Two Proportions: Videos & Practice Problems

In hypothesis testing for two samples, the goal is to determine if there is a significant difference between the proportions of two groups. The process closely mirrors that of a one-sample test, involving the formulation of hypotheses, calculation of test statistics, and interpretation of results.

To begin, it is essential to establish the null hypothesis (H₀) and the alternative hypothesis (H_A). For two proportions, the null hypothesis typically states that the two population proportions are equal: H₀: p₁ = p₂. The alternative hypothesis can be two-tailed, indicating that the proportions are not equal: H_A: p₁ ≠ p₂.

Before proceeding with calculations, certain conditions must be verified. The samples should be random and independent, and there should be at least five successes and five failures in each sample to ensure normality. For example, if we have a study on the effectiveness of a nicotine patch, we might find that 11 out of 20 participants using a placebo quit smoking, while 17 out of 23 participants using the patch quit. Both groups meet the criteria for normality.

Next, we calculate the test statistic using the formula for the z-score for two proportions:

$$ z = \frac{(p_1 - p_2) - (P_1 - P_2)}{\sqrt{P_{bar} \cdot (1 - P_{bar}) \left( \frac{1}{n_1} + \frac{1}{n_2} \right)}} $$

Here, p₁ and p₂ are the sample proportions, and P_bar is the pooled proportion calculated as:

$$ P_{bar} = \frac{x_1 + x_2}{n_1 + n_2} $$

where x₁ and x₂ are the number of successes in each group, and n₁ and n₂ are the sample sizes. In our example, P_bar would be calculated as:

$$ P_{bar} = \frac{11 + 17}{20 + 23} = \frac{28}{43} \approx 0.65 $$

Substituting the values into the z-score formula allows us to find the z-score, which in this case is approximately -1.3.

After calculating the z-score, the next step is to determine the p-value associated with this z-score. For a two-tailed test, the p-value is calculated as:

$$ p\text{-value} = 2 \cdot P(Z < z) $$

Using statistical tables or software, we find that the p-value corresponding to a z-score of -1.3 is approximately 0.193.

Finally, we compare the p-value to the significance level (α = 0.05). Since the p-value (0.193) is greater than α, we fail to reject the null hypothesis. This indicates that there is not enough evidence to conclude that there is a significant difference in the proportions of individuals quitting smoking between the two methods.

In summary, conducting a hypothesis test for two proportions involves formulating hypotheses, ensuring sample conditions are met, calculating the z-score using pooled proportions, determining the p-value, and drawing conclusions based on the comparison of the p-value to the significance level.

In this analysis, we explore the effectiveness of seat belts in reducing injuries through a hypothesis test involving two independent samples of drivers. The first group consists of drivers not wearing seat belts, where 50 out of 2,400 were injured, while the second group includes drivers wearing seat belts, with 15 out of 1,600 injured. We aim to test the claim that not wearing a seat belt results in a greater proportion of injuries, using a significance level (alpha) of 0.01.

To begin, we establish our null hypothesis (H₀): p₁ = p₂, indicating that the proportions of injuries in both groups are equal. The alternative hypothesis (H_a) is p₁ > p₂, suggesting that the proportion of injuries among those not wearing seat belts is greater than those wearing them.

Next, we calculate the sample proportions: for the first group (not wearing seat belts), p₁ = 50/2400 = 0.0208, and for the second group (wearing seat belts), p₂ = 15/1600 = 0.0094. We then compute the pooled sample proportion (p̄) as follows:

p̄ = (x₁ + x₂) / (n₁ + n₂) = (50 + 15) / (2400 + 1600) = 65 / 4000 = 0.0163.

The complement of the pooled proportion, q̄, is calculated as q̄ = 1 - p̄ = 0.9837. With these values, we can now compute the z-score using the formula:

z = (p₁ - p₂) / √[p̄ * q̄ * (1/n₁ + 1/n₂)].

Substituting the values, we find:

z = (0.0208 - 0.0094) / √[0.0163 * 0.9837 * (1/2400 + 1/1600)] ≈ 2.81.

To determine the p-value, we look up the z-score in the standard normal distribution. For z = 2.81, the p-value is approximately 0.0025. Since this p-value is less than our significance level of 0.01, we reject the null hypothesis.

In conclusion, there is sufficient evidence to support the claim that not wearing a seat belt results in a greater proportion of injuries compared to wearing one. This reinforces the importance of seat belt use for driver safety.

In statistical analysis, constructing a confidence interval for the difference in proportions between two samples is a crucial skill. This process closely mirrors the steps taken for a single sample, with a few modifications to the equations used for the point estimator and margin of error.

