10. Hypothesis Testing for Two Samples

Two Proportions

10. Hypothesis Testing for Two Samples

Two Proportions: Videos & Practice Problems

Topic summary

In hypothesis testing for two samples, the null hypothesis states that the proportions are equal (p₁ = p₂), while the alternative hypothesis indicates a difference (p₁ ≠ p₂). The test statistic is calculated using the formula: $z^{2} = \frac{p ₁ - p ₂}{\sqrt{\frac{p ₁ (1 - p ₁)}{n₁} + \frac{p ₂ (1 - p ₂)}{n₂}}} .$ The p-value is then compared to the significance level (α) to draw conclusions. For confidence intervals, the point estimate is the difference in sample proportions, and the margin of error is calculated using critical z-values and sample proportions.

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Difference in Proportions: Hypothesis Tests

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Difference in Proportions: Hypothesis Tests Video Summary

In hypothesis testing involving two samples, the goal is to determine if there is a significant difference between the proportions of two groups. The process mirrors that of a one-sample test, with a few key adjustments. Initially, you will formulate your null hypothesis (H₀) and alternative hypothesis (H_a). For two proportions, the null hypothesis typically states that the two population proportions are equal (H₀: p₁ = p₂), while the alternative hypothesis indicates that they are not equal (H_a: p₁ ≠ p₂), suggesting a two-tailed test.

Before proceeding, it is essential to verify that the samples are random and independent, and that there are at least five successes and five failures in each sample to satisfy the normality condition. For example, if you have a study on the effectiveness of a nicotine patch, you might find that 11 out of 20 participants using a placebo quit smoking, while 17 out of 23 participants using the patch quit. This data allows you to calculate the sample proportions: p₁ = 11/20 = 0.55 and p₂ = 17/23 ≈ 0.74.

The next step involves calculating the test statistic, which is a z-score. The formula for the z-score in the context of two proportions is given by:

z = \frac{(p_1 - p_2) - (p_1 - p_2)_{0}}{\sqrt{p_{bar}(1 - p_{bar})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}

Here, (p₁ - p₂)₀ is the hypothesized difference in proportions, which is zero under the null hypothesis. The pooled proportion (p_bar) is calculated as:

p_{bar} = \frac{x_1 + x_2}{n_1 + n_2}

In this case, p_bar = (11 + 17) / (20 + 23) = 28 / 43 ≈ 0.65. The complement, q_bar, is 1 - p_bar = 0.35. Plugging these values into the z-score formula allows you to compute the z-score, which in this example is approximately -1.3.

Once the z-score is determined, the next step is to find the p-value associated with this z-score. Since this is a two-tailed test, the p-value is calculated as twice the probability of observing a z-score less than -1.3. Using statistical tables or software, you can find that the p-value is approximately 0.193.

Finally, you compare the p-value to the significance level (α = 0.05). In this case, since 0.193 > 0.05, you 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 being tested.

In summary, conducting a hypothesis test for two proportions involves formulating hypotheses, calculating a z-score using pooled proportions, determining the p-value, and making a conclusion based on the comparison of the p-value to the significance level. This structured approach allows for a clear analysis of differences between two groups.

Problem

A human resources department is comparing two employee training programs to see if they lead to different pass rates on a required certification exam. They randomly select two groups of employees. In Program A, 16 out of 20 employees passed the exam. In Program B, 30 out of 40 employees passed. Are the basic conditions met to conduct a 2-proportion hypothesis test?

Yes, the basic conditions are met.

No, the basic conditions are not met.

There is not enough information to answer the question.

Problem

A school administrator wants to compare the proportion of students who passed a standardized math exam in two different schools by taking samples from 2 classes. Assume the samples are random and independent. Calculate the $z$ -score for testing whether there is a significant difference in the population proportions of student passing rates, but do not find a $P$ -value or draw a conclusion for the hypothesis test.
Class A: 72 out of 120 students passed.
Class B: 65 out of 100 students passed.

-0.76

1.07

-1.07

0.76

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Difference in Proportions: Hypothesis Tests Example 1

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Difference in Proportions: Hypothesis Tests Example 1 Video Summary

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. 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 of 0.01.

