1}{2231}\right)}\right) $$Step 5 — Compute. The estimated proportion for women is 0.0719 and for men 0.0820, so the observed difference is \(-0.0101\). The standard deviation of the difference is 0.00804, giving \(z = -1.256\) and a left-tail p-value of 0.1045.
Step 6 — Decide. \(\alpha = 0.01\) and the p-value is 0.1045, so \(\alpha <\) p-value. Do not reject \(H_{0}\).
Answer: At the 1% level of significance, from the sample data, there is not sufficient evidence to conclude that the proportion of women not wearing seat belts is less than the proportion of men not wearing seat belts. This is worth contrasting with Try It Now 9.3.2: here the data do lean the way the claim predicted, and with more than 4,000 people surveyed — yet a gap of one percentage point on a base rate near 8% still is not enough at the 1% level.
On a TI-83/83+/84/84+: press STAT, arrow over to TESTS, and press 6:2-PropZTest. Enter 156 for x1, 2169 for n1, 183 for x2, and 2231 for n2. Arrow down to p1: and arrow to the \(<p2\) option. Press ENTER, arrow down to Calculate and press ENTER. The calculator returns p = 0.1045 and the test statistic \(z = -1.256\).
Try it in rāSHio
A one-tailed proportion test is where the order you enter the two samples starts to matter. In rāSHio choose Stats → Prop Stats, enter this study's counts — 156 of 2169 for women as the first sample, 183 of 2231 for men as the second — then set the alternative to less-than. It returns \(z = -1.256\) and the left-tail p-value 0.1045. Swap the two samples and leave the alternative alone, and the panel will hand you 0.8955 instead: the direction you claim is read against the order you typed, which is the same trap the example's \(p_{F} - p_{M}\) ordering is warning you about.
Dr. Hana Sato, a government researcher who rides with her wife most weekends, is investigating whether there is a difference in the use of helmets by motorcyclists in different geographic regions, among those states where helmet use is required by law. The survey records that for motorcyclists in the northeast U.S., 7,622 out of 113,231 motorcyclists did not wear helmets. In the southeast U.S., 7,439 out of 104,873 motorcyclists did not wear helmets. Test at a 5% significance level, and answer the following questions.
a. Is this a test of two means or two proportions?
b. Which distribution do you use to perform the test?
c. What is the random variable?
d. What are the null and alternative hypotheses? Write them in symbols.
e. Is this test right-, left-, or two-tailed?
f. What is the p-value?
g. Do you reject or not reject the null hypothesis?
h. Write the conclusion.
Solution — helmet use in two regions
a. Two proportions. Each region is summarized by a rate of non-use, not an average.
b. The normal distribution.
c. \(P^{\prime}_{NE} - P^{\prime}_{SE}\), the difference in the proportions of motorcyclists not wearing helmets in the northeast and the southeast.
d. \(H_{0}: p_{NE} = p_{SE}\) and \(H_{a}: p_{NE} \neq p_{SE}\).
e. Two-tailed — "whether there is a difference" gives no direction.
f. The sample proportions are \(\frac{7622}{113231} = 0.0673\) and \(\frac{7439}{104873} = 0.0709\), and the pooled proportion is \(p_{c} = \frac{7622 + 7439}{113231 + 104873} = 0.0691\). The standard deviation of the difference is 0.00109, so \(z = -3.33\) and the two-tailed p-value is 0.00087.
g. Reject the null hypothesis, since \(0.00087 < 0.05\).
Answer (h): At the 5% level of significance, from the sample data, there is sufficient evidence to conclude that the proportion of motorcyclists who do not wear helmets differs between the northeast and the southeast.
A note on what significance does and does not tell you. The difference here is 0.36 of a percentage point — 6.73% against 7.09% — which is tiny, and it is highly significant only because the samples run to six figures. With enough data almost any real difference becomes detectable. Whether a gap that small matters for policy is a separate question that no p-value answers, and it is the same point Cohen's \(d\) made for means back in §9.1.
A marketing firm claims that the proportion of younger adults who own electric vehicles is greater than the proportion of older adults who own electric vehicles. A random sample of U.S. adults was taken, and the survey indicates the following:
- Out of a sample of 232 older adults (aged 35 or older), 5% own electric vehicles.
- Out of a sample of 1,343 younger adults (aged 34 or younger), 10% own electric vehicles.
