Carrying Out a Test for the Difference of Two Population Proportions - AP Statistics Study Guide

Quick Summary

This guide will equip you to perform a complete significance test for the difference between two population proportions. You will learn to state appropriate hypotheses, verify the necessary conditions, calculate the pooled standard error and the z-test statistic, and find the corresponding p-value. Ultimately, you will be able to draw a statistically sound conclusion in the context of a real-world problem, determining if there is convincing evidence of a difference between two groups.

Key Concepts

The goal of a two-sample z-test for a difference in proportions is to compare a categorical variable across two independent populations or two treatment groups in an experiment. We want to know if the true proportion of "successes" in the first group ($p_1$) is different from the true proportion in the second group ($p_2$).

The Four-Step Inference Process (SPDC)

1. Hypotheses

We start by assuming there is no difference between the two population proportions. This is our null hypothesis.

• Null Hypothesis (H₀):$p_1-p_2=0$ (or $p_1=p_2$)

  • This hypothesis claims that the true proportions for the two groups are equal.

• Alternative Hypothesis (Hₐ): This is what we are trying to find evidence for. It can be one of three forms:

  • Two-sided: `Hₐ: p₁ - p₂ \neq 0$ (or $p_1\ne p_2$) — We are looking for any difference.

  • One-sided (greater than): `Hₐ: p₁ - p₂ > 0$ (or $p_1>p_2$) — We are looking for evidence that the first group's proportion is larger.

  • One-sided (less than): `Hₐ: p₁ - p₂ < 0$ (or $p_1<p_2$) — We are looking for evidence that the first group's proportion is smaller.

2. Conditions

Before we can perform calculations, we must check three conditions to ensure our methods are valid.

• Random: The data must come from two independent random samples OR from two groups in a randomized experiment. This is crucial for generalizing results.

• 10% Condition (Independence within samples): When sampling without replacement, the sample sizes should be no more than 10% of their respective population sizes ($n_1\le 0.10N_1$ and $n_2\le 0.10N_2$). This allows us to treat the samples as independent even though we are sampling without replacement.

• Large Counts Condition (Normality): The sampling distribution of the difference in sample proportions, ${\hat{p}}_1-{\hat{p}}_2$, is approximately Normal if we have enough "successes" and "failures" in all groups. For a significance test, we use the pooled proportion to check this condition.

  • $n_1{\hat{p}}_c\ge 10$

  • $n_1(1-{\hat{p}}_c)\ge 10$

  • $n_2{\hat{p}}_c\ge 10$

  • $n_2(1-{\hat{p}}_c)\ge 10$

3. Calculations

If the conditions are met, we can proceed with calculating the test statistic and p-value.

• Pooled Sample Proportion (p̂_c): Because our null hypothesis assumes $p_1=p_2$, we combine, or "pool," our two samples to get a single, better estimate of this common proportion.

  • Formula:${\hat{p}}_c=(Numberofsuccessesinbothsamples)/(Totalindividualsinbothsamples)=(x_1+x_2)/(n_1+n_2)$

• Standard Error of the Difference (using p̂_c): This measures the typical variation in the difference between sample proportions, assuming the null hypothesis is true.

  • Formula:$S{E}_{p}ooled=sqrt(({\hat{p}}_c(1-{\hat{p}}_c)/n_1)+({\hat{p}}_c(1-{\hat{p}}_c)/n_2))$

• Test Statistic (z-score): This measures how many standard errors the observed difference in sample proportions is from the hypothesized difference of 0.

  • Formula:$z=(({\hat{p}}_1-{\hat{p}}_2)-0)/S{E}_{p}ooled$

• P-value: The p-value is the probability of observing a difference in sample proportions as extreme or more extreme than what we actually observed, assuming the null hypothesis is true. We find this using the standard Normal distribution (z-distribution).

  • [Image: Normal distribution curves showing left-tailed, right-tailed, and two-tailed p-values.]

  • For Hₐ: p₁ - p₂ > 0, the p-value is the area to the right of z.

  • For Hₐ: p₁ - p₂ < 0, the p-value is the area to the left of z.