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

Next, we calculate the sample proportions: for the first group, p₁ = 50 / 2400 = 0.0208, and for the second group, p₂ = 15 / 1600 = 0.0094. The z-score is then computed using the formula:

Z = \frac{(p₁ - p₂) - 0}{\sqrt{p_bar(1 - p_bar)\left(\frac{1}{n₁} + \frac{1}{n₂}\right)}}

Where p_bar is the pooled sample proportion, calculated as:

p_bar = \frac{x₁ + x₂}{n₁ + n₂} = \frac{50 + 15}{2400 + 1600} = \frac{65}{4000} = 0.0163

Substituting the values into the z-score formula yields a z-score of 2.81. This z-score indicates how far our sample data deviates from the null hypothesis.

To determine the p-value, we assess the probability of observing a z-score greater than 2.81 in a standard normal distribution. This results in a p-value of approximately 0.0025. Since this p-value is less than our significance level of 0.01, we reject the null hypothesis.

In conclusion, the evidence suggests that there is a statistically significant difference in the proportions of injuries between drivers who wear seat belts and those who do not. Specifically, not wearing a seat belt is associated with a greater proportion of injuries, reinforcing the importance of seat belt use for safety.

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Difference in Proportions: Confidence Intervals

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Difference in Proportions: Confidence Intervals Video Summary

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 using the formula:

$$\hat{p_1} - \hat{p_2}$$

where $$\hat{p_1}$$ and $$\hat{p_2}$$ are the sample proportions from the two groups being compared. For instance, if the success rates of two treatments are 0.55 and 0.74, the point estimate would be:

$$0.55 - 0.74 = -0.19$$

This indicates a 19% difference in success rates, suggesting that the first treatment is less effective than the second.

Next, the margin of error (E) is calculated using a modified formula that incorporates both sample proportions:

$$E = z \cdot \sqrt{\frac{\hat{p_1} \cdot (1 - \hat{p_1})}{n_1} + \frac{\hat{p_2} \cdot (1 - \hat{p_2})}{n_2}}$$

Here, $$z$$ is the critical z-value corresponding to the desired confidence level. For a 90% confidence level, the critical z-value is approximately 1.645. The sample sizes, $$n_1$$ and $$n_2$$, are also essential in this calculation. After substituting the values into the equation, the margin of error can be determined.

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

Lower Bound: $$\hat{p_1} - \hat{p_2} - E$$

Upper Bound: $$\hat{p_1} - \hat{p_2} + E$$

For example, if the margin of error is 0.24, the confidence interval would be:

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 from -0.43 to 0.05.

Interpreting the confidence interval is vital for hypothesis testing. If the interval includes zero, it suggests that there is no significant difference between the two proportions, leading to a failure to reject the null hypothesis. Conversely, if the interval does not include zero, it indicates a significant difference, allowing for the rejection of the null hypothesis. In this case, since the interval from -0.43 to 0.05 includes zero, we conclude that there is insufficient evidence to assert a difference in success rates between the two treatments.

In summary, constructing a confidence interval for the difference in proportions involves calculating the point estimate, determining the margin of error, and interpreting the results in the context of hypothesis testing. This process is essential for making informed decisions based on statistical data.

Problem

A researcher using a survey constructs a 90% confidence interval for a difference in two proportions. According to the data, they calculate $p̂_1-p̂_2=0.09$ with a margin of error of 0.07. Should they reject or fail to reject the claim that there is no difference in these two proportions?

Reject

Fail to reject

There is not enough information to answer the question.

Problem

The data below is taken from two random, independent samples. Calculate the margin of error for a 99% confidence interval for the difference in population proportions.
$x_1=87$ , $x_2=68$
$n_1=120$ , $n_2=115$

0.061

0.062

0.158

0.060

example

Difference in Proportions: Confidence Intervals Example 2

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Difference in Proportions: Confidence Intervals Example 2 Video Summary

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 there is a significant difference in attendance rates between these two groups.

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 for the difference in proportions, we first verify that the samples are random and independent, and that there are more than five successes and failures in each group, which allows us to assume normal distribution.

The next step involves calculating the critical z-value for a 95% confidence level. This is determined using the formula:

$$ z_{\alpha/2} = 1.96 $$

Next, we calculate the sample proportions for each group:

For on-campus students:

$$ p_1 = \frac{x_1}{n_1} = \frac{102}{50} = 0.68 $$

For off-campus students:

$$ p_2 = \frac{x_2}{n_2} = \frac{74}{30} \approx 0.569 $$

We then find the point estimate for the difference in proportions:

$$ \hat{p_1} - \hat{p_2} = 0.68 - 0.569 = 0.111 $$

To calculate the margin of error, we use the formula:

$$ E = z_{\alpha/2} \sqrt{\frac{\hat{p_1} \hat{q_1}}{n_1} + \frac{\hat{p_2} \hat{q_2}}{n_2}} $$

where $ \hat{q_1} = 1 - \hat{p_1} $ and $ \hat{q_2} = 1 - \hat{p_2} $. After substituting the values, we find:

$$ E \approx 0.113 $$

Finally, we construct the confidence interval by adding and subtracting the margin of error from the point estimate:

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:

$$ (-0.002, 0.224) $$

In the second part of the analysis, we assess the claim that on-campus students are more likely to attend events than off-campus students. This involves checking if the confidence interval includes zero. Since our interval does include zero, we fail to reject the null hypothesis, which states that there is no difference in attendance rates between the two groups. Therefore, we conclude that there is not enough evidence to support the claim that on-campus students are more likely to attend events than their off-campus counterparts.

Here’s what students ask on this topic:

What is the null hypothesis in a two-proportion hypothesis test?

In a two-proportion hypothesis test, the null hypothesis (H₀) assumes that the two population proportions are equal. Mathematically, it is expressed as:

H_{0} : p_{1} = p_{2}

Alternatively, this can be written as:

H_{0} : p_{1} - p_{2} = 0

This means that there is no difference between the two proportions being compared. The null hypothesis serves as the default assumption, and the goal of the test is to determine whether there is enough evidence to reject this assumption in favor of the alternative hypothesis.

How do you calculate the test statistic for a two-proportion z-test?

The test statistic for a two-proportion z-test is calculated using the formula:

z = \frac{(p_{1} - p_{2})}{\sqrt{\frac{p q}{n_{1}} + \frac{p q}{n_{2}}}}

Here:

$p$ is the pooled sample proportion, calculated as:

p = \frac{x_{1} + x_{2}}{n_{1} + n_{2}}

$q$ is the complement of $p$ , calculated as $q = 1 - p$ .
$n_{1}$ and $n_{2}$ are the sample sizes for the two groups.

This formula helps determine how far the observed difference in sample proportions is from the null hypothesis, in terms of standard errors.

What does it mean if the confidence interval for the difference in proportions includes zero?

If the confidence interval for the difference in proportions includes zero, it means that there is no statistically significant difference between the two population proportions at the given confidence level. This is because zero represents the possibility that the two proportions are equal (i.e., $p_{1} - p_{2} = 0$ ).

For example, if the confidence interval is $[- 0.05, 0.10]$ , the interval includes zero, suggesting that the true difference in proportions could be zero. In this case, we fail to reject the null hypothesis, as there is insufficient evidence to conclude a significant difference between the two proportions.

However, if the interval does not include zero, it indicates a significant difference between the proportions, and we would reject the null hypothesis.

How do you interpret the p-value in a two-proportion hypothesis test?

The p-value in a two-proportion hypothesis test represents the probability of observing a difference in sample proportions as extreme as (or more extreme than) the one calculated, assuming the null hypothesis is true. A small p-value (typically less than the significance level, $α$ , such as 0.05) indicates strong evidence against the null hypothesis, suggesting that the observed difference is unlikely to have occurred by random chance.

For example, if the p-value is 0.03 and $α$ is 0.05, we reject the null hypothesis and conclude that there is a significant difference between the two proportions. Conversely, if the p-value is greater than $α$ , we fail to reject the null hypothesis, indicating insufficient evidence to support a significant difference.

What is the difference between a pooled proportion and individual sample proportions?

The pooled proportion is a weighted average of the sample proportions from two groups, used in hypothesis testing for two proportions. It is calculated as:

p = \frac{x_{1} + x_{2}}{n_{1} + n_{2}}

Here, $x_{1}$ and $x_{2}$ are the number of successes, and $n_{1}$ and $n_{2}$ are the sample sizes.

Individual sample proportions, on the other hand, are calculated separately for each group as:

p_{1} = \frac{x_{1}}{n_{1}}

and

p_{2} = \frac{x_{2}}{n_{2}}

The pooled proportion is used in the test statistic formula for hypothesis testing, while individual sample proportions are used in confidence interval calculations.