Is the proportion of younger adults greater than the proportion of older adults with respect to owning electric vehicles? Test at the 5% level of significance.
Solution — a right-tailed test from reported percentages
Step 1 — Turn the percentages back into counts. The test needs successes, not rates: 10% of 1,343 is 134 younger adults, and 5% of 232 is 12 older adults. Let Y and O be the subscripts for younger and older adults.
Step 2 — Name the random variable. \(P^{\prime}_{Y} - P^{\prime}_{O}\), the difference in the proportions of younger and older adults who own electric vehicles.
Step 3 — Write the hypotheses.
$$ H_{0}: p_{Y} = p_{O} \quad \text{or} \quad p_{Y} - p_{O} = 0 $$ $$ H_{a}: p_{Y} > p_{O} \quad \text{or} \quad p_{Y} - p_{O} > 0 $$The words "greater than" indicate that the test is right-tailed.
Step 4 — Choose the distribution. The distribution is approximately normal. Pool:
$$ p_{c} = \frac{x_{Y} + x_{O}}{n_{Y} + n_{O}} = \frac{134 + 12}{1343 + 232} = 0.0927 \qquad 1 - p_{c} = 0.9073 $$ $$ P^{\prime}_{Y} - P^{\prime}_{O} \sim N\left(0,\ \sqrt{(0.0927)(0.9073)\left(\frac{1}{1343} + \frac{1}{232}\right)}\right) $$Step 5 — Compute. The estimated proportions are \(p^{\prime}_{Y} = 0.10\) and \(p^{\prime}_{O} = 0.05\). The standard deviation of the difference is 0.0206, so \(z = 2.33\) and the right-tail p-value is 0.0099.
Step 6 — Decide. \(\alpha = 0.05 > 0.0099\), so reject \(H_{0}\).
Answer: At the 5% level of significance, from the sample data, there is sufficient evidence to conclude that a larger proportion of younger adults own electric vehicles as compared to older adults.
Where the rounding bites. The problem gives you percentages, not counts, and 10% of 1,343 is 134.3 — not a whole number of people. Rounding to 134 gives \(z = 2.33\) and p = 0.0099; rounding to 135 gives \(z = 2.36\) and p = 0.0092. Both reject at the 5% level, so the conclusion is safe either way, but the two answers differ in the third decimal place. When a problem hands you a percentage of an odd-sized sample, expect that kind of wobble and do not treat the last digit of your p-value as exact.
On a TI-83/83+/84/84+: press STAT, arrow over to TESTS, and press 6:2-PropZTest. Enter 134 for x1, 1343 for n1, 12 for x2, and 232 for n2. Arrow down to p1: and arrow to the \(>p2\) option. Press ENTER, arrow down to Calculate and press ENTER. The calculator returns \(z = 2.33\) and p = 0.0099.
Try it in rāSHio
Stats → Prop Stats in rāSHio asks for successes and sample sizes, never percentages, so this example's 10% of 1,343 and 5% of 232 have to become 134 and 12 before you can enter anything — the conversion Step 1 does by hand is the panel's entry requirement. With the alternative set to greater-than it returns \(z = 2.33\) and p = 0.0099. Then re-enter it with 135 instead of 134: the p-value moves to 0.0092, which makes the rounding wobble the example warns about something you can watch rather than take on trust.
Problem Set 9.3
Use the following information to answer the next five exercises. Two types of phone operating system are being tested to determine if there is a difference in the proportions of system failures (crashes). Fifteen out of a random sample of 150 phones with \(OS_{1}\) had system failures within the first eight hours of operation. Nine out of another random sample of 150 phones with \(OS_{2}\) had system failures within the first eight hours of operation. \(OS_{2}\) is believed to be more stable (have fewer crashes) than \(OS_{1}\).
Problem 1. Is this a test of means or proportions?
Problem 1 Solution
Step 1 — Identify what each sample measures: Each operating system is summarized by a failure rate — the fraction of phones that crashed within the first eight hours — not by an average of some measured quantity.
Answer: This is a test of two proportions.
Problem 2. What is the random variable?
Problem 2 Solution
Step 1 — Name the quantity the test tracks: The two samples each give a rate of system failures, so the random variable is the difference between those two sample proportions.
