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Chapter 8: Problem 26

In Exercises 25-28, determine whether a normal sampling distribution can be used. If it can be used, test the claim about the difference between two population proportions \(p_{1}\) and \(p_{2}\) at the level of significance \(\alpha\). Assume the samples are random and independent. Claim: \(p_{1} \leq p_{2} ; \alpha=0.01\) Sample statistics: \(x_{1}=36, n_{1}=100\) and \(x_{2}=46, n_{2}=200\)

Short Answer

Expert verified
Fail to reject the null hypothesis; insufficient evidence to support \(p_1 < p_2\).

Step by step solution

01

Check Conditions for Normal Approximation

We need to ensure that the sample sizes are large enough for the normal approximation. The conditions are: \(n_1 p_1 \geq 5, \ n_1 (1 - p_1) \geq 5, \ n_2 p_2 \geq 5, \ and \ n_2 (1 - p_2) \geq 5\). First estimate \(p_1\) and \(p_2\) as \(\hat{p}_1 = \frac{x_1}{n_1} = \frac{36}{100} = 0.36\) and \(\hat{p}_2 = \frac{x_2}{n_2} = \frac{46}{200} = 0.23\). Calculate \(n_1 \hat{p}_1 = 100 \times 0.36 = 36\), \(n_1 (1 - \hat{p}_1) = 64\), \(n_2 \hat{p}_2 = 46\), and \(n_2 (1 - \hat{p}_2) = 154\). All values are greater than 5.
02

State Hypotheses

The null hypothesis \(H_0\) is that \(p_1 = p_2\), and the alternative hypothesis \(H_a\) is that \(p_1 < p_2\). We are testing whether the first proportion is less than or equal to the second.
03

Calculate Test Statistic

First, find the pooled sample proportion \(\hat{p} = \frac{x_1 + x_2}{n_1 + n_2} = \frac{36 + 46}{100 + 200} = \frac{82}{300} \approx 0.2733\). The standard error is \(SE = \sqrt{\hat{p}(1-\hat{p})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)} = \sqrt{0.2733 \times (1-0.2733) \times \left(\frac{1}{100} + \frac{1}{200}\right)} \approx 0.0585\). The test statistic is \(z = \frac{\hat{p}_1 - \hat{p}_2}{SE} = \frac{0.36 - 0.23}{0.0585} \approx 2.22\).
04

Determine Critical Value and Make Decision

Using a significance level of \(\alpha = 0.01\), find the critical value for a one-tailed test, which is \(-z_{\alpha} = -2.33\) (from z-table). Since our test statistic \(z = 2.22\) is greater than \(-z_{\alpha}\), we fail to reject the null hypothesis.