  • For `Hₐ: p₁ - p₂ \neq 0$, the p-value is twice the area in the tail beyond z (e.g., $2\ast P(Z\ge |z|)$)

4. Conclusion

Our conclusion has two parts: a statistical decision and a contextual interpretation.

• Statistical Decision: Compare the p-value to the significance level, α (alpha).

  • If p-value \le α, we reject the null hypothesis (H₀).

  • If p-value > α, we fail to reject the null hypothesis (H₀).

• Contextual Interpretation: Write a sentence that explains what your decision means in the context of the problem.

  • If you reject H₀: "Because our p-value of [p-value] is less than α = [alpha level], we reject H₀. We have convincing evidence that [state the alternative hypothesis in words]."

  • If you fail to reject H₀: "Because our p-value of [p-value] is greater than α = [alpha level], we fail to reject H₀. We do not have convincing evidence that [state the alternative hypothesis in words]."

Key Vocabulary

• Two-Sample z-test for p₁ - p₂: The formal name of the significance test used to compare two proportions from two independent groups.

• Pooled Sample Proportion (p̂_c): An estimate of the common population proportion, calculated by combining the data from two samples. It is used in significance tests because the null hypothesis assumes the two population proportions are equal.

• Standard Error of the Difference: An estimate of the standard deviation of the sampling distribution of ${\hat{p}}_1-{\hat{p}}_2$. For a hypothesis test, it is calculated using the pooled sample proportion.

• Test Statistic (z): A standardized score that measures how many standard errors an observed statistic (in this case, ${\hat{p}}_1-{\hat{p}}_2$) is from the value claimed in the null hypothesis.

• P-value: The probability of obtaining a test statistic as extreme or more extreme than the one calculated from the sample data, assuming the null hypothesis is true. A small p-value provides evidence against the null hypothesis.

Calculator Tech (TI-84)

The entire two-proportion z-test can be performed efficiently on the TI-84 calculator.

Function:$2-PropZTest$

Keystrokes:

1. Press STAT.

2. Arrow over to the TESTS menu.

3. Select 6:2-PropZTest...

Inputs:

You will be prompted to enter the following information:

• $x1$: The number of successes in the first sample (must be an integer).

• $n1$: The sample size of the first sample.

• $x2$: The number of successes in the second sample (must be an integer).

• $n2$: The sample size of the second sample.

• $p1$: Choose the form of your alternative hypothesis ($\text{neqp}2$, $<p2$, $>p2$).

• $Calculate$: Press ENTER to run the test.

Output Screen:

The calculator will display:

• $z=...$: The calculated z-test statistic.

• $p=...$: The calculated p-value.

• $\hat{p}1=...$: The sample proportion for the first group ($x1/n1$).

• $\hat{p}2=...$: The sample proportion for the second group ($x2/n2$).

• $\hat{p}=...$: The pooled sample proportion, ${\hat{p}}_c$.

• $n1=...$ and $n2=...$: Your sample sizes.

How to Show Work on the FRQ

To earn full credit on an inference FRQ, you must clearly communicate every step of the process. Use the State, Plan, Do, Conclude framework.

STATE:

1. Define the parameters of interest in context.

   • $p_1=$ the true proportion of [context for group 1].

   • $p_2=$ the true proportion of [context for group 2].

2. State the hypotheses using correct symbols and context.

   • H₀: p₁ - p₂ = 0 (The true proportion of... is the same for both groups).

   • `Hₐ: p₁ - p₂ \neq 0$ (or $>$ or $<$) (The true proportion of... is different for the two groups).

3. State the significance level, α. If not given, use α = 0.05.

PLAN:

1. Name the procedure: "We will perform a two-sample z-test for a difference in population proportions."

2. Check the conditions for inference.

   • Random: "The data come from two independent random samples of [group 1] and [group 2]" OR "The treatments were randomly assigned to [experimental units]."

   • 10% Condition: "The sample of $n_1$ is less than 10% of all [population 1], and the sample of $n_2$ is less than 10% of all [population 2]. (e.g., $100<0.10\ast (allUSadults)$). It is reasonable to assume this."

   • Large Counts: "We must check the counts of successes and failures using the pooled proportion, ${\hat{p}}_c$."

     • First, calculate ${\hat{p}}_c=(x_1+x_2)/(n_1+n_2)$.