To begin, the point estimator for the difference in proportions is calculated by subtracting the sample proportions: \( \hat{p}_1 - \hat{p}_2 \). For instance, if \( \hat{p}_1 = 0.55 \) and \( \hat{p}_2 = 0.74 \), the point estimate would be \( 0.55 - 0.74 = -0.19 \). This indicates a 19% difference in success rates between the two groups being studied.

Next, the margin of error (E) is determined using the critical z-value corresponding to the desired confidence level, which in this case is 90%. The critical z-value for a 90% confidence level is approximately 1.645. The formula for the margin of error for two sample proportions is given by:

$$ E = z \cdot \sqrt{\frac{\hat{p}_1(1 - \hat{p}_1)}{n_1} + \frac{\hat{p}_2(1 - \hat{p}_2)}{n_2}} $$

Here, \( n_1 \) and \( n_2 \) are the sample sizes for the two groups. For example, if \( n_1 = 20 \) and \( n_2 = 23 \), and using the previously calculated proportions, the margin of error can be computed. After performing the calculations, suppose the margin of error is found to be 0.24.

With the point estimate and margin of error, the confidence interval can be constructed. The lower and upper bounds of the confidence interval are calculated as follows:

Lower Bound: \( -0.19 - 0.24 = -0.43 \)

Upper Bound: \( -0.19 + 0.24 = 0.05 \)

Thus, the 90% confidence interval for the difference in proportions is \( (-0.43, 0.05) \).

Interpreting this confidence interval in relation to hypothesis testing is essential. The null hypothesis typically posits that there is no difference in proportions (i.e., the difference is zero). If the confidence interval includes zero, as it does in this case, we fail to reject the null hypothesis, indicating insufficient evidence to claim a difference in success rates between the two groups. Conversely, if the interval did not include zero, we would reject the null hypothesis, suggesting a significant difference exists.

In summary, constructing a confidence interval for the difference in proportions involves calculating the point estimate, determining the margin of error using the critical z-value, and interpreting the results in the context of hypothesis testing. This process is vital for making informed conclusions based on sample data.

In this analysis, we explore how to construct a confidence interval for the difference in proportions between two groups of students: those living on campus and those living off campus, regarding their attendance at campus events. The goal is to determine if on-campus students are more likely to attend events than off-campus students.

We begin by gathering data from two random samples: 50 on-campus students, of which 102 attended at least one event, and 30 off-campus students, with 74 attending at least one event. To construct a 95% confidence interval, we first verify the conditions for using a normal approximation, ensuring that the samples are random and independent, and that there are more than five successes and failures in each group.

The next step involves calculating the critical z-value for a 95% confidence level, which is found to be 1.96. This value is derived from the standard normal distribution, representing the z-score that captures the central 95% of the distribution.

We then calculate the sample proportions for both groups. The proportion of on-campus students who attended events, denoted as \( \hat{p}_1 \), is calculated as:

\( \hat{p}_1 = \frac{x_1}{n_1} = \frac{102}{150} = 0.68 \)

For off-campus students, the proportion \( \hat{p}_2 \) is:

\( \hat{p}_2 = \frac{x_2}{n_2} = \frac{74}{130} \approx 0.569 \)

The point estimate for the difference in proportions is then calculated as:

\( \hat{p}_1 - \hat{p}_2 = 0.68 - 0.569 = 0.111 \)

Next, we compute the margin of error using the formula:

\( E = z_{\alpha/2} \sqrt{\frac{\hat{p}_1(1 - \hat{p}_1)}{n_1} + \frac{\hat{p}_2(1 - \hat{p}_2)}{n_2}} \)

Substituting the values, we find:

\( E = 1.96 \sqrt{\frac{0.68 \times 0.32}{50} + \frac{0.569 \times 0.431}{30}} \approx 0.113 \)

With the point estimate and margin of error, we can now establish the confidence interval:

Lower bound: \( 0.111 - 0.113 = -0.002 \)

Upper bound: \( 0.111 + 0.113 = 0.224 \)

Thus, the 95% confidence interval for the difference in proportions is approximately \( (-0.002, 0.224) \).

To evaluate the claim that on-campus students are more likely to attend events, we analyze whether zero is included in our confidence interval. Since the interval does include zero, we fail to reject the null hypothesis, which posits that there is no difference in attendance between the two groups. This indicates that there is insufficient evidence to support the claim that on-campus students are more likely to attend events than their off-campus counterparts.

In summary, the constructed confidence interval and subsequent hypothesis testing reveal that while there is a point estimate suggesting on-campus students may attend more events, the evidence does not strongly support this claim, as indicated by the inclusion of zero in the confidence interval.