Answer: \(P^{\prime}_{OS_{1}} - P^{\prime}_{OS_{2}}\), the difference in the proportions of phones that had system failures within the first eight hours of operation with \(OS_{1}\) and \(OS_{2}\).
Problem 3. State the null and alternative hypotheses.
Problem 3 Solution
Step 1 — Translate the claim into symbols: "\(OS_{2}\) is believed to be more stable (have fewer crashes) than \(OS_{1}\)" is a claim that \(OS_{1}\)'s failure rate is the larger one, so the alternative hypothesis carries that direction.
Answer:
$$ H_{0}: p_{OS_{1}} = p_{OS_{2}} \quad \text{or} \quad p_{OS_{1}} - p_{OS_{2}} = 0 $$ $$ H_{a}: p_{OS_{1}} > p_{OS_{2}} \quad \text{or} \quad p_{OS_{1}} - p_{OS_{2}} > 0 $$Problem 4. What is the p-value?
Problem 4 Solution
Step 1 — Check the conditions and pool: Successes and failures are 15 and 135 for \(OS_{1}\), 9 and 141 for \(OS_{2}\) — all at least five. Pool the two samples:
$$ p_{c} = \frac{15 + 9}{150 + 150} = 0.08 $$Step 2 — Standardize: \(p^{\prime}_{OS_{1}} = \frac{15}{150} = 0.10\) and \(p^{\prime}_{OS_{2}} = \frac{9}{150} = 0.06\). The standard deviation of the difference is
$$ \sqrt{(0.08)(0.92)\left(\frac{1}{150} + \frac{1}{150}\right)} = 0.031326 $$so \(z = \frac{0.10 - 0.06}{0.031326} = 1.2769\). This is a right-tailed test, so the p-value is the area above \(z = 1.2769\).
Answer: p-value = 0.1008.
Problem 5. What can you conclude about the two operating systems?
Problem 5 Solution
Step 1 — Compare the p-value to a significance level: The p-value, 0.1008, is larger than any conventional \(\alpha\) — 0.10, 0.05, or 0.01 — so \(H_{0}\) is not rejected at any of them.
Answer: There is not sufficient evidence to conclude that \(OS_{1}\)'s failure rate is higher than \(OS_{2}\)'s. In other words, this sample does not establish that \(OS_{2}\) is more stable than \(OS_{1}\).
Use the following information to answer the next twelve exercises. In a recent Census, three percent of the U.S. population reported being of two or more races. However, the percent varies tremendously from state to state. Suppose that two random surveys are conducted. In the first random survey, out of 1,000 North Dakotans, only nine people reported being of two or more races. In the second random survey, out of 500 Nevadans, 17 people reported being of two or more races. Conduct a hypothesis test to determine if the population percents are the same for the two states or if the percent for Nevada is statistically higher than for North Dakota.
Problem 6. Is this a test of means or proportions?
Problem 6 Solution
Step 1 — Identify what each survey measures: Each state reported a percent of respondents who identify as two or more races, not an average of a measured quantity.
Answer: This is a test of two proportions.
Problem 7. State the null and alternative hypotheses.
a) \(H_{0}\): \(\underline{\hspace{6em}}\)
b) \(H_{a}\): \(\underline{\hspace{6em}}\)
Problem 7 Solution
Step 1 — Translate the claim into symbols: The question asks whether Nevada's percent is statistically higher than North Dakota's, so the alternative hypothesis carries that direction. Let \(NV\) and \(ND\) be the subscripts for Nevada and North Dakota.
Answer:
a) \(H_{0}: p_{NV} = p_{ND}\) (or \(p_{NV} - p_{ND} = 0\))
b) \(H_{a}: p_{NV} > p_{ND}\) (or \(p_{NV} - p_{ND} > 0\))
Problem 8. Is this a right-tailed, left-tailed, or two-tailed test? How do you know?
Problem 8 Solution
Step 1 — Read the direction in the claim: "Statistically higher" points one way — it is not merely asking whether the two percents differ, but whether Nevada's is greater.
Answer: Right-tailed. The claim specifies a direction (Nevada higher, not just "different"), and the difference in the alternative hypothesis, \(p_{NV} - p_{ND} > 0\), is written to test the upper tail.
Problem 9. What is the random variable of interest for this test?
Problem 9 Solution
Step 1 — Name the statistic being tested: The test compares the two sample proportions of respondents who are two or more races.