     • Then show all four calculations: $n_1{\hat{p}}_c=...\ge 10$, $n_1(1-{\hat{p}}_c)=...\ge 10$, $n_2{\hat{p}}_c=...\ge 10$, $n_2(1-{\hat{p}}_c)=...\ge 10$. Conclude: "Since all counts are at least 10, the sampling distribution of ${\hat{p}}_1-{\hat{p}}_2$ is approximately Normal."

DO:

1. Calculate the sample proportions ${\hat{p}}_1$ and ${\hat{p}}_2$.

2. Write the general formula for the test statistic: $z=({\hat{p}}_1-{\hat{p}}_2)/sqrt(({\hat{p}}_c(1-{\hat{p}}_c)/n_1)+({\hat{p}}_c(1-{\hat{p}}_c)/n_2))$.

3. Plug in the values and show the calculated z-statistic. It is acceptable to write "From calculator..." for the final values.

   • ${\hat{p}}_c=...$

   • $z=([value]-[value])/sqrt(([val]([val])/[val])+([val]([val])/[val]))=...$

4. State the p-value. A sketch of a Normal curve with the p-value shaded is highly recommended.

   • $p-value=...$

CONCLUDE:

1. Make a decision about the null hypothesis by comparing the p-value to alpha.

   • "Because the p-value of [value] is [less than / greater than] α = [value]..."

2. State whether you reject or fail to reject H₀.

   • "...we [reject / fail to reject] H₀."

3. Interpret the conclusion in the context of the problem, referencing the alternative hypothesis.

   • "We [have / do not have] convincing statistical evidence that the true proportion of [context for group 1] is [different from / greater than / less than] the true proportion of [context for group 2]."

Practice Problems

Problem 1:

A polling agency wants to investigate if there is a difference in the proportion of men and women in the U.S. who support a certain piece of legislation. They take a random sample of 250 men and find that 135 support it. They take a separate independent random sample of 300 women and find that 174 support it. Is there convincing evidence at the α = 0.05 level of a difference in the proportion of all U.S. men and women who support this legislation?

Solution:

STATE:

• $p_M$ = the true proportion of all U.S. men who support the legislation.

• $p_W$ = the true proportion of all U.S. women who support the legislation.

• H₀: p_M - p_W = 0 (The true proportions of support are the same for men and women).

• Hₐ: p_M - p_W \neq 0 (The true proportions of support are different for men and women).

• Significance level: $\alpha =0.05$.

PLAN:

• Procedure: We will perform a two-sample z-test for a difference in population proportions.

• Conditions:

  1. Random: The data come from "a random sample of 250 men" and a "separate independent random sample of 300 women." The condition is met.

  2. 10% Condition: 250 is less than 10% of all U.S. men, and 300 is less than 10% of all U.S. women. The condition is met.

  3. Large Counts: We first find the pooled proportion.

     $x_M=135$, $n_M=250$. $x_W=174$, $n_W=300$.

     ${\hat{p}}_c=(135+174)/(250+300)=309/550\approx 0.5618$.

     • $n_M\ast {\hat{p}}_c=250\ast 0.5618=140.45\ge 10$

     • $n_M\ast (1-{\hat{p}}_c)=250\ast (1-0.5618)=109.55\ge 10$

     • $n_W\ast {\hat{p}}_c=300\ast 0.5618=168.54\ge 10$

     • $n_W\ast (1-{\hat{p}}_c)=300\ast (1-0.5618)=131.46\ge 10$

     Since all expected counts are at least 10, the normality condition is met.

DO:

• Sample proportions: ${\hat{p}}_M=135/250=0.54$, ${\hat{p}}_W=174/300=0.58$.

• Test Statistic:

  $z=(({\hat{p}}_M-{\hat{p}}_W)-0)/sqrt(({\hat{p}}_c(1-{\hat{p}}_c)/n_M)+({\hat{p}}_c(1-{\hat{p}}_c)/n_W))$

  $z=((0.54-0.58)-0)/sqrt((0.5618(0.4382)/250)+(0.5618(0.4382)/300))$

  $z=-0.04/sqrt(0.000984+0.000820)=-0.04/0.04247\approx -0.942$

• P-value (from $2-PropZTest$ or normalcdf):

  Since this is a two-tailed test, $p-value=2\ast P(Z\le -0.942)=2\ast normalcdf(-1E99,-0.942,0,1)\approx 0.346$.

