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Independent random samples of 280 and 350 observations were selected from binomial populations 1 and 2 , respectively. Sample 1 had 132 successes, and sample 2 had 178 successes. Do the data present sufficient evidence to indicate that the proportion of successes in population 1 is smaller than the proportion in population \(2 ?\) Use one of the two methods of testing presented in this section, and explain your conclusions.

Short Answer

Expert verified
Explain. Answer: To determine if we have enough evidence, we must compare the calculated test statistic (Z) to the critical value (Z伪). If Z < Z伪, we reject the null hypothesis and conclude that the proportion of successes in population 1 is smaller than the proportion in population 2. If Z 鈮 Z伪, we fail to reject the null hypothesis, meaning we cannot conclude that the proportion of successes in population 1 is smaller based on the sample data.

Step by step solution

01

Set up null and alternative hypotheses

The null and alternative hypotheses are as follows: H鈧: p鈧 - p鈧 = 0 (The proportions are equal) H鈧: p鈧 - p鈧 < 0 (The proportion of successes in population 1 is smaller than the proportion in population 2) Here, p鈧 is the proportion of successes in population 1, and p鈧 is the proportion of successes in population 2.
02

Determine the significance level (alpha)

We will assume a common significance level, 伪 = 0.05. This means that we will accept a 5% chance of committing a Type I error - rejecting the null hypothesis when it is actually true.
03

Calculate the test statistic (Z)

We will use the following formula for the z-test statistic for the difference in proportions: Z = (P鈧 - P鈧 - 0) / sqrt(P(1 - P)((1/n鈧) + (1/n鈧))) where P鈧 is the proportion of successes in sample 1, P鈧 is the proportion of successes in sample 2, P is the pooled proportion, and n鈧 and n鈧 are the sample sizes. First, calculate the sample proportions: P鈧 = 132 / 280 P鈧 = 178 / 350 Next, calculate the pooled proportion, which is the total successes divided by the total sample size: P = (132 + 178) / (280 + 350) Finally, compute the z-test statistic using the formula: Z = (P鈧 - P鈧 - 0) / sqrt(P(1 - P)((1/n鈧) + (1/n鈧)))
04

Determine the critical value and compare to the test statistic

Since we are performing a one-tailed test with 伪 = 0.05, we need to find the critical value (Z伪) that corresponds to the 5% significance level in the standard normal distribution. Using a Z table or calculator, we find that: Z伪 = -1.645 Now we will compare the calculated test statistic (Z) to the critical value (Z伪): If Z < Z伪, we reject the null hypothesis, otherwise, we fail to reject the null hypothesis.
05

Draw a conclusion

Based on the comparison between the test statistic (Z) and the critical value (Z伪), we can either reject or fail to reject the null hypothesis. If we reject the null hypothesis, we conclude that there is sufficient evidence to indicate that the proportion of successes in population 1 is smaller than the proportion in population 2. Otherwise, if we fail to reject the null hypothesis, we cannot conclude that the proportion of successes in population 1 is smaller than the proportion in population 2 based on the sample data.

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Key Concepts

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

Binomial Proportion Test
The binomial proportion test is used to compare proportions between two groups or populations. It's especially useful when you want to determine if a significant difference exists between the success rates of two independent samples. Such a test uses sample data to infer if the proportions vary at the population level. Here, two sample groups from separate binomial populations were considered. One sample had a total of 280 with 132 successes and another had 350 with 178 successes.

This kind of test can unveil whether the difference in proportions hints at a real disparity in the populations from which these samples were drawn. In our context, the focus was on checking if one population's success rate was smaller than the other's. These kinds of tests grow essential in fields like medicine, quality control, and social sciences, as they furnish statistically-backed conclusions about comparative performance or behavior.
Null and Alternative Hypotheses
In hypothesis testing, one of the first steps is to formulate the null and alternative hypotheses. The null hypothesis, denoted by H鈧, represents a default statement or position that indicates no effect or difference. Contrastingly, the alternative hypothesis, H鈧 or Ha, is what the test aims to provide evidence for. It's a statement suggesting that there is indeed an effect or difference.

For our case, the null hypothesis was posed as H鈧: \( p鈧 - p鈧 = 0 \), implying that both population proportions are equal. Meanwhile, the alternative hypothesis was stated as H鈧: \( p鈧 - p鈧 < 0 \), suggesting that the proportion of successes in population 1 is less than that in population 2. Laying down these hypotheses sets the stage for other statistical procedures used to analyze the data and draw conclusions.
Significance Level
Setting a significance level is paramount to conducting a hypothesis test. It essentially defines the threshold for making a decision about the hypothesis. Most commonly, a significance level, 伪, of 0.05 is adopted. This implies a 5% risk of making a Type I error, which would occur if one incorrectly rejects the null hypothesis when it is true.

By establishing this level, one delineates the likelihood of accepting erroneous conclusions. In the given exercise, using 伪 = 0.05 meant there was a 5% chance of concluding that one population's success rate was lesser when it was actually not. Deciding on this level earlier ensures clarity and accuracy in results interpretation, giving assertions more reliability.
Z-Test Statistic
The Z-test statistic is a crucial component in determining whether to reject or accept the null hypothesis when examining differences between proportions. Calculated using sample data, it measures how far and in what direction the sample proportion deviates from the expected under the null hypothesis. For this case, it is computed using the formula:
  • Z = \( \frac{(P鈧 - P鈧)}{\sqrt{P(1-P)(\frac{1}{n鈧亇 + \frac{1}{n鈧倉)}} \)
Here, \( P鈧 \) and \( P鈧 \) are sample proportions, and P is the pooled proportion. Once calculated, the Z-statistic is compared to a critical value from the standard normal distribution. If the calculated Z is less than the critical Z value (Z伪), the null hypothesis is rejected. In our example, the critical value was Z伪 = -1.645, given a one-tailed test at 5% significance.

Such analyses help ascertain whether the observed differences in sample proportions strongly suggest variations within overall populations.

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Most popular questions from this chapter

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