Answer: \(P^{\prime}_{NV} - P^{\prime}_{ND}\)
Problem 10. In words, define the random variable for this test.
Problem 10 Solution
Step 1 — Put the random variable into words: It is the same quantity as 9.3.9, described rather than symbolized.
Answer: The random variable is the difference in proportions (percents) of the populations that are of two or more races in Nevada and North Dakota.
Problem 11. Which distribution (normal or Student's t) would you use for this hypothesis test?
Problem 11 Solution
Step 1 — Identify the setting: This is a hypothesis test comparing two independent population proportions.
Answer: The normal distribution.
Problem 12. Explain why you chose the distribution you did for Exercise 9.3.11.
Problem 12 Solution
Step 1 — Check the success–failure condition: North Dakota has 9 successes and 991 failures; Nevada has 17 successes and 483 failures. All four counts clear five by a wide margin.
Answer: Our sample sizes are much greater than five each, so we use the normal distribution for two proportions for this hypothesis test.
Problem 13. Calculate the test statistic.
Problem 13 Solution
Step 1 — Compute the sample and pooled proportions:
$$ p^{\prime}_{NV} = \frac{17}{500} = 0.0340 \qquad p^{\prime}_{ND} = \frac{9}{1000} = 0.0090 \qquad p_{c} = \frac{17 + 9}{500 + 1000} = 0.017333 $$Step 2 — Standardize: The standard deviation of the difference is
$$ \sqrt{(0.017333)(0.982667)\left(\frac{1}{500} + \frac{1}{1000}\right)} = 0.007148 $$so \(z = \frac{0.0340 - 0.0090}{0.007148} = 3.4973\).
Answer: \(z \approx 3.50\)
Problem 14. Sketch a graph of the situation. Mark the hypothesized difference and the sample difference. Shade the area corresponding to the p-value.
Problem 14 Solution
Step 1 — Describe the sketch: Draw a normal curve centered at 0, since the hypothesized difference under \(H_{0}\) is \(p_{NV} - p_{ND} = 0\). Mark the sample difference, \(0.034 - 0.009 = 0.025\), to the right of center. Because the test is right-tailed, shade the area under the curve to the right of the sample difference — the region at or beyond \(z = 3.50\) — and label that shaded area as the p-value.
Answer: Answers may vary. The sketch should show a normal curve centered at 0 (the hypothesized difference), the sample difference of 0.025 marked to its right, and the tail to the right of 0.025 shaded to represent the p-value.
Problem 15. Find the p-value.
Problem 15 Solution
Step 1 — Read the tail area: With \(z = 3.4973\) and a right-tailed test, the p-value is the area under the standard normal curve above \(z\).
Answer: p-value = 0.000235 (approximately 0.0002).
Problem 16. At a pre-conceived \(\alpha = 0.05\), what is your:
a) Decision:
b) Reason for the decision:
c) Conclusion (write out in a complete sentence):
Problem 16 Solution
Step 1 — Compare the p-value to \(\alpha\): \(\alpha = 0.05\) and the p-value is 0.000235, so p-value \(< \alpha\).
Answer:
a) Decision: Reject the null hypothesis.
b) Reason for the decision: p-value \(< \alpha\).
c) Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the proportion of the population that is of two or more races in Nevada is statistically higher than that in North Dakota.
Problem 17. Does it appear that the proportion of Nevadans who are two or more races is higher than the proportion of North Dakotans? Why or why not?
Problem 17 Solution
Step 1 — Weigh the evidence: The test rejected \(H_{0}\) with a p-value of 0.000235, far below 0.05 — strong evidence against equal proportions, not the kind of result that shows up from sampling variation alone.
Answer: Yes. The hypothesis test rejected the null hypothesis of equal proportions at the 5% level, so the sample data support the conclusion that a higher percentage of Nevadans than North Dakotans are of two or more races.
Key Terms
independent population proportions — two proportions estimated from separate random samples, where neither sample's membership depends on the other's.
pooled proportion — \(p_{c} = \frac{x_{A} + x_{B}}{n_{A} + n_{B}}\), the single success rate estimated from both samples combined; used because the null hypothesis asserts the two population proportions are equal.
two-proportion z-test — the test of two independent population proportions, using the normal distribution and the pooled proportion in its standard deviation.
success–failure condition — the requirement that each sample have at least five successes and at least five failures before the normal approximation may be used.