CONCLUDE:

• Because the p-value of 0.346 is greater than α = 0.05, we fail to reject H₀.

• We do not have convincing statistical evidence of a difference in the true proportion of all U.S. men and women who support this legislation.

Problem 2:

A pharmaceutical company has developed a new allergy drug. To test its effectiveness, they recruit 120 volunteers who suffer from allergies. They randomly assign 60 volunteers to receive the new drug and 60 to receive a placebo. After one week, 38 of the volunteers who received the drug reported a significant reduction in symptoms, while 29 of those who received the placebo reported a reduction. Does this experiment provide convincing evidence that the new drug is more effective than the placebo? Use a significance level of α = 0.01.

Solution:

STATE:

• $p_D$ = the true proportion of all allergy sufferers who would experience symptom reduction from the new drug.

• $p_P$ = the true proportion of all allergy sufferers who would experience symptom reduction from the placebo.

• H₀: p_D - p_P = 0 (The drug is not more effective than the placebo).

• Hₐ: p_D - p_P > 0 (The drug is more effective than the placebo).

• Significance level: $\alpha =0.01$.

PLAN:

• Procedure: We will perform a two-sample z-test for a difference in population proportions.

• Conditions:

  1. Random: The 120 volunteers were "randomly assigned" to the drug and placebo groups. The condition is met.

  2. 10% Condition: This is a randomized experiment, not random sampling from a population, so this condition is not applicable.

  3. Large Counts: We find the pooled proportion.

     $x_D=38$, $n_D=60$. $x_P=29$, $n_P=60$.

     ${\hat{p}}_c=(38+29)/(60+60)=67/120\approx 0.5583$.

     • $n_D\ast {\hat{p}}_c=60\ast 0.5583=33.5\ge 10$

     • $n_D\ast (1-{\hat{p}}_c)=60\ast (1-0.5583)=26.5\ge 10$

     • $n_P\ast {\hat{p}}_c=60\ast 0.5583=33.5\ge 10$

     • $n_P\ast (1-{\hat{p}}_c)=60\ast (1-0.5583)=26.5\ge 10$

     Since all expected counts are at least 10, the normality condition is met.

DO:

• Sample proportions: ${\hat{p}}_D=38/60\approx 0.6333$, ${\hat{p}}_P=29/60\approx 0.4833$.

• Test Statistic (using $2-PropZTest$ on calculator with x1=38, n1=60, x2=29, n2=60, p1:>p2):

  • $z\approx 1.66$

• P-value:

  • $p-value\approx 0.0485$

CONCLUDE:

• Because the p-value of 0.0485 is greater than α = 0.01, we fail to reject H₀.

• We do not have convincing statistical evidence that the true proportion of allergy sufferers who experience symptom reduction is greater for the new drug than for the placebo.

Common Mistakes to Avoid

• Forgetting to Pool for the Test: The biggest error is using the standard error formula from a two-proportion confidence interval for a significance test. For a test, the null hypothesis assumes $p_1=p_2$, so we must combine the samples to calculate a single pooled proportion (${\hat{p}}_c$) for both the Large Counts check and the standard error calculation.

• Incorrect Large Counts Check: Checking the Large Counts condition using the individual sample proportions (${\hat{p}}_1$ and ${\hat{p}}_2$) instead of the pooled proportion (${\hat{p}}_c$). The condition check must align with the assumption made in the null hypothesis.

• Calculator Input Error: Entering proportions (e.g., 0.65) into the $x1$ or $x2$ fields in the $2-PropZTest$ function on the calculator. These inputs must be the integer counts of successes. If you are given a percentage, you must first calculate the count by multiplying $\hat{p}\ast n$.

• Misinterpreting the P-value: Do not say "The p-value is the probability that the null hypothesis is true." The correct interpretation is: "The p-value is the probability of getting a sample result as or more extreme than ours, assuming the null hypothesis is true."

• Stating "Accept H₀": We never "accept" or "prove" the null hypothesis. We only find that we don't have enough evidence to reject it. Always use the phrase "fail to reject H₀."