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Key Concepts

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

Sampling Distribution
Understanding sampling distribution is crucial when conducting statistical analysis. It is the probability distribution of a given statistic based on a random sample.
Sampling distributions assist us in making statistical inferences about a larger population from which the sample is drawn.
For instance, if you repeatedly take samples from a population and calculate a statistic (such as the sample mean or proportion) each time, the distribution of these statistics is your sampling distribution.
  • Central Limit Theorem: With a large enough sample size, the sampling distribution of the mean will be approximately normally distributed, regardless of the original population distribution.
  • The more data points included in your sample size, the closer this distribution will resemble a normal distribution.
  • In hypothesis testing, this concept helps determine if observed differences are likely due to chance or represent actual differences in the population.
Population Proportions
Population proportions compare the characteristic of groups within a population, defined as the ratio for each characteristic compared to the entire population. Understanding and estimating population proportions is vital in creating a foundation for hypothesis tests involving categorical data.
  • Proportions are often estimated using sample data, represented as \(\hat{p}\) for each sample group.
  • We utilize these sample proportions to infer the actual proportions of the broader population.
  • In the given problem, we observed sample proportions of \(36\%\) and \(23\%\) respectively for different samples, labeled \(\hat{p}_1\) and \(\hat{p}_2\).
By understanding the concept of population proportions, one can ascertain the likelihood that sample findings are reflective of the larger population.
Hypothesis Testing
Hypothesis testing is a statistical method that uses sample data to evaluate a hypothesis about a population parameter. It is an essential process in statistical inference.
In this problem, hypothesis testing is used to determine if there is a significant difference between two population proportions.
  • The null hypothesis \(H_0\) states that there is no effect or no difference, such as \(p_1 = p_2\).
  • The alternative hypothesis \(H_a\) suggests a specific direction of effect, like \(p_1 < p_2\).
  • We calculate a test statistic (such as a \(z\)-score) to determine whether to reject the null hypothesis.
By comparing your calculated statistic with a critical value from a statistical distribution (like the normal distribution), you can make decisions about the hypothesis.
Significance Level
The significance level, denoted as \(\alpha\), is a threshold used in hypothesis testing. It represents the probability of rejecting the null hypothesis when it is true, known as a Type I error.
In the exercise, a significance level of \(0.01\) was used. This indicates a very strict criterion for concluding that there is an effect or difference.
  • If your calculated \(p\)-value or test statistic fall below this level, you reject the null hypothesis.
  • A lower \(\alpha\) reduces the chance of Type I error but increases the chance of a Type II error (failing to reject a false null hypothesis).
  • The choice of \(\alpha\) reflects the study's tolerance for risk and the importance of avoiding errors.
  • In decision-making, using \(\alpha = 0.01\), as in the exercise, shows high confidence is needed to support claims of differences.
Understanding the significance level helps in interpreting the results from hypothesis testing confidently.

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Most popular questions from this chapter

In Exercises 9 and 10 , (a) identify the claim and state \(H_{0}\) and \(H_{a}\), \((b)\) find the critical value(s) and identify the rejection region(s), (c) find the standardized test statistic \(z\), (d) decide whether to reject or fail to reject the null hypothesis, and (e) interpret the decision in the context of the original claim. Assume the samples are random and independent, and the populations are normally distributed. A career counselor claims that the mean annual salary of entry-level paralegals in Peoria, Illinois, and Gary, Indiana, is the same. The mean annual salary of 40 randomly selected entry-level paralegals in Peoria is \(\$ 50,410\). Assume the population standard deviation is \(\$ 9320\). The mean annual salary of 35 randomly selected entry-level paralegals in Gary is \(\$ 47,350\). Assume the population standard deviation is \(\$ 9330\). At \(\alpha=0.10\), is there enough evidence to reject the counselor's claim? (Adapted from Salary.com)

In Exercises 11-16, test the claim about the difference between two population means \(\mu_{1}\) and \(\mu_{2}\) at the level of significance \(\alpha\). Assume the samples are random and independent, and the populations are normally distributed. Claim: \(\mu_{1} \geq \mu_{2} ; \alpha=0.01\). Assume \(\sigma_{1}^{2}=\sigma_{2}^{2}\) Sample statistics: \(\bar{x}_{1}=44.5, s_{1}=5.85, n_{1}=17\) and \(\bar{x}_{2}=49.1, s_{2}=5.25, n_{2}=18\)

In Exercises 11-16, test the claim about the difference between two population means \(\mu_{1}\) and \(\mu_{2}\) at the level of significance \(\alpha\). Assume the samples are random and independent, and the populations are normally distributed. Claim: \(\mu_{1} \neq \mu_{2} ; \alpha=0.01\). Assume \(\sigma_{1}^{2}=\sigma_{2}^{2}\) Sample statistics: \(\bar{x}_{1}=61, s_{1}=3.3, n_{1}=5\) and \(\bar{x}_{2}=55, s_{2}=1.2, n_{2}=7\)

In Exercises 1-4, classify the two samples as independent or dependent and justify your answer. Sample 1: The weights of 43 adults Sample 2: The weights of the same 43 adults after participating in a diet and exercise program

In Exercises 19-22, test the claim about the mean of the differences for a population of paired data at the level of significance \(\alpha\). Assume the samples are random and dependent, and the populations are normally distributed. Claim: \(\mu_{d}=0 ; \alpha=0.01\). Sample statistics: \(\bar{d}=8.5, s_{d}=10.7, n=16\)
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