91Ó°ÊÓ

For one binomial experiment, 200 binomial trials produced 60 successes. For a second independent binomial experiment, 400 binomial trials produced 156 successes. At the \(5 \%\) level of significance, test the claim that the probability of success for the second binomial experiment is greater than that for the first. (a) Compute the pooled probability of success for the two experiments. (b) Check Requirements What distribution does the sample test statistic follow? Explain. (c) State the hypotheses. (d) Compute \(\hat{p}_{1}-\hat{p}_{2}\) and the corresponding sample distribution value. (e) Find the \(P\) -value of the sample test statistic. (f) Conclude the test. (g) Interpret the results.

Step by step solution

01

Pooled Probability of Success

Compute the pooled probability of success (\( \hat{p} \)) for the experiments. The formula for pooled probability is:\[ \hat{p} = \frac{x_1 + x_2}{n_1 + n_2} \]where \( x_1 = 60 \), \( n_1 = 200 \), \( x_2 = 156 \), and \( n_2 = 400 \). Substitute these values into the formula:\[ \hat{p} = \frac{60 + 156}{200 + 400} = \frac{216}{600} = 0.36 \]

02

Distribution of Test Statistic

The test statistic follows a normal distribution under the null hypothesis, due to the large sample sizes. This allows the use of the Z-test for two proportions, which is appropriate as both sample sizes and the number of successes are sufficiently large for normal approximation conditions to hold.

03

State Hypotheses

Define the null and alternative hypotheses: \( H_0 : p_1 = p_2 \)\( H_a : p_1 < p_2 \)where \( p_1 \) and \( p_2 \) are the probabilities of success for the first and second experiments, respectively.

04

Compute Proportion Difference

Calculate the sample proportions and their difference:\[ \hat{p}_1 = \frac{x_1}{n_1} = \frac{60}{200} = 0.30 \]\[ \hat{p}_2 = \frac{x_2}{n_2} = \frac{156}{400} = 0.39 \]The difference in sample proportions is:\[ \hat{p}_2 - \hat{p}_1 = 0.39 - 0.30 = 0.09 \]

05

Sample Distribution Value

Calculate the standard error (SE) for the difference in proportions:\[ \text{SE} = \sqrt{\hat{p}(1-\hat{p})(\frac{1}{n_1} + \frac{1}{n_2})} \]\[ \text{SE} = \sqrt{0.36 \times 0.64 \times (\frac{1}{200} + \frac{1}{400})} \approx 0.032 \]Compute the Z-score:\[ Z = \frac{\hat{p}_2 - \hat{p}_1}{\text{SE}} = \frac{0.09}{0.032} \approx 2.8125 \]

06

Find P-value

Based on the Z-score from Step 5, we can find the P-value from the standard normal distribution. Using a Z-table or calculator, determine the P-value for \( Z = 2.8125 \). The P-value is approximately 0.0025.

07

Conclusion of Test

Since the P-value (0.0025) is less than the significance level \( \alpha = 0.05 \), we reject the null hypothesis \( H_0 \). This provides evidence to support the claim that \( p_2 \) is greater than \( p_1 \).

08

Interpret Results

The analysis shows there is statistically significant evidence at the 5% significance level to conclude that the probability of success for the second experiment is greater than that for the first experiment.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Binomial Distribution

A binomial distribution is a probability distribution that models the number of successes in a fixed number of independent trials. Each trial has two possible outcomes: success or failure. In this context:

• We have two separate experiments with a certain number of trials and successes. The first experiment has 200 trials resulting in 60 successes, while the second has 400 trials with 156 successes.
• The probabilities of success in each experiment can be different but are constant within each experiment.
• The binomial distribution is defined by two parameters: the number of trials (n) and the probability of success (p).

An essential feature of the binomial model is that it fits scenarios where outcomes are binary. In this exercise, calculating the proportions provides a fundamental basis for testing whether the success probability differs across the experiments.

Z-test for Two Proportions

The Z-test for two proportions is used to test if there is a difference between the success probabilities of two independent samples. Here are key aspects:

• We have two samples with calculated success proportions: 0.30 for the first and 0.39 for the second.
• The goal is to determine if the observed difference of 0.09 in proportions is statistically significant.
• The Z-test involves computing a test statistic that follows a normal distribution under the null hypothesis. For this, we compute standard error and then find the Z-score.

The formula for the test statistic (Z-score) is based on the difference in sample proportions divided by the standard error. The resulting Z-score helps in determining the P-value, which tells us the probability of observing such a difference if the null hypothesis is true.

Pooled Probability

Pooled probability is a combined estimate of the success probability across both samples under the assumption that the null hypothesis is true:

• The formula to compute pooled probability (\( \hat{p} \)) is \( \frac{x_1 + x_2}{n_1 + n_2} \), where \( x_1 \) and \( x_2 \) are the number of successes, and \( n_1 \) and \( n_2 \) are the number of trials for each sample.
• Here, \( \hat{p} \) is calculated as \( \frac{60 + 156}{200 + 400} = 0.36 \).
• Using the pooled probability, we are able to calculate the standard error for the difference in proportions.

The pooled probability simplifies the process of testing the equality of proportions by providing a single measure that reflects the combined success rate across all trials. This plays a crucial role in determining the standard error for the Z-test, ultimately influencing the test's sensitivity and